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17Calculus Integrals - Volume of Revolution Using The Washer-Disc Method

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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 washer (disc) method. For other ways to calculate volume, see the links in the related topics panel.

If you want a full video lecture on this topic, we recommend this video and this instructor.

Prof Leonard - Volume of Solids By Disks and Washers Method [2hr-47mins-48secs]

video by Prof Leonard

Alternate Names

Disc Method

Disk Method

Ring Method

We choose to use the term washer-disc method to refer to this technique. We think it covers the two most commonly used and most descriptive names. This, of course, is a personal preference for this site and you need to check with your instructor to see what they require.

What Is A Washer?

If you are not familiar with a washer (other than to wash clothes), this wiki page has pictures and explains what a washer is. In short, it is a disc with a circular hole in it whose center is the same as the full disc.

Overview

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, we discuss the washer-disc method where 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. This is a critical step to setting up your integral correctly. If you didn't completely understand this from the main volume integrals page, you can go over it again here.

Describing A Region In The xy-Plane

To describe an area in the xy-plane, the first step is to plot the boundaries and determine the actual region that needs to be described. There are several graphing utilities listed on the tools page. Our preference is to use the free program winplot (used to plot these graphs; we used gimp to add labels and other graphics). However, graphing by hand is usually the best and quickest way.

We use the graph to the right to facilitate this discussion. A common way to describe this area is the area bounded by \(f(x)\) (red line), \(g(x)\) (blue line) and \(x=a\) (black line).
[Remember that an equation like \(x=a\) can be interpreted two ways, either the point x whose value is a or the vertical line. You should be able to tell what is meant by the context.]

Okay, so we plotted the boundaries and shaded the area to be described. Now, we need to choose a direction to start, either vertically or horizontally. We will show both ways, starting with vertically, since it is more natural and what you are probably used to seeing. Also, this area is easier to describe vertically than horizontally (you will see why as you read on).

Vertically

Our first step is to draw a vertical arrow on the graph somewhere within the shaded area, like we have done here. Some books draw an example rectangle with the top on the upper graph and the bottom on the lower graph. That is the same idea as we have done with the arrow.

Now we need to think of this arrow as starting at the left boundary and sweeping across to the right boundary of the area. This sweeping action is important since it will sweep out the area. As we think about this sweeping, we need to think about where the arrow enters and leaves the shaded area. Let's look our example graph to demonstrate. Think about the arrow sweeping left to right. Notice that it always enters the area by crossing \(g(x)\), no matter where we draw it. Similarly, the arrow always exits the area by crossing \(f(x)\), no matter where we draw it. Do you see that?

But wait, how far to the right does it go? We are not given that information. What we need to do is find the x-value where the functions \(f(x)\) and \(g(x)\) intersect. You should be able to do that. We will call that point \((b,f(b))\). Also, we will call the left boundary \(x=a\). So now we have everything we need to describe this area. We give the final results below.

Vertical Arrow

\( g(x) \leq y \leq f(x) \)

arrow leaves through \(f(x)\) and enters through \(g(x)\)

\( a \leq x \leq b \)

arrow sweeps from left (\(x=a\)) to right (\(x=b\))

Horizontally

We can also describe this area horizontally (or using a horizontal arrow). We will assume that we can write the equations of \(f(x)\) and \(g(x)\) in terms of \(y\). ( This is not always possible, in which case we cannot describe the area in this way. ) For the sake of this discussion, we will call the corresponding equations \(f(x) \to F(y)\) and \(g(x) \to G(y)\).

Let's look at the graph. Notice we have drawn a horizontal arrow. Just like we did with the vertical arrow, we need to determine where the arrow enters and leaves the shaded area. In this case, the arrow sweeps from the bottom up. As it sweeps, we can see that it always crosses the vertical line \(x=a\). However, there is something strange going on at the point \((b,f(b))\). Notice that when the arrow is below \(f(b)\), the arrow exits through \(g(x)\) but when the arrow is above \(f(b)\), the arrow exits through \(f(x)\). This is a problem. To overcome this, we need to break the area into two parts at \(f(b)\).

Lower Section - - This section is described by the arrow leaving through \(g(x)\). So the arrow sweeps from \(g(a)\) to \(g(b)\).
Upper Section - - This section is described by the arrow leaving through \(f(x)\). The arrow sweeps from \(f(b)\) to \(f(a)\).
The total area is the combination of these two areas. The results are summarized below.

Horizontal Arrow

lower section

\( a \leq x \leq G(y) \)

arrow leaves through \(G(y)\) and enters through \(x=a\)

\( g(a) \leq y \leq g(b) \)

arrow sweeps from bottom (\(y=g(a)\)) to top (\(y=g(b)\))

upper section

\( a \leq x \leq F(y) \)

arrow leaves through \(F(y)\) and enters through \(x=a\)

\( f(b) \leq y \leq f(a) \)

arrow sweeps from bottom (\(y=f(b)\)) to top (\(y=f(a)\))

Type 1 and Type 2 Regions

Some instructors may describe regions in the plane as either Type 1 or Type 2 (you may see II instead of 2). As you know from the above discussion, some regions are better described vertically or horizontally. Type 1 regions are regions that are better described vertically, while Type 2 regions are better described horizontally. The example above was a Type 1 region.

Here is a quick video clip going into more detail on Type 1 and Type 2 regions.

Krista King Math - type I and type 2 regions [1min-39secs]

video by Krista King Math

Okay, so now that you know how to describe an area in the plane, we will use that knowledge to calculate a volume of revolution. The technique we discuss on this page is the washer-disc 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.

washer-disc method

x-axis rotation

y-axis rotation

Now we will discuss each of these plots separately and explain each part of the plots.

Getting Started

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.

Once those steps are done, you are ready to set up your integral.

The volume integral using the washer-disc method is based on the volume of a washer or disc. Let's think a bit about the volume of a washer-disc. If we start with a full disc (no hole in the middle), the volume is the surface area times the thickness. Since the disc is a circle, the area of a circle is \(\pi R^2\) where \(R\) is the radius of the circle. The volume is \(\pi R^2 t\) where \(t\) is the thickness. We choose to use a capital R here as the radius of the disc.

Now, with a washer, we take the disc we just discussed and put a circular hole in it with it's center the same as the full disc. (Think of a CD or DVD disc.) Now the volume is reduced by what we have taken out of the center. This empty space has volume \(\pi r^2 t\), where \(r\) is the radius of the small hole. The thickness, \(t\), is the same as the full disc.

So now we have what we need to put together an equation for the washer-disc with a hole in the middle. We take the volume of the full disc and subtract the volume of the hole to get \(V = \pi R^2 t - \pi r^2 t = \pi t(R^2-r^2)\). [Notice the special case when there is no hole in the middle, can be thought of as \(r=0\) giving the volume of the disk as just \(V=\pi R^2 t\).]

summary of the washer-disc method

the representative rectangle is perpendicular to axis of revolution

\(R\) is the distance from the axis of rotation to the far end of the representative rectangle

\(r\) is the distance from the axis of rotation to the closest end of the representative rectangle

x-axis rotation equation

\(\displaystyle{ V = \pi \int_{a}^{b}{R^2-r^2~dx} }\)

y-axis rotation equation

\(\displaystyle{ V = \pi \int_{c}^{d}{R^2-r^2~dy} }\)

Note - Notice that \(R\) and \(r\) are distances, so they are always positive (although since we square them, the sign doesn't make any difference in the equations).

washer-disc method with x-axis rotation

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. In this video, notice that the axis of rotation runs along one side of the figure and, consequently, \(r=0\).

Khan Academy - Disc Method [10min-4secs]

video by Khan Academy

Okay, now let's work some problems using the washer-disc method revolving an area about the x-axis.

x-axis rotation practice

area: \( y=1/x, x=1, x=3 \)

axis of rotation: x-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=1/x, x=1, x=3 \) revolved about the x-axis. Give your answer in exact terms.

Final Answer

\( 2\pi/3 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=1/x, x=1, x=3 \) revolved about the x-axis. Give your answer in exact terms.

Solution

3487 video

Final Answer

\( 2\pi/3 \)

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area: \( y=x^2, y=4x-x^2 \)

axis of rotation: x-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, y=4x-x^2 \) revolved about the x-axis. Give your answer in exact terms.

Final Answer

\( V = 32\pi/3 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, y=4x-x^2 \) revolved about the x-axis. Give your answer in exact terms.

Solution

3494 video

Final Answer

\( V = 32\pi/3 \)

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area: \(y=\sqrt{x-1}, y=0, x=5\)

axis of rotation: x-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=\sqrt{x-1}, y=0, x=5\) revolved about the x-axis. Give your answer in exact terms.

Final Answer

\(V=8\pi\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=\sqrt{x-1}, y=0, x=5\) revolved about the x-axis. Give your answer in exact terms.

Solution

324 video

video by Krista King Math

Final Answer

\(V=8\pi\)

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area: \(y=x^2, x=y^2\)

axis of rotation: x-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=x^2, x=y^2\) revolved about the x-axis. Give your answer in exact terms.

Final Answer

\(\displaystyle{V=\frac{3\pi}{10}}\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=x^2, x=y^2\) revolved about the x-axis. Give your answer in exact terms.

Solution

This problem is solved by two different instructors.

877 video

877 video

video by Krista King Math

Final Answer

\(\displaystyle{V=\frac{3\pi}{10}}\)

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area: \(y=x^3, y=x, x \geq 0\)

axis of rotation: x-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=x^3, y=x, x \geq 0\) revolved about the x-axis. Give your answer in exact terms.

Final Answer

\(V=4\pi/21\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=x^3, y=x, x \geq 0\) revolved about the x-axis. Give your answer in exact terms.

Solution

1174 video

video by PatrickJMT

Final Answer

\(V=4\pi/21\)

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area: \(y=\sqrt{x}, y \geq 0, x=4\)

axis of rotation: x-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=\sqrt{x}, y \geq 0, x=4\) revolved about the x-axis. Give your answer in exact terms.

Final Answer

\(8\pi\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=\sqrt{x}, y \geq 0, x=4\) revolved about the x-axis. Give your answer in exact terms.

Solution

This problem is solved by two different instructors.

In the second video, he doesn't finish the integration, so here are the details.
\(\displaystyle{ \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi [(4^2)/2 - (0^2)/2] = 8\pi}\)

1359 video

1359 video

video by PatrickJMT

Final Answer

\(8\pi\)

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area: \(f(x)=2-\sin(x)\), \(x=0\), \(x=2\pi\), \(y=0\)

axis of rotation: x-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(f(x)=2-\sin(x)\), \(x=0\), \(x=2\pi\), \(y=0\) revolved about the x-axis. Give your answer in exact terms.

Final Answer

\(V = 9\pi^2\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(f(x)=2-\sin(x)\), \(x=0\), \(x=2\pi\), \(y=0\) revolved about the x-axis. Give your answer in exact terms.

Solution

2003 video

video by Dr Chris Tisdell

Final Answer

\(V = 9\pi^2\)

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area: \(y^2=x-2, x=5\) in the first quadrant

axis of rotation: x-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y^2=x-2, x=5\) in the first quadrant revolved about the x-axis. Give your answer in exact terms.

Final Answer

\( 4.5\pi \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y^2=x-2, x=5\) in the first quadrant revolved about the x-axis. Give your answer in exact terms.

Solution

2272 video

video by Michel vanBiezen

Final Answer

\( 4.5\pi \)

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area: \( y=x^2, y=x \)

axis of rotation: x-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, y=x \) revolved about the x-axis. Give your answer in exact terms.

Final Answer

\( 2\pi/15 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, y=x \) revolved about the x-axis. Give your answer in exact terms.

Solution

2273 video

video by Michel vanBiezen

Final Answer

\( 2\pi/15 \)

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area: \(y=\sqrt{9-x^2}\) in the first quadrant

axis of rotation: x-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=\sqrt{9-x^2}\) in the first quadrant revolved about the x-axis. Give your answer in exact terms.

Final Answer

\( V=18\pi \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=\sqrt{9-x^2}\) in the first quadrant revolved about the x-axis. Give your answer in exact terms.

Solution

2276 video

video by MIP4U

Final Answer

\( V=18\pi \)

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area: \(y=\sqrt{x}, y=x^2\)

axis of rotation: x-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=\sqrt{x}, y=x^2\) revolved about the x-axis. Give your answer in exact terms.

Solution

1184 video

video by Khan Academy

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area: \(y=\sqrt{x}, y\geq0, x=1\)

axis of rotation: x-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=\sqrt{x}, y\geq0, x=1\) revolved about the x-axis. Give your answer in exact terms.

Solution

1182 video

video by Khan Academy

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Derive the equation for the volume of a sphere of radius r using the washer-disc method.

Problem Statement

Derive the equation for the volume of a sphere of radius r using the washer-disc method.

Solution

1183 video

video by Khan Academy

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Looking back at the plots, you will notice that the axis of revolution is always a coordinate axis, either the x-axis or the y-axis. A twist you will see is when the axis of revolution is another line. On this site, we will discuss only axes that are parallel to one of the coordinate axes.
In this case, the equations that will change are the ones that describe the distance from the axes of rotation. We suggest that you set up a sum from the parallel coordinate axis to the axis of rotation and then solve for whatever variable you need. This concept, especially, requires you to think over in your mind several times and look at examples.

parallel to x-axis rotation practice

area: \( y=x^2, x=0, y=4 \)

axis of rotation: \( y=4 \)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, x=0, y=4 \) revolved about the \( y=4 \). Give your answer in exact terms.

Final Answer

\( V = 256\pi/15 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, x=0, y=4 \) revolved about the \( y=4 \). Give your answer in exact terms.

Solution

3490 video

Final Answer

\( V = 256\pi/15 \)

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area: \( y=2-x^2, y=1 \)

axis of rotation: \( y=1 \)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=2-x^2, y=1 \) revolved about the \( y=1 \). Give your answer in exact terms.

Final Answer

\( V = 16\pi/15 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=2-x^2, y=1 \) revolved about the \( y=1 \). Give your answer in exact terms.

Solution

3491 video

Final Answer

\( V = 16\pi/15 \)

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area: \( y=x, y=x^2 \)

axis of rotation: \( y=2 \)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x, y=x^2 \) revolved about \( y=2 \). Give your answer in exact terms.

Final Answer

\( V = 8\pi/15 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x, y=x^2 \) revolved about \( y=2 \). Give your answer in exact terms.

Solution

3497 video

Final Answer

\( V = 8\pi/15 \)

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area: \( y=1/x, y=0, x=1, x=3 \)

axis of rotation: \( y=-1 \)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=1/x, y=0, x=1, x=3 \) revolved about \( y=-1 \). Give your answer in exact terms.

Final Answer

\( V = 2\pi/3 + 2\pi\ln(3) = 2\pi(\ln(3)+1/3) \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=1/x, y=0, x=1, x=3 \) revolved about \( y=-1 \). Give your answer in exact terms.

Solution

3498 video

Final Answer

\( V = 2\pi/3 + 2\pi\ln(3) = 2\pi(\ln(3)+1/3) \)

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area: \( f(x)=x, g(x)=x^2-3x \)

axis of rotation: \(y=5\)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( f(x)=x, g(x)=x^2-3x \) revolved about \(y=5\). Give your answer in exact terms.

Hint

Click here to see the dynamic GeoGebra plot.

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( f(x)=x, g(x)=x^2-3x \) revolved about \(y=5\). Give your answer in exact terms.

Final Answer

\(\displaystyle{ V = \frac{1472\pi}{15} }\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( f(x)=x, g(x)=x^2-3x \) revolved about \(y=5\). Give your answer in exact terms.

Hint

Click here to see the dynamic GeoGebra plot.

Solution

1. Draw the plot and the example rectangle and label r and R. See the hint or the animation. We drew the example rectangle perpendicular to the axis of revolution since we were told to use the washer-disc method. The distance r is the distance from the axis of revolution to closest end of the rectangle. The distance R is from the axis of revolution to the farthest end of the example rectangle.

2. Choose The Integral - The volume integral we need is \( V = \pi\int_a^b{R^2-r^2~dx} \). We chose this integral because we are told to use the washer-disc method, so we need an integral with r and R. We integrate with respect to x since the example rectangle is vertical and consequently it moves horizontally, sweeping in the x-direction.

3. Determine r and R - From the plot, let's start on the axis of revolution and move down to the x-axis of revolution. This distance is 5 units. Moving back up to the end of the rectangle that lands on \(y=x\), we move y units. What we are left with is r, so \(r=5-y\). However, we need to replace y with expression for y in terms of x. Since \(y=x\) is the line that we are working with, we have \(r=5-x\).
Determining the expression for R is a bit trickier. Starting on the axis of revolution, we move down to the x-axis which is 5 units. However, when \(x < 3\) we need to go a little further in the same direction to get the full distance R. Let's put that aside for a minute and think about the part of the graph for \(x > 3\). In this case, we need to go back up y units, so \(R=5-y\). This looks the same as r but in this case, we are landing on the plot \(y=x^2-3x\). So \(R=5-(x^2-3x) = -x^2+3x+5\). Let's plug in a few values and compare the values to graph to see if they match.

\(x=3\)

\(R=-3^2+3(3)+5=5\) 𞀄

\(x=4\)

\(R=-4^2+3(4)+5=1\) 𞀄

So far, so good. Let's plug in a few values \(x < 3\) and see what we get.

\(x=0\)

\(R=-0^2+3(0)+5=5\) 𞀄

\(x=2\)

\(R=-2^2+3(2)+5=7\) 𞀄

So it looks like we have the correct equation for R. We can do the same with r to check our equation. This does not guarantee that we have the right equations but it may give an indication if they are incorrect. I usually check both endpoints and at least one other point, two other points is even better.

4. Set up and evaluate the integral - If we look at the animation above, we can tell that the rectangle sweeps across the area from \(x=0\) to \(x=4\). So our integral is \(\displaystyle{ V = \pi \int_0^4{ (-x^2+3x+5)^2 - } }\) \(\displaystyle{ (5-x)^2 ~ dx }\). Let's evaluate it.

\(\displaystyle{ V = \pi\int_0^4{ (-x^2+3x+5)^2 - (5-x)^2 ~ dx } }\)

\(\displaystyle{ V = \pi\int_0^4{ (x^4-3x^3-5x^2-3x^3+ } }\) \(9x^2+15x-5x^2+15x+25) - \) \( (25-10x+x^2) ~dx \)

\(\displaystyle{ V = \pi\int_0^4{ x^4-6x^3-2x^2+40x ~dx } }\)

\(\displaystyle{ V = \pi\left[ \frac{x^5}{5} - \frac{6x^4}{4} -\frac{2x^3}{3}+\frac{40x^2}{2} \right]_0^4 }\)

\(\displaystyle{ V = \pi\left[ \frac{4^5}{5} - \frac{6(4)^4}{4} -\frac{2(4)^3}{3}+\frac{40(4)^2}{2} \right] - 0 }\)

\(\displaystyle{ V = \pi\left[ \frac{1024}{5} - \frac{1536}{4} -\frac{128}{3}+\frac{640}{2} \right] }\)

\(\displaystyle{ V = \pi \frac{1472}{15} }\)

Final Answer

\(\displaystyle{ V = \frac{1472\pi}{15} }\)

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area: \(y=x^2, x=y^2\)

axis of rotation: \(y=1\)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=x^2, x=y^2\) revolved about \(y=1\). Give your answer in exact terms.

Final Answer

\(\displaystyle{V=11\pi/30}\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=x^2, x=y^2\) revolved about \(y=1\). Give your answer in exact terms.

Solution

1172 video

video by Krista King Math

Final Answer

\(\displaystyle{V=11\pi/30}\)

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area: \( y=x^3, y=x, x\geq0 \)

axis of rotation: \(y=-2\)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^3, y=x, x\geq0 \) revolved about \(y=-2\). Give your answer in exact terms.

Final Answer

\(V=25\pi/21\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^3, y=x, x\geq0 \) revolved about \(y=-2\). Give your answer in exact terms.

Solution

In the video, he set up the integral but did not evaluate it. His integral was \(\displaystyle{ \int_{0}^{1}{\pi[(2+x)^2 - \pi(2+x^3)^2]dx} }\). This evaluates to \(\displaystyle{ \pi \left[ x^2 + \frac{x^3}{3} - x^4 - \frac{x^7}{7} \right]_{0}^{1} }\)

1175 video

video by PatrickJMT

Final Answer

\(V=25\pi/21\)

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area: \( y=x^3, y=x, x\geq0 \)

axis of rotation: \(y=5\)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^3, y=x, x\geq0 \) revolved about \(y=5\). Give your answer in exact terms.

Final Answer

\(V=97\pi/42\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^3, y=x, x\geq0 \) revolved about \(y=5\). Give your answer in exact terms.

Solution

1176 video

video by PatrickJMT

Final Answer

\(V=97\pi/42\)

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area: \( y=x^2, y=x \)

axis of rotation: \(y=5\)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, y=x \) revolved about \(y=5\). Give your answer in exact terms.

Final Answer

\( 23\pi/15 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, y=x \) revolved about \(y=5\). Give your answer in exact terms.

Solution

2275 video

video by Michel vanBiezen

Final Answer

\( 23\pi/15 \)

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Now, let's work some problems with the y-axis as the axis of rotation. Here is the plot that contains all the information you need to work these problems.

washer-disc method with y-axis rotation

y-axis rotation practice

area: \( y=x^2, x=0, y=4 \)

axis of rotation: y-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, x=0, y=4 \) revolved about the y-axis. Give your answer in exact terms.

Final Answer

\( V = 8\pi \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, x=0, y=4 \) revolved about the y-axis. Give your answer in exact terms.

Solution

3488 video

Final Answer

\( V = 8\pi \)

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area: \( y=x^{2/3}, x=0, y=1 \)

axis of rotation: y-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^{2/3}, x=0, y=1 \) revolved about the y-axis. Give your answer in exact terms.

Final Answer

\( V = \pi/4 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^{2/3}, x=0, y=1 \) revolved about the y-axis. Give your answer in exact terms.

Solution

3489 video

Final Answer

\( V = \pi/4 \)

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area: \( y^2 = x, x=2y \)

axis of rotation: y-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y^2 = x, x=2y \) revolved about the y-axis. Give your answer in exact terms.

Final Answer

\( V = 64\pi/15 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y^2 = x, x=2y \) revolved about the y-axis. Give your answer in exact terms.

Solution

3496 video

Final Answer

\( V = 64\pi/15 \)

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area: \( y=\sqrt{x}, y=0, x=3 \)

axis of rotation: y-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=\sqrt{x}, y=0, x=3 \) revolved about the y-axis. Give your answer in exact terms.

Final Answer

\( V = 36\pi\sqrt{3}/5 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=\sqrt{x}, y=0, x=3 \) revolved about the y-axis. Give your answer in exact terms.

Solution

3495 video

Final Answer

\( V = 36\pi\sqrt{3}/5 \)

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area: \(x=2\sqrt{y}, x=0, y=9\)

axis of rotation: y-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(x=2\sqrt{y}, x=0, y=9\) revolved about the y-axis. Give your answer in exact terms.

Final Answer

\(V=162\pi\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(x=2\sqrt{y}, x=0, y=9\) revolved about the y-axis. Give your answer in exact terms.

Solution

343 video

video by Krista King Math

Final Answer

\(V=162\pi\)

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area: \(y=1-x^2, y=0\)

axis of rotation: y-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=1-x^2, y=0\) revolved about the y-axis. Give your answer in exact terms.

Final Answer

\(\displaystyle{V=\frac{\pi}{2}}\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=1-x^2, y=0\) revolved about the y-axis. Give your answer in exact terms.

Solution

878 video

video by Krista King Math

Final Answer

\(\displaystyle{V=\frac{\pi}{2}}\)

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area: \(y=x^2, y=5, x=0\)

axis of rotation: y-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=x^2, y=5, x=0\) revolved about the y-axis. Give your answer in exact terms.

Final Answer

\( 25\pi/2 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=x^2, y=5, x=0\) revolved about the y-axis. Give your answer in exact terms.

Solution

2271 video

video by Michel vanBiezen

Final Answer

\( 25\pi/2 \)

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area: \( y=x^2, y=x \)

axis of rotation: y-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, y=x \) revolved about the y-axis. Give your answer in exact terms.

Final Answer

\( \pi/6\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, y=x \) revolved about the y-axis. Give your answer in exact terms.

Solution

2274 video

video by Michel vanBiezen

Final Answer

\( \pi/6\)

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area: \( y=\sqrt{x} \) on \([0,4]\)

axis of rotation: y-axis

method: washer-disc

Problem Statement

Unless otherwise instructed, use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=\sqrt{x} \) on \([0,4]\) revolved about the y-axis. Give your answer in exact terms.

Final Answer

\( V=32\pi/5 \)

Problem Statement

Unless otherwise instructed, use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=\sqrt{x} \) on \([0,4]\) revolved about the y-axis. Give your answer in exact terms.

Solution

2277 video

video by MIP4U

Final Answer

\( V=32\pi/5 \)

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area: \( x^2+y^2=1, y \geq 1/2, x \geq 0 \)

axis of rotation: y-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( x^2+y^2=1, y \geq 1/2, x \geq 0 \) revolved about the y-axis. Give your answer in exact terms.

Final Answer

\( V=5\pi/24 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( x^2+y^2=1, y \geq 1/2, x \geq 0 \) revolved about the y-axis. Give your answer in exact terms.

Solution

2280 video

video by Michel vanBiezen

Final Answer

\( V=5\pi/24 \)

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area: \( y=-3x+6 \), x-axis, y-axis

axis of rotation: y-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=-3x+6 \), the x-axis and the y-axis revolved about the y-axis. Give your answer in exact terms.

Final Answer

\( V=8\pi \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=-3x+6 \), the x-axis and the y-axis revolved about the y-axis. Give your answer in exact terms.

Solution

2281 video

video by Michel vanBiezen

Final Answer

\( V=8\pi \)

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area: \(y=x^2, y=0, x=1\)

axis of rotation: y-axis

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=x^2, y=0, x=1\) revolved about the y-axis. Give your answer in exact terms.

Solution

1185 video

video by Khan Academy

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parallel to y-axis rotation practice

area: \( y=\sqrt{x}, y=0, x=3\)

axis of rotation: \( x=3 \)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=\sqrt{x}, y=0, x=3 \) revolved about the \( x=3 \). Give your answer in exact terms.

Final Answer

\( V = 24\pi\sqrt{3}/5 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=\sqrt{x}, y=0, x=3 \) revolved about the \( x=3 \). Give your answer in exact terms.

Solution

3492 video

Final Answer

\( V = 24\pi\sqrt{3}/5 \)

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area: \( x=y^2, x=1 \)

axis of rotation: \( x=1 \)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( x=y^2, x=1 \) revolved about \( x=1 \). Give your answer in exact terms.

Final Answer

\( V = 16\pi/15 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( x=y^2, x=1 \) revolved about \( x=1 \). Give your answer in exact terms.

Solution

3493 video

Final Answer

\( V = 16\pi/15 \)

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area: \( y=\sqrt{x}, y=0, x=3 \)

axis of rotation: \( x=6 \)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=\sqrt{x}, y=0, x=3 \) revolved about \( x=6 \). Give your answer in exact terms.

Final Answer

\( V = 84\pi\sqrt{3}/5 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=\sqrt{x}, y=0, x=3 \) revolved about \( x=6 \). Give your answer in exact terms.

Solution

3499 video

Final Answer

\( V = 84\pi\sqrt{3}/5 \)

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area: \( y=x^2, x=y^2 \)

axis of rotation: \( x=-1 \)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, x=y^2 \) revolved about \( x=-1 \). Give your answer in exact terms.

Final Answer

\( V = 29\pi\/30 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( y=x^2, x=y^2 \) revolved about \( x=-1 \). Give your answer in exact terms.

Solution

3500 video

Final Answer

\( V = 29\pi\/30 \)

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area: \(y=x^3, y=0, x=1\)

axis of rotation: \(x=2\)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=x^3, y=0, x=1\) revolved about \(x=2\). Give your answer in exact terms.

Final Answer

\( V = 3\pi/5 \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=x^3, y=0, x=1\) revolved about \(x=2\). Give your answer in exact terms.

Solution

1173 video

video by Krista King Math

Final Answer

\( V = 3\pi/5 \)

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area: \(y=2\sqrt{x}, y=0, x=4\)

axis of rotation: \(x=5\)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=2\sqrt{x}, y=0, x=4\) revolved about \(x=5\). Give your answer in exact terms.

Final Answer

\(V=832\pi/15\)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \(y=2\sqrt{x}, y=0, x=4\) revolved about \(x=5\). Give your answer in exact terms.

Solution

1916 video

video by MIP4U

Final Answer

\(V=832\pi/15\)

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area: \( x^2+y^2=1 \), x-axis, y-axis

axis of rotation: \(x=2\)

method: washer-disc

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( x^2+y^2=1 \), the x-axis and the y-axis, revolved about \(x=2\). Give your answer in exact terms.

Final Answer

\( V=(\pi/12)[3\pi-4] \)

Problem Statement

Use the washer-disc method to calculate the volume of rotation of the area bounded by \( x^2+y^2=1 \), the x-axis and the y-axis, revolved about \(x=2\). Give your answer in exact terms.

Solution

2279 video

video by Michel vanBiezen

Final Answer

\( V=(\pi/12)[3\pi-4] \)

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You CAN Ace Calculus

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Topics Listed Alphabetically

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Semiconductors

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