You CAN Ace Calculus
related topics on other pages |
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external links you may find helpful |
Larson Calculus (Google books) Volume - Disk, Washer Methods and Shell, Cylinder Methods pgs 458-477 |
Pauls Online Notes - Volumes of Solids of Revolution / Method of Rings |
Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |
Arc Length (Vector Functions) |
Arc Length Function |
Arc Length Parameter |
Conservative Vector Fields |
Cross Product |
Curl |
Curvature |
Cylindrical Coordinates |
Lagrange Multipliers |
Line Integrals |
Partial Derivatives |
Partial Integrals |
Path Integrals |
Potential Functions |
Principal Unit Normal Vector |
Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
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Help Keep 17Calculus Free |
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This page covers single volume integrals when an area is rotated about a vertical or horizontal line. The area is defined by equations in the form \(y=f(x)\) or \(x=f(y)\) and we use the the cylinder (shell) method. For other ways to calculate volume, see the links in the related topics panel.
If you want a full-length lecture on this topic, we recommend this video and this instructor.
video by Prof Leonard
Overview |
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When calculating the volume of rotation, there are 3 factors that determine how to set up the integral.
1. method (washer-disc or cylinder-shell)
2. axis of rotation
3. function (graph and form of the equations)
On this page, the axis of rotation will always be either an axis or a straight line that is parallel to one of the axes. However, before we discuss the rotation of an area, we need to know how to describe an area in the plane. Go to the main volume integrals page to review that material.
The technique we discuss on this page is the cylinder-shell method. We will use these figures extensively in the discussion of these techniques. Click here to download a one page pdf document of these pictures, with space to write notes. Feel free to copy this page to use while studying, working practice problems, in exams (if allowed by your instructor) and to give to fellow students. If you are an instructor and you think it will help your students, you may make as many copies as you need to use in your classes. For everyone, we ask that you keep the information that you got it from this website at the bottom of the sheet.
cylinder-shell method | |
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x-axis rotation |
y-axis rotation |
Getting Started |
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Here are some key things that you need to do and know to get started.
1. Draw a rough plot of the area that is being rotated. This is usually best done by hand since you will need to label it.
2. Decide what method you will use, washer-disc or cylinder-shell.
3. On the rough plot from point 1, label the axis of rotation and draw a representative rectangle somewhere in the area.
4. Label R and r or p and h (depending on your method; details below).
Once those steps are done, you are ready to set up your integral. Let's look at each technique separately.
We choose to use the term cylinder-shell method since it seems to bring to mind a cylinder more readily than the term shell, which in general usage could be almost any shape, and it covers the two most common references to this technique. This, of course, is a personal preference for this site and you need to check with your instructor to see what they require.
summary of the cylinder-shell method | |
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the representative rectangle is parallel to axis of revolution | |
\(p\) is the distance from the axis of rotation to the representative rectangle | |
\(h\) is the height of the representative rectangle | |
x-axis rotation equation | \(\displaystyle{ V = 2\pi \int_{c}^{d}{ph~dy} }\) |
y-axis rotation equation | \(\displaystyle{ V = 2\pi \int_{a}^{b}{ph~dx} }\) |
Notes
1. Notice that \(p\) and \(h\) are distances, so they are always positive.
2. The distance \(p\) is from the axis of rotation to the representative rectangle. Where on the rectangle doesn't matter since the rectangle is infinitely thin. If it helps, you can think of the distance to the center of the rectangle.
cylinder-shell method with x-axis rotation |
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If you feel like you need further explanation of this, here is a video that tries to explain this method by drawing in three dimensions. He also works a simple example using this technique.
video by Khan Academy
Additional Things To Watch For |
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It would be nice if, after learning these two techniques, that was all there was to it. However, as usual in mathematics, that is not the case. Here are a couple twists you will probably see in your homework and maybe even on an exam. We will just touch on them and how to work with them and allow you to fill in the details.
Twist 1. Multiple Integrals |
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Looking back at the plots used in the discussions, you will notice that no matter where we draw the representative rectangle, each end will always be on the same curve. That will not always be the case. Here is how you handle that situation.
Look back at the discussion on describing area in the xy-plane. When one end of the rectangle 'jumps' from one graph to another, you need to break the integral at that point. Then you will have more than one integral and you just add the result of each together.
Twist 2. Axis of Revolution Not A Coordinate Axis |
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We touch on this on the washer-disc method page but let's look at a quick example that may help you understand this more deeply. We will just set up part of this problem and you can take it from there. The complete solution is found in practice problem 2287.
Problem Statement - - Calculate the volume of revolution of the area defined by \(y=x^2-3x\) and \(y=x\) rotated about the line \(y=5\).
Solution Overview - - We have set up the plot shown on the right with all the pertinent data. Everything should look familiar to you except that we added an additional line from the x-axis to the rectangle and the new variable y, both in red. The trick here is to find the equation for p (assuming we use the cylinder-shell method, which is not the best way here). We know that the distance from the x-axis to the axis of rotation is always 5. So we build an equation using that information, \(y+p=5\) and solve for p, \(p=5-y\).
We deliberatedly chose y for that distance since the distance is parallel to the y-axis. This allows us to use the equation straight-away. If the distance had been parallel to the x-axis, we would have used x instead.
Take some time to think about this and then look as some examples and practice problems. You will be able to comprehend it soon.
If you decide to work this problem, notice that setting this up as we have shown will produce an integral that is difficult to solve. Try using the washer-disc method instead.
x-axis rotation practice |
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
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Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new. |
Cylinder-Shell Method - Practice Problems Conversion |
[1-324] - [2-877] - [3-1174] - [4-1359] - [5-1184] - [6-2003] - [7-1182] - [8-1183] - [9-1172] |
[10-1175] - [11-1176] - [12-343] - [13-878] - [14-1185] - [15-1173] - [16-1916] - [17-313] - [18-1179] |
[19-2193] - [20-1177] - [21-1180] - [22-303] - [23-1178] - [24-1181] - [25-1186] |
Please update your notes to this new numbering system. The display of this conversion information is temporary. |
area: \( y=x, y=2-x, y=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Problem Statement |
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area: \( y=x, y=2-x, y=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Hint |
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Problem Statement |
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area: \( y=x, y=2-x, y=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Final Answer |
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\(2\pi/3\) |
Problem Statement |
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area: \( y=x, y=2-x, y=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Hint |
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Solution |
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1.Set up the plot with labels - See the hint or the animation for the plot and enclosed area. Since we were asked to use the cylinder-shell method, the representative rectangle is horizontal.
2. Choose The Integral - The volume integral we need is \( V = 2\pi\int_c^d{ph~dy} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to y since the example rectangle is horizontal and consequently it moves vertically, sweeping in the y-direction.
3. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is the length of the example rectangle. Remember that both of these are positive. The expression for p is \(p=y\) and for h we have \(h=2-2y\).
4. Set Up the Integral - If you play with the animation, you can tell that the example rectangle sweeps from \(y=0\) to \(y=1\). So our integral is \( V = 2\pi\int_0^1{y(2-2y) ~dy} \).
5. Evaluate the Integral
\(\displaystyle{ V = 2\pi\int_0^1{y(2-2y) ~dy} }\) |
\(\displaystyle{ V = 4\pi\int_0^1{y-y^2 ~dy} }\) |
\(\displaystyle{ V = 4\pi\left[ \frac{y^2}{2} - \frac{y^3}{3} \right]_0^1 }\) |
\(\displaystyle{ V = 4\pi\left[ \frac{1}{2} - \frac{1}{3} \right] }\) |
\(\displaystyle{ V=2\pi/3 }\) |
Final Answer |
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\(2\pi/3\) |
close solution |
area: \( x=y^2, x=0, y=3 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Problem Statement |
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area: \( x=y^2, x=0, y=3 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Hint |
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Problem Statement |
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area: \( x=y^2, x=0, y=3 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Final Answer |
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\(\displaystyle{ V = \frac{81\pi}{2} }\) |
Problem Statement |
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area: \( x=y^2, x=0, y=3 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Hint |
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Solution |
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1.Set up the plot with labels - See the hint or the animation for the plot and enclosed area. Since we were asked to use the cylinder-shell method, the representative rectangle is horizontal.
2. Choose The Integral - The volume integral we need is \( V = 2\pi\int_c^d{ph~dy} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to y since the example rectangle is horizontal and consequently it moves vertically, sweeping in the y-direction.
3. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is the length of the example rectangle. Remember that both of these are positive. The expression for p is \(p=y\) and for h we have \(h=y^2\).
4. Set Up the Integral - If you play with the animation, you can tell that the example rectangle sweeps from \(y=0\) to \(y=3\). So our integral is \( V = 2\pi\int_0^3{y^3 ~dy} \).
5. Evaluate the Integral
\(\displaystyle{ V = 2\pi\int_0^3{y^3 ~dy} }\) |
\(\displaystyle{ V = 2\pi \left. \frac{y^4}{4} \right|_0^3 }\) |
\(\displaystyle{ V = \frac{\pi}{2} (3)^4 = \frac{81\pi}{2} }\) |
Final Answer |
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\(\displaystyle{ V = \frac{81\pi}{2} }\) |
close solution |
area: \( y=2x^{-3/2}, y=2, y=16, x=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Problem Statement |
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area: \( y=2x^{-3/2}, y=2, y=16, x=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Hint |
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Problem Statement |
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area: \( y=2x^{-3/2}, y=2, y=16, x=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Final Answer |
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\( V = 90\pi \) |
Problem Statement |
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area: \( y=2x^{-3/2}, y=2, y=16, x=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Hint |
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Solution |
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1.Set up the plot with labels - See the hint or the animation for the plot and enclosed area. Since we were asked to use the cylinder-shell method, the representative rectangle is horizontal.
2. Choose The Integral - The volume integral we need is \( V = 2\pi\int_c^d{ph~dy} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to y since the example rectangle is horizontal and consequently it moves vertically, sweeping in the y-direction.
3. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is the length of the example rectangle. Remember that both of these are positive. The expression for p is \(p=y\) and for h we have \(h=(2/y)^{2/3}\). To get the expression for h, notice that h is from the y-axis to the plot \(y=2x^{-3/2}\), so we have \(h=x\). However, we need h in terms of y since we are integrating with respect to y. Since the x lands on the curve \(y=2x^{-3/2}\), that's the equation we need to solve for x, which is \(x=(2/y)^{2/3}\).
4. Set Up the Integral - If you play with the animation, you can tell that the example rectangle sweeps from \(y=2\) to \(y=16\). So our integral is \( V = 2\pi\int_2^{16}{y(2/y)^{2/3} ~dy} \).
5. Evaluate the Integral
\(\displaystyle{ V = 2\pi\int_2^{16}{y(2/y)^{2/3} ~dy} }\) |
\(\displaystyle{ V = 2\pi (2)^{2/3} \int_2^{16}{ y^{1/3} ~dy} }\) |
\(\displaystyle{ V = \pi (2)^{5/3} \left[ \frac{y^{4/3}}{4/3} \right]_2^{16} }\) |
\(\displaystyle{ V = \frac{3}{4} \pi (2)^{5/3} \left[ 16^{4/3} - 2^{4/3} \right] = 90\pi }\) |
Final Answer |
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\( V = 90\pi \) |
close solution |
area: \( y=\sqrt{\arccos x}, y=0, x=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Problem Statement |
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area: \( y=\sqrt{\arccos x}, y=0, x=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Hint |
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Problem Statement |
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area: \( y=\sqrt{\arccos x}, y=0, x=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Final Answer |
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\( V = \pi \) |
Problem Statement |
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area: \( y=\sqrt{\arccos x}, y=0, x=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Hint |
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Solution |
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1.Set up the plot with labels - See the hint or the animation for the plot and enclosed area. Since we were asked to use the cylinder-shell method, the representative rectangle is horizontal.
2. Choose The Integral - The volume integral we need is \( V = 2\pi\int_c^d{ph~dy} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to y since the example rectangle is horizontal and consequently it moves vertically, sweeping in the y-direction.
3. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is the length of the example rectangle. Remember that both of these are positive. The expression for p is \(p=y\) and for h we have \(h=\cos(y^2)\). To get the expression for h, we notice that \(h=x\) and since we need the expression for x in terms of y, we use \(y=\sqrt{\arccos x}\) and solve for x.
4. Set Up the Integral - If you play with the animation, you can tell that the example rectangle sweeps from \(y=0\) to somewhere near 1 but not exactly 1. At \(x=0\), \(y=\sqrt{\arccos 0} = \sqrt{\pi/2}\). So our integral is \( V = 2\pi\int_0^{\sqrt{\pi/2}}{ y\cos(y^2) ~dy} \).
5. Evaluate the Integral
\(\displaystyle{ V = 2\pi\int_0^{\sqrt{\pi/2}}{ y\cos(y^2) ~dy} }\) |
Use integration by substitution where \(u=y^2\). |
\(u=y^2 \to du = 2y~dy\) |
Now convert the limits of integration in terms of u. |
\(y=0 \to u=0\) and \(y=\sqrt{\pi/2} \to u=\pi/2\) |
\(\displaystyle{ V = 2\pi\int_0^{\pi/2}{ \cos u ~\frac{du}{2} } }\) |
\(\displaystyle{ V = \pi \left[ \sin u \right]_0^{\pi/2} }\) |
\( V = \pi [\sin(\pi/2) - \sin 0] = \pi \) |
Final Answer |
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\( V = \pi \) |
close solution |
area: \( y=4-x, x=0, y=2 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Problem Statement |
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area: \( y=4-x, x=0, y=2 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Hint |
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built with GeoGebra |
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Problem Statement |
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area: \( y=4-x, x=0, y=2 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Final Answer |
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\( V=32\pi/3 \) |
Problem Statement |
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area: \( y=4-x, x=0, y=2 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Hint |
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built with GeoGebra |
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Solution |
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1. Choose The Integral - The volume integral we need is \( V = 2\pi\int_c^d{ph~dy} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to y since the example rectangle is horizontal and consequently it moves vertically, sweeping in the y-direction.
2. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is from the y-axis to the function. So for the point \((x,y)\) on the plot, \(p=y\) and \(x=h\). Since the integral is in terms of y, we need to convert \(h=x\) into y. Since the point sits on the function, we know that \(y=4-x \to x=4-y\). This gives us \(h=4-y\).
3. Determine the Limits of Integration - Now we are almost ready to set up our integral. We need to know the limits of integration. We think of the rectangle as sweeping from bottom to top, always in the positive direction, across the shaded area. The bottom of the rectangle is at \(y=2\), the top is at \(y=4\), so we need to integrate from 2 to 4.
4. Evaluate the Integral
\(\displaystyle{ V = 2\pi\int_c^d{ph~dy} = }\) \(\displaystyle{ 2\pi\int_2^4{y(4-y)~dy} }\) |
\(\displaystyle{ V = 2\pi\int_2^4{4y-y^2 ~dy} }\) |
\(\displaystyle{ V = 2\pi \left[ \frac{4y^2}{2} - \frac{y^3}{3} \right]_2^4 }\) |
\(\displaystyle{ V = 2\pi \left[ 32-8-\frac{64}{3}+\frac{8}{3} \right] }\) |
\(\displaystyle{ V = 2\pi \left[ 24-\frac{56}{3} \right] }\) |
\(\displaystyle{ V = 2\pi \left[ \frac{72-56}{3} \right] }\) |
\(\displaystyle{ V = \frac{2\pi}{3} \left[ 16 \right] = \frac{32\pi}{3} }\) |
Final Answer |
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\( V=32\pi/3 \) |
close solution |
area: \( y=\sqrt{x}, x=4, y=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Problem Statement |
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area: \( y=\sqrt{x}, x=4, y=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Hint |
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built with GeoGebra |
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Problem Statement |
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area: \( y=\sqrt{x}, x=4, y=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Final Answer |
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\( V=8\pi \) |
Problem Statement |
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area: \( y=\sqrt{x}, x=4, y=0 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Hint |
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built with GeoGebra |
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Solution |
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1. Choose The Integral - The volume integral we need is \( V = 2\pi\int_c^d{ph~dy} \).
We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to y since the example rectangle is horizontal and consequently it moves vertically, sweeping in the y-direction.
2. Determine p and h - As labeled in the plot, distance p goes from the axis of rotation to the example rectangle. This distance is always y, so \(p=y\).
The distance h is a little more difficult to determine. If we think about adding h to the length of the gap to the left of the curve, the distance is always 4. So, can we determine the length of the gap? Yes, we can. Let's think about the left endpoint of the rectangle and label it \((x,y)\). We know that since it is on the curve \(y=\sqrt{x}\), we can write \((x,y) = (x,\sqrt{x})\). Another way to think about this point is that the distance from the y-axis to the curve is x. That's great! However, since our integral is in terms of y, we need to know what x is in terms of y. That's easy, since the point is on the curve \(y=\sqrt{x}\) or \(y^2=x\) (as long as x is positive, which it is in this problem). So the length of the gap between the y-axis and the plot is x or \(y^2\). So now we have an equation, \(x+h=4 \to y^2+h=4 \to h=4-y^2\). In summary, \(p=y\) and \(h=4-y^2\).
Note: You may need to go through this explanation again since the concept is probably new to you. Do not hesitate to make a rough plot and label everything until you understand it. Also, try to put the explanation into your own words.
3. Determine the Limits of Integration - Now we are almost ready to set up our integral. We need to know the limits of integration. We think of the rectangle as sweeping from bottom to top, always in the positive direction, across the shaded area. So, the rectangle starts at zero and ends at \(2\).
Now, we have everything we need to set up the integral.
4. Evaluate the Integral
\(\displaystyle{ V = 2\pi\int_c^d{ph~dy} = }\) \(\displaystyle{ 2\pi\int_0^2{y(4-y^2)~dy} }\) |
\(\displaystyle{ V = 2\pi\int_0^2{ 4y-y^3 ~dy} }\) |
\(\displaystyle{ V = 2\pi \left[ \frac{4y^2}{2} - \frac{y^4}{4} \right]_0^2 }\) |
\(\displaystyle{ V = 2\pi \left[ 8 - \frac{16}{4} \right] - 2\pi[0-0] = 8\pi }\) |
Final Answer |
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\( V=8\pi \) |
close solution |
area: \( y=x^2, y=0, x=2 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Problem Statement |
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area: \( y=x^2, y=0, x=2 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Final Answer |
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\( V=32\pi/5 \) |
Problem Statement |
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area: \( y=x^2, y=0, x=2 \) |
axis of rotation: x-axis |
method: cylinder-shell |
Solution |
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video by MIP4U
Final Answer |
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\( V=32\pi/5 \) |
close solution |
parallel to x-axis rotation practice |
When the axis of rotation is not the x-axis, the same rules apply, i.e. p is the distance (always positive) from the axis of rotation to the example rectangle and h is the length (also positive) of the example rectangle.
In these practice problems, we describe another way of determining p and h than we talked about in the previous problem set. Both techniques produce the same result but one may be easier for you to use. So use the one that makes more sense to you, unless your instructor says otherwise.
area: \( y=x^2, y=0, x=1 \) |
axis of rotation: \(y=-2\) |
method: cylinder-shell |
Problem Statement |
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area: \( y=x^2, y=0, x=1 \) |
axis of rotation: \(y=-2\) |
method: cylinder-shell |
Hint |
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Problem Statement |
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area: \( y=x^2, y=0, x=1 \) |
axis of rotation: \(y=-2\) |
method: cylinder-shell |
Final Answer |
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\( V = 23\pi/15 \) |
Problem Statement |
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area: \( y=x^2, y=0, x=1 \) |
axis of rotation: \(y=-2\) |
method: cylinder-shell |
Hint |
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Solution |
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1.Set up the plot with labels - See the hint or the animation for the plot and enclosed area. Since we were asked to use the cylinder-shell method, the representative rectangle is horizontal.
2. Choose The Integral - The volume integral we need is \( V = 2\pi\int_c^d{ph~dy} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to y since the example rectangle is horizontal and consequently it moves vertically, sweeping in the y-direction.
3. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is the length of the example rectangle. Remember that both of these are positive. Since the axis of rotation is not the x-axis, we need take into account the added distance. So we start with the distance y which would be the expression for p if the x-axis was the axis of rotation and then we notice that \(y=-2\) is another 2 units away from the example rectangle. So we add 2 to y to get \(p=y+2\).
The axis of rotation does not affect h, so we think about moving from \(x=1\) to the y-axis then turn around and move back x units to get back to the example rectangle. This gives us \(h=1-x\). But we need the expression for h in terms of y, and since x lands on the curve \(y=x^2\), we need to solve this for x to get \(x=\pm\sqrt{y}\). We choose positive since we are in the first quadrant (so y is positive), giving us \(h=1-\sqrt{y}\).
4. Set Up the Integral - If you play with the animation, you can tell that the example rectangle sweeps from \(y=0\) to \(y=1\). (You should confirm this with the equations.) So our integral is \(\displaystyle{ V = 2\pi \int_0^1{ (2+y)(1-\sqrt{y}) ~dy } }\)
5. Evaluate the Integral
\(\displaystyle{ V = 2\pi \int_0^1{ (2+y)(1-\sqrt{y}) ~dy } }\) |
\(\displaystyle{ V = 2\pi \int_0^1{ 2-2y^{1/2}+y-y^{3/2} ~dy } }\) |
\(\displaystyle{ V = 2\pi \left[ 2y - \frac{2y^{3/2}}{3/2} + \frac{y^2}{2} - \frac{y^{5/2}}{5/2} \right]_0^1 }\) |
\(\displaystyle{ V = 2\pi \left[ 2 - \frac{4}{3} + \frac{1}{2} - \frac{2}{5} \right] - 0 = \frac{23\pi}{15} }\) |
Final Answer |
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\( V = 23\pi/15 \) |
close solution |
area: \( y=x^2, y=0, x=1 \) |
axis of rotation: \(y=2\) |
method: cylinder-shell |
Problem Statement |
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area: \( y=x^2, y=0, x=1 \) |
axis of rotation: \(y=2\) |
method: cylinder-shell |
Hint |
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Problem Statement |
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area: \( y=x^2, y=0, x=1 \) |
axis of rotation: \(y=2\) |
method: cylinder-shell |
Final Answer |
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\( V = 17\pi/15 \) |
Problem Statement |
---|
area: \( y=x^2, y=0, x=1 \) |
axis of rotation: \(y=2\) |
method: cylinder-shell |
Hint |
---|
Solution |
---|
1. Set up the plot with labels - See the hint or the animation for the plot and enclosed area. Since we were asked to use the cylinder-shell method, the representative rectangle is horizontal.
2. Choose The Integral - The volume integral we need is \( V = 2\pi\int_c^d{ph~dy} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to y since the example rectangle is horizontal and consequently it moves vertically, sweeping in the y-direction.
3. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is the length of the example rectangle. Remember that both of these are positive. To determine p we think about moving from \(y=2\), the axis of rotation, to the x-axis, which is 2 units. Then we turn around and move back up y unitls. The result is \(p=2-y\). Similarly for h, we go from \(x=2\) to the y-axis, we have gone 2 units. Then we turn around and move x units back to curve \(y=x^2\) to give us \(h=2-x\). However, the integral is in terms of y, so we need to convert the x to a function of y. Since the x we are talking about lands on the curve \(y=x^2\), we need to solve for x using that equation giving us \(x=\pm\sqrt{y}\). Since we are in the first quadrant and y is positive, we choose positive from this equation. Putting \(x=\sqrt{y}\) in the equation of h, we have \(h=2-\sqrt{y}\).
4. Set Up the Integral - If you play with the animation, you can tell that the example rectangle sweeps from \(y=0\) to \(y=1\). (You should confirm this with the equations.) So our integral is \(\displaystyle{ V = 2\pi \int_0^1{ (2-y)(1-\sqrt{y}) ~dy } }\)
5. Evaluate the Integral
\(\displaystyle{ V = 2\pi \int_0^1{ (2-y)(1-\sqrt{y}) ~dy } }\) |
\(\displaystyle{ V = 2\pi \int_0^1{ 2-2y^{1/2}-y+y^{3/2} ~dy } }\) |
\(\displaystyle{ V = 2\pi \left[ 2y-\frac{2y^{3/2}}{3/2}-\frac{y^2}{2}+\frac{y^{5/2}}{5/2} \right]_0^1 }\) |
\(\displaystyle{ V = 2\pi \left[ 2-\frac{4}{3}-\frac{1}{2}+\frac{2}{5} \right] = \frac{17\pi}{15} }\) |
Final Answer |
---|
\( V = 17\pi/15 \) |
close solution |
area: \(y=x^3, y=0, x=1\) |
axis of rotation: \(y=1\) |
method: cylinder-shell |
Problem Statement |
---|
area: \(y=x^3, y=0, x=1\) |
axis of rotation: \(y=1\) |
method: cylinder-shell |
Final Answer |
---|
\(\displaystyle{V=\frac{5\pi}{14}}\) |
Problem Statement |
---|
area: \(y=x^3, y=0, x=1\) |
axis of rotation: \(y=1\) |
method: cylinder-shell |
Solution |
---|
video by Krista King Math
Final Answer |
---|
\(\displaystyle{V=\frac{5\pi}{14}}\) |
close solution |
area: \(y=\sqrt{x}, y=x\) |
axis of rotation: \(y=-3\) |
method: cylinder-shell |
Problem Statement |
---|
area: \(y=\sqrt{x}, y=x\) |
axis of rotation: \(y=-3\) |
method: cylinder-shell |
Solution |
---|
video by PatrickJMT
close solution |
area: \( f(x)=x, g(x)=x^2-3x \) |
axis of rotation: \(y=5\) |
method: cylinder-shell | ||
Set up but do not evaluate the integral(s). |
---|
Problem Statement |
---|
area: \( f(x)=x, g(x)=x^2-3x \) |
axis of rotation: \(y=5\) |
method: cylinder-shell | ||
Set up but do not evaluate the integral(s). |
---|
Hint |
---|
Problem Statement |
---|
area: \( f(x)=x, g(x)=x^2-3x \) |
axis of rotation: \(y=5\) |
method: cylinder-shell | ||
Set up but do not evaluate the integral(s). |
---|
Final Answer |
---|
\(\displaystyle{ V = 2\pi \int_0^4{ (5-y)[ 3/2+(y+9/4)^{1/2} - y ] ~ dy } + }\) \(\displaystyle{ 2\pi \int_{-9/4}^0{ 2(5-y)(y+9/4)^{1/2} ~ dy } }\) |
Problem Statement |
---|
area: \( f(x)=x, g(x)=x^2-3x \) |
axis of rotation: \(y=5\) |
method: cylinder-shell | ||
Set up but do not evaluate the integral(s). |
---|
Hint |
---|
Solution |
---|
1. Choose The Integral - The volume integral we need is \( V = 2\pi\int_c^d{ph~dy} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to y since the example rectangle is horizontal and consequently it moves vertically, sweeping in the y-direction.
2. Determine p and h - The distance p is from the axis of rotation to the example rectangle. However, in this problem we need two integrals. You can see way from the plot above. If you think about the example rectangle sweeping through the entire shaded area, the right end is always on the plot of \(g(x)\). However, the left end jumps at \(x=0\) between \(f(x)\) and \(g(x)\). So we need to set up two integrals, one from \(y=-9/4\) to \(y=0\), the second from \(y=0\) to \(y=4\).
But first, we need to solve for \(x\) in each of the functions. The first one is done for us, \(y=x\). For the second one, we need to complete the square.
We start with this equation. |
\( y=x^2-3x \) |
Take the coefficient of \(x\) (assuming the coefficient of \(x^2\) is 1). |
Form \(\displaystyle{ \left( \frac{-3}{2} \right)^2 = \frac{9}{4}}\) |
Take the result and add it and subtract from the right side of the original equation. |
\(\displaystyle{ y=x^2-3x+\frac{9}{4}-\frac{9}{4} }\) |
The first three terms are now a perfect square. |
\(\displaystyle{ y=(x-3/2)^2-9/4 }\) |
Now we have only one \(x\), so it is easy to solve. |
\( y+9/4 = (x-3/2)^2 \) |
\( \pm(y+9/4) = (x-3/2) \) |
\( x=3/2\pm(y+9/4) \) |
A few comments are in order.
First, we need the \(\pm\) when we take the square root. The expression \( x=3/2-(y+9/4) \) is the part of the graph to the left of \(x=3/2\) and \( x=3/2+(y+9/4) \) is to the right of \(x=3/2\).
Second, if you have any difficulty understanding these steps, review completing the square on this precalculus page.
Now we need to determine p and h for each part of the graph. Let's start with the top part, i.e. \(0 \leq y \leq 4\). To find an expression for p, let's think about walking a path in the y-direction, starting at the axis of rotation, \(y=5\). If we walking from 5 down to the x-axis, we have moved 5 units. Then we turn around and walk back up to the representative rectangle. This last distance is y. So we have moved 5 units then gone back y. The net distance we have moved from the axis of rotation to the representative rectangle is \(5-y\) units. This is the distance p, so \(p=5-y\).
Let's use the same technique to find the length h. Start at the far end of the representative rectangle. Moving from there to the x-axis is a distance x, where the x falls on the curve \(g(x)\). We solved for x earlier and got \( x=3/2+(y+9/4) \). So, we moved a distance \( 3/2+(y+9/4) \). Then we turn around and move another distance x landing on the plot \(y=x\). In terms of y, we backtracked y. The net distance we moved is \( h = 3/2+(y+9/4) - y \).
Using the same ideas, we come up with \(p=5-y\) and \(h=2(y+9/4)^{1/2}\) for the range \(-9/4\leq y \leq 0\).
3. Set Up The Integrals - \(V=V_1+V_2\) where
\(\displaystyle{ V_1 = 2\pi \int_0^4{ (5-y)[ 3/2+(y+9/4)^{1/2} - y ] ~ dy } }\) |
\(\displaystyle{ V_2 = 2\pi \int_{-9/4}^0{ 2(5-y)(y+9/4)^{1/2} ~ dy } }\) |
We were told to not evaluate the integrals, so \(V=V_1+V_2\) is the final answer.
These integrals would be very difficult to evaluate, if even possible. For this problem, it is much easier to use the washer-disc method. Practice problem 2288 shows the washer-disc method to solve this problem.
Final Answer |
---|
\(\displaystyle{ V = 2\pi \int_0^4{ (5-y)[ 3/2+(y+9/4)^{1/2} - y ] ~ dy } + }\) \(\displaystyle{ 2\pi \int_{-9/4}^0{ 2(5-y)(y+9/4)^{1/2} ~ dy } }\) |
close solution |
y-axis rotation practice |
cylinder-shell method with y-axis rotation |
---|
Okay, now let's look at some problems with y-axis rotation. Here is the plot with all the information you need to work these problems. Notice that the representative rectangle is vertical since, with the cylinder-shell method, we need the rectangle parallel to the axis of rotation. Since the rectangle sweeps in the x direction, we integrate with respect to x. So our volume integral is \(V=2\pi\int_a^b{ph~dx}\).
area: \( y=3x, x=0, y=3 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Problem Statement |
---|
area: \( y=3x, x=0, y=3 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Hint |
---|
Problem Statement |
---|
area: \( y=3x, x=0, y=3 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Final Answer |
---|
\( V = \pi \) |
Problem Statement |
---|
area: \( y=3x, x=0, y=3 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Hint |
---|
Solution |
---|
1. Set up the plot with labels - See the hint for the plot and enclosed area. Since we were asked to use the cylinder-shell method, the representative rectangle is vertical.
2. Choose The Integral - The volume integral we need is \( V = 2\pi\int_a^b{ph~dx} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to x since the example rectangle is vertical and consequently it moves horizontally, sweeping in the x-direction.
3. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is the length of the example rectangle. Remember that both of these are positive. The distance p is just x, i.e. \(p=x\). We can think of the distance h as the equation of the top end minus the equation at the bottom end of the rectangle, just like we did when we calculated the area between two curves. In this case \(h=3-3x\).
4. Set Up the Integral - Looking at the plot, you can tell that the example rectangle sweeps from \(x=0\) to \( x=1 \). (You should confirm this with the equations.) So our integral is \(\displaystyle{ V = 2\pi \int_0^1{ x(3-3x) ~dx } }\)
5. Evaluate the Integral
\(\displaystyle{ V = 2\pi \int_0^1{ x(3-3x) ~dx } }\) |
\(\displaystyle{ V = 2\pi \int_0^1{ 3x-3x^2 ~dx } }\) |
\(\displaystyle{ V = 2\pi \left[ \frac{3x^2}{2} - \frac{3x^3}{3} \right]_0^1 }\) |
\(\displaystyle{ V = 2\pi \left[ \frac{3}{2} - 1 \right] - 2\pi [0] = \pi }\) |
Final Answer |
---|
\( V = \pi \) |
close solution |
area: \( x^3-x^8+1, y=1 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Problem Statement |
---|
area: \( x^3-x^8+1, y=1 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Hint |
---|
Problem Statement |
---|
area: \( x^3-x^8+1, y=1 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Final Answer |
---|
\( V = \pi/5 \) |
Problem Statement |
---|
area: \( x^3-x^8+1, y=1 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Hint |
---|
Solution |
---|
1. Set up the plot with labels - See the hint for the plot and enclosed area. Since we were asked to use the cylinder-shell method, the representative rectangle is vertical.
2. Choose The Integral - The volume integral we need is \( V = 2\pi\int_a^b{ph~dx} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to x since the example rectangle is vertical and consequently it moves horizontally, sweeping in the x-direction.
3. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is the length of the example rectangle. Remember that both of these are positive. The distance p is just x, i.e. \(p=x\). We can think of the distance h as the equation of the top end minus the equation at the bottom end of the rectangle, just like we did when we calculated the area between two curves. In this case \( h=x^3-x^8+1-1 = x^3-x^8 \).
4. Set Up the Integral - Looking at the plot, you can tell that the example rectangle sweeps from \(x=0\) to \( x=1 \). (You should confirm this with the equations.) So our integral is \(\displaystyle{ V = 2\pi \int_0^1{ x(x^3-x^8) ~dx } }\)
5. Evaluate the Integral
\(\displaystyle{ V = 2\pi \int_0^1{ x(x^3-x^8) ~dx } }\) |
\(\displaystyle{ V = 2\pi \int_0^1{ x^4-x^9 ~dx } }\) |
\(\displaystyle{ V = 2\pi \left[ \frac{x^5}{5}-\frac{x^{10}}{10} \right]_0^1 }\) |
\(\displaystyle{ V = 2\pi \left[ \frac{1}{5}-\frac{1}{10} \right] = \frac{\pi}{5} }\) |
Final Answer |
---|
\( V = \pi/5 \) |
close solution |
area: \(y=\sin(x^2), y=0, x:[0,\sqrt{\pi}]\) |
axis of rotation: y-axis |
method: cylinder-shell |
Problem Statement |
---|
area: \(y=\sin(x^2), y=0, x:[0,\sqrt{\pi}]\) |
axis of rotation: y-axis |
method: cylinder-shell |
Final Answer |
---|
\(V=2\pi\) |
Problem Statement |
---|
area: \(y=\sin(x^2), y=0, x:[0,\sqrt{\pi}]\) |
axis of rotation: y-axis |
method: cylinder-shell |
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(V=2\pi\) |
close solution |
area: \( y=\cos(x^2), x=0, y=0 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Problem Statement |
---|
area: \( y=\cos(x^2), x=0, y=0 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Hint |
---|
Problem Statement |
---|
area: \( y=\cos(x^2), x=0, y=0 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Final Answer |
---|
\( V = \pi \) |
Problem Statement |
---|
area: \( y=\cos(x^2), x=0, y=0 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Hint |
---|
Solution |
---|
1. Set up the plot with labels - See the hint for the plot and enclosed area. Since we were asked to use the cylinder-shell method, the representative rectangle is vertical.
2. Choose The Integral - The volume integral we need is \( V = 2\pi\int_a^b{ph~dx} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to x since the example rectangle is vertical and consequently it moves horizontally, sweeping in the x-direction.
3. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is the length of the example rectangle. Remember that both of these are positive. The distance p is just x, i.e. \(p=x\). The distance h is \(y=\cos(x^2)\), i.e. \(h=\cos(x^2)\).
4. Set Up the Integral - Looking at the plot, you can tell that the example rectangle sweeps from \(x=0\) to somewhere between 1 and 2. We need to solve for this value.
\(0 = \cos(x^2) \to \arccos(0) = x^2\)
The angle when cosine is zero is \(\pi/2\), so \(x^2=\pi/2 \to x=\pm \sqrt{\pi/2}\). We can tell from our plot that \(x > 0\), so we choose \(x=\sqrt{\pi/2}\). So our integral is \(\displaystyle{ V = 2\pi \int_0^{\sqrt{\pi/2}}{ x\cos(x^2) ~dx } }\)
5. Evaluate the Integral
\(\displaystyle{ V = 2\pi \int_0^{\sqrt{\pi/2}}{ x\cos(x^2) ~dx } }\) |
We need to use integration by substitution. We choose \(u=x^2\). |
\(u=x^2 \to du = 2x~dx\) |
Converting the limits of integration from x to u, we have |
\(x=0 \to u=0^2=0\) and \(x=\sqrt{\pi/2} \to u=\pi/2\) |
\(\displaystyle{ V = 2\pi \int_0^{\pi/2}{ \cos(u) ~du/2 } }\) |
\(\displaystyle{ V = \pi \left[ \sin u \right]_0^{\pi/2} }\) |
\(\displaystyle{ V = \pi \left[ \sin(\pi/2) - \sin(0) \right] = \pi }\) |
Final Answer |
---|
\( V = \pi \) |
close solution |
area: \(y=-x^2+x, y=0\) |
axis of rotation: y-axis |
method: cylinder-shell |
Problem Statement |
---|
area: \(y=-x^2+x, y=0\) |
axis of rotation: y-axis |
method: cylinder-shell |
Solution |
---|
video by PatrickJMT
close solution |
area: \(\displaystyle{ y=\frac{1}{1+x^2}, y=0, x=0, x=2 }\) |
axis of rotation: y-axis |
method: cylinder-shell |
Problem Statement |
---|
area: \(\displaystyle{ y=\frac{1}{1+x^2}, y=0, x=0, x=2 }\) |
axis of rotation: y-axis |
method: cylinder-shell |
Hint |
---|
Problem Statement |
---|
area: \(\displaystyle{ y=\frac{1}{1+x^2}, y=0, x=0, x=2 }\) |
axis of rotation: y-axis |
method: cylinder-shell |
Final Answer |
---|
\( V = \pi \ln 5 \) |
Problem Statement |
---|
area: \(\displaystyle{ y=\frac{1}{1+x^2}, y=0, x=0, x=2 }\) |
axis of rotation: y-axis |
method: cylinder-shell |
Hint |
---|
Solution |
---|
1. Set up the plot with labels - See the hint or the animation for the plot and enclosed area. Since we were asked to use the cylinder-shell method, the representative rectangle is vertical.
2. Choose The Integral - The volume integral we need is \( V = 2\pi\int_a^b{ph~dx} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to x since the example rectangle is vertical and consequently it moves horizontally, sweeping in the x-direction.
3. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is the length of the example rectangle. Remember that both of these are positive. The distance p is just x, i.e. \(p=x\). The distance h is just y but we need h in terms of x and this is just \(h=1/(1+x^2)\).
4. Set Up the Integral - If you play with the animation, you can tell that the example rectangle sweeps from \(x=0\) to \(x=2\). (You should confirm this with the equations.) So our integral is \(\displaystyle{ V = 2\pi \int_0^2{ \frac{x}{1+x^2}~dx } }\)
5. Evaluate the Integral
\(\displaystyle{ V = 2\pi \int_0^2{ \frac{x}{1+x^2}~dx } }\) |
Use integration by substitution where \(u=1+x^2\). |
\(u=1+x^2 \to du = 2x~dx\) |
Converting the limits of integration, we get |
\(x=0 \to u=1\) and \(x=2 \to u=1+4=5\) |
So now our integral in terms of u is |
\(\displaystyle{ V = 2\pi \int_1^5{ \frac{1}{u}\frac{du}{2} } }\) |
\(\pi \ln u |_1^5\) |
\(\pi \ln 5 - \pi \ln 1 = \pi \ln 5 \) |
Final Answer |
---|
\( V = \pi \ln 5 \) |
close solution |
area: \(f(x)=-2x^2+16x+18\), \( x=0 \), \(y=0\) |
axis of rotation: y-axis |
method: cylinder-shell |
Problem Statement |
---|
area: \(f(x)=-2x^2+16x+18\), \( x=0 \), \(y=0\) |
axis of rotation: y-axis |
method: cylinder-shell |
Final Answer |
---|
\(2673\pi\) |
Problem Statement |
---|
area: \(f(x)=-2x^2+16x+18\), \( x=0 \), \(y=0\) |
axis of rotation: y-axis |
method: cylinder-shell |
Solution |
---|
video by MIP4U
Final Answer |
---|
\(2673\pi\) |
close solution |
area: \( y=x^2+2, y=0, x=0, x=2 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Problem Statement |
---|
area: \( y=x^2+2, y=0, x=0, x=2 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Final Answer |
---|
\( V=16\pi \) |
Problem Statement |
---|
area: \( y=x^2+2, y=0, x=0, x=2 \) |
axis of rotation: y-axis |
method: cylinder-shell |
Solution |
---|
video by MIP4U
Final Answer |
---|
\( V=16\pi \) |
close solution |
parallel to y-axis rotation practice |
As before, the same rules apply for these problems. The rectangle is parallel to the axis of rotation, in this case it is vertical. The distance h is not affected by the location of the axis but p is. If we can find an expression for p a distance from the y-axis, we usually just add or subtract a constant for the shift of axis. Let's work some practice problems to clarify this idea.
area: \(y=4x-x^2, y=3\) |
axis of rotation: \(x=1\) |
method: cylinder-shell |
Problem Statement |
---|
area: \(y=4x-x^2, y=3\) |
axis of rotation: \(x=1\) |
method: cylinder-shell |
Final Answer |
---|
\(\displaystyle{V=\frac{8\pi}{3}}\) |
Problem Statement |
---|
area: \(y=4x-x^2, y=3\) |
axis of rotation: \(x=1\) |
method: cylinder-shell |
Solution |
---|
video by Krista King Math
Final Answer |
---|
\(\displaystyle{V=\frac{8\pi}{3}}\) |
close solution |
area: \( y=x^2, x=1, y=0 \) |
axis of rotation: \(x=-2\) |
method: cylinder-shell |
Problem Statement |
---|
area: \( y=x^2, x=1, y=0 \) |
axis of rotation: \(x=-2\) |
method: cylinder-shell |
Hint |
---|
Problem Statement |
---|
area: \( y=x^2, x=1, y=0 \) |
axis of rotation: \(x=-2\) |
method: cylinder-shell |
Final Answer |
---|
\( V = 11\pi/6 \) |
Problem Statement |
---|
area: \( y=x^2, x=1, y=0 \) |
axis of rotation: \(x=-2\) |
method: cylinder-shell |
Hint |
---|
Solution |
---|
1. Set up the plot with labels - See the hint for the plot and enclosed area. Since we were asked to use the cylinder-shell method, the representative rectangle is vertical.
2. Choose The Integral - The volume integral we need is \( V = 2\pi\int_a^b{ph~dx} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to x since the example rectangle is vertical and consequently it moves horizontally, sweeping in the x-direction.
3. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is the length of the example rectangle. Remember that both of these are positive. The distance h is \(y=x^2 \), i.e. \(h=x^2\). The distance from the y-axis to the example rectangle is just x. To this we add 2 to compensate for the shift of the axis to \(-2\) giving us \(p=2+x\).
4. Set Up the Integral - Looking at the plot, you can tell that the example rectangle sweeps from \(x=0\) to \(x=1\). So our integral is \(\displaystyle{ V = 2\pi \int_0^1{ (2+x)(x^2) ~dx } }\)
5. Evaluate the Integral
\(\displaystyle{ V = 2\pi \int_0^1{ (2+x)(x^2) ~dx } }\) |
\(\displaystyle{ V = 2\pi \int_0^1{ 2x^2+x^3 ~dx } }\) |
\(\displaystyle{ V = 2\pi \left[ \frac{2x^3}{3}+\frac{x^4}{4} \right]_0^1 }\) |
\(\displaystyle{ V = 2\pi \left[ \frac{2}{3}+\frac{1}{4} \right] = \frac{11\pi}{6} }\) |
Final Answer |
---|
\( V = 11\pi/6 \) |
close solution |
area: \( y=x^2, x=1, y=0 \) |
axis of rotation: \(x=1\) |
method: cylinder-shell |
Problem Statement |
---|
area: \( y=x^2, x=1, y=0 \) |
axis of rotation: \(x=1\) |
method: cylinder-shell |
Hint |
---|
Problem Statement |
---|
area: \( y=x^2, x=1, y=0 \) |
axis of rotation: \(x=1\) |
method: cylinder-shell |
Final Answer |
---|
\( V = \pi/6 \) |
Problem Statement |
---|
area: \( y=x^2, x=1, y=0 \) |
axis of rotation: \(x=1\) |
method: cylinder-shell |
Hint |
---|
Solution |
---|
1. Set up the plot with labels - See the hint for the plot and enclosed area. Since we were asked to use the cylinder-shell method, the representative rectangle is vertical.
2. Choose The Integral - The volume integral we need is \( V = 2\pi\int_a^b{ph~dx} \). We chose this integral because we are told to use the cylinder-shell method, so we need an integral with p and h. We integrate with respect to x since the example rectangle is vertical and consequently it moves horizontally, sweeping in the x-direction.
3. Determine p and h - The distance p is from the axis of rotation to the example rectangle. The distance h is the length of the example rectangle. Remember that both of these are positive. The distance h is \(y=x^2 \), i.e. \(h=x^2\). The distance from the y-axis to the example rectangle is x. We subtract that distance from 1 to get p. So \(p=1-x\).
4. Set Up the Integral - Looking at the plot, you can tell that the example rectangle sweeps from \(x=0\) to \(x=1\). So our integral is \(\displaystyle{ V = 2\pi \int_0^1{ x^2(1-x) ~dx } }\)
5. Evaluate the Integral
\(\displaystyle{ V = 2\pi \int_0^1{ x^2(1-x) ~dx } }\) |
\(\displaystyle{ V = 2\pi \int_0^1{ x^2(1-x) ~dx } }\) |
\(\displaystyle{ V = 2\pi \int_0^1{ x^2-x^3 ~dx } }\) |
\(\displaystyle{ V = 2\pi \left[ \frac{x^3}{3}- \frac{x^4}{4} \right]_0^1 }\) |
\(\displaystyle{ V = 2\pi \left[ \frac{1}{3}- \frac{1}{4} \right] = \frac{\pi}{6} }\) |
Final Answer |
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\( V = \pi/6 \) |
close solution |
area: \(y=\sqrt{x}, y=x\) |
axis of rotation: \(x=-4\) |
method: cylinder-shell |
Problem Statement |
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area: \(y=\sqrt{x}, y=x\) |
axis of rotation: \(x=-4\) |
method: cylinder-shell |
Solution |
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He also sets up the integral for rotating about \(x=10\).
video by PatrickJMT
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area: \(y=-x^2+x, y=0\) |
axis of rotation: \(x=-3\) |
method: cylinder-shell |
Problem Statement |
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area: \(y=-x^2+x, y=0\) |
axis of rotation: \(x=-3\) |
method: cylinder-shell |
Solution |
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video by PatrickJMT
close solution |
area: \(y=x^2, y=4\) |
axis of rotation: \(x=5\) |
method: cylinder-shell |
Problem Statement |
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area: \(y=x^2, y=4\) |
axis of rotation: \(x=5\) |
method: cylinder-shell |
Final Answer |
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\( V=320\pi/3 \) |
Problem Statement |
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area: \(y=x^2, y=4\) |
axis of rotation: \(x=5\) |
method: cylinder-shell |
Solution |
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video by Michel vanBiezen
Final Answer |
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\( V=320\pi/3 \) |
close solution |
area: \( y=\sqrt{x}, x=4, y=0 \) |
axis of rotation: \(x=6\) |
method: cylinder-shell |
Problem Statement |
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area: \( y=\sqrt{x}, x=4, y=0 \) |
axis of rotation: \(x=6\) |
method: cylinder-shell |
Final Answer |
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\( V=192\pi/5 \) |
Problem Statement |
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area: \( y=\sqrt{x}, x=4, y=0 \) |
axis of rotation: \(x=6\) |
method: cylinder-shell |
Solution |
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In the video, she sets up the integral but does not solve it. Here is the rest of the solution.
\(\displaystyle{ V = 2\pi \int_0^4{ (6-x)\sqrt{x} ~dx } }\) |
\(\displaystyle{ V = 2\pi \int_0^4{ 6x^{1/2} - x^{3/2} ~dx } }\) |
\(\displaystyle{ V = 2\pi \left[ \frac{6x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2} \right]_0^4 }\) |
\(\displaystyle{ V = 2\pi \left[ 4x^{3/2} - \frac{2}{5}x^{5/2} \right]_0^4 }\) |
\(\displaystyle{ V = 2\pi [ 4(4)^{3/2} - (2/5)(4)^{5/2} ] }\) |
\(\displaystyle{ V = 2\pi [ 32 - 64/5 ] = 192\pi/5 }\) |
video by MIT OCW
Final Answer |
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\( V=192\pi/5 \) |
close solution |
area: \(y=x^2, x=1, x=2, y=4\) |
axis of rotation: \(x=-2\) |
method: cylinder-shell |
Problem Statement |
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area: \(y=x^2, x=1, x=2, y=4\) |
axis of rotation: \(x=-2\) |
method: cylinder-shell |
Solution |
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This problem is solved in two consecutive videos.
video by Khan Academy
video by Khan Academy
close solution |