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This page contains practice problems that require you to know both the washer-disc method and the cylinder-shell method.

For these problems, we assume that you have worked lots of problems on the washer-disc and cylinder-shell pages. So you are comfortable working problems that tell you which technique to use. For these problems, we may not tell you which technique to use or you may be asked to use both. Sometimes you will need to set up both integrals and work one or both. Keep in mind that your answer should be the same regardless of which technique you use.
We do not go into a lot of explanation of the basics in these problems. If you don't understand something, go back to the page and section with the technique that you are struggling with and go through those problems again. They have a lot of explanation that will help you with these problems.
These problems are meant to help you hone your skills before your exam. Probably, your exam will not tell which technique to use, so you may have to decide, or you will be asked to use both techniques on the same problem.

Notes

1. The first 4 problems have the same area rotated about different axes. These are good to work in a group so that you can get a feel for how the different axes affect solutions.
2. Many of these problems have limits of integration that are the same for both methods. This is unusual and does not always happen. In these problems, the reason they are equal is due to the choice of the enclosed areas. So do not expect it to happen for every problem.
3. As you work through these problems, compare the two techniques and start to notice patterns. For example, since the rectangles of the two methods are perpendicular, you will always integrate one with respect to x and the other with respect to y.

Practice

Instructions - - Unless otherwise instructed, use both the washer-disc and cylinder-shell methods to solve these problems. Evaluate both integrals, if possible, to confirm that your answers are the same.

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

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

method: both

Problem Statement

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

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

method: both

Hint

Problem Statement

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

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

method: both

Final Answer

\( V = 52\pi/15 \)

Problem Statement

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

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

method: both

Hint

Solution

1. Set up the plots with labels - See the hint for the plots and enclosed area. Although we could have put everything on one plot, we decided to do two separate plots for clarity. Check with your instructor to see what they require. It may help to use colour to separate each technique.

2. Choose The Integrals - The two integrals we need are \(V=\pi\int_a^b{R^2-r^2~dx}\) for the washer-disc method and \(V=2\pi\int_c^d{ph~dy}\) for the cylinder-shell method.

3. Determine p, h, r and R - By now, you should be able to know how to determine the distances, since you worked the practice problems on the washer-disc and cylinder-shell pages. So we just present the answers here.

washer-disc method

\(r=2+x^2\)

\(R=3\)

cylinder-shell method

\(p=2+y\)

\(h=\sqrt{y}\)

4. Set Up the Integrals

washer-disc method

\(\displaystyle{ V = \pi\int_0^1{ 3^2 - (2+x^2)^2 ~ dx } }\)

cylinder-shell method

\(\displaystyle{ V = 2\pi\int_0^1{ (2+y)\sqrt{y}~dy }}\)

Note: The limits of integration for these two integrals are equal. However, that will not usually occur.

5. Evaluate the Integrals

washer-disc method

\(\displaystyle{ V = \pi\int_0^1{ 3^2 - (2+x^2)^2 ~ dx } }\)

\(\displaystyle{ V = \pi\int_0^1{ 9-4-4x^2-x^4 ~ dx } }\)

\(\displaystyle{ V = \pi \left[ 5x - \frac{4x^3}{3} - \frac{x^5}{5} \right]_0^1 }\)

\(\displaystyle{ V = \pi \left[ 5 - \frac{4}{3} - \frac{1}{5} \right] = \frac{52\pi}{15} }\)

cylinder-shell method

\(\displaystyle{ V = 2\pi\int_0^1{ (2+y)\sqrt{y}~dy }}\)

\(\displaystyle{ V = 2\pi\int_0^1{ 2y^{1/2}+y^{3/2} ~dy }}\)

\(\displaystyle{ V = 2\pi \left[ \frac{2y^{3/2}}{3/2} + \frac{y^{5/2}}{5/2} \right]_0^1 }\)

\(\displaystyle{ V = 2\pi \left[ \frac{4}{3} + \frac{2}{5} \right] = \frac{52\pi}{15} }\)

Final Answer

\( V = 52\pi/15 \)

close solution

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

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

method: both

Problem Statement

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

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

method: both

Hint

plots built with GeoGebra

Problem Statement

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

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

method: both

Final Answer

\( V = 11\pi/6 \)

Problem Statement

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

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

method: both

Hint

plots built with GeoGebra

Solution

1. Set up the plots with labels - See the hint for the plots and enclosed area. Although we could have put everything on one plot, we decided to do two separate plots for clarity. Check with your instructor to see what they require. It may help to use colour to separate each technique.

2. Choose The Integrals - The two integrals we need are \(V=\pi\int_c^d{R^2-r^2~dy}\) for the washer-disc method and \(V=2\pi\int_a^b{ph~dx}\) for the cylinder-shell method.

3. Determine p, h, r and R - By now, you should be able to know how to determine the distances, since you worked the practice problems on the washer-disc and cylinder-shell pages. So we just present the answers here.

washer-disc method

\(r=1\)

\(R=1+\sqrt{y}\)

cylinder-shell method

\(p=1+x\)

\(h=1-x^2\)

4. Set Up the Integrals

washer-disc method

\(\displaystyle{ V = \pi\int_0^1{ (1+\sqrt{y})^2 - 1^2 ~ dy } }\)

cylinder-shell method

\(\displaystyle{ V = 2\pi\int_0^1{ (1+x)(1-x^2)~dx } }\)

Note: The limits of integration for these two integrals are the same. However, that will not usually occur.

5. Evaluate the Integrals

washer-disc method

\(\displaystyle{ V = \pi\int_0^1{ (1+\sqrt{y})^2 - 1^2 ~ dy } }\)

\(\displaystyle{ V = \pi\int_0^1{ 1+2\sqrt{y}+y-1 ~ dy } }\)

\(\displaystyle{ V = \pi \left[ \frac{2y^{3/2}}{3/2} + \frac{y^2}{2} \right]_0^1 }\)

\(\displaystyle{ V = \pi \left[ \frac{4}{3} + \frac{1}{2} \right] = \frac{11\pi}{6} }\)

cylinder-shell method

\(\displaystyle{ V = 2\pi\int_0^1{ (1+x)(1-x^2)~dx } }\)

\(\displaystyle{ V = 2\pi\int_0^1{ 1-x^2+x-x^3 ~dx } }\)

\(\displaystyle{ V = 2\pi \left[ x - \frac{x^3}{3} + \frac{x^2}{2} - \frac{x^4}{4} \right]_0^1 }\)

\(\displaystyle{ V = 2\pi \left[ 1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} \right] = \frac{11\pi}{6} }\)

Final Answer

\( V = 11\pi/6 \)

close solution

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

axis of rotation: \(y=6\)

method: both

Problem Statement

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

axis of rotation: \(y=6\)

method: both

Hint

plots built with GeoGebra

Problem Statement

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

axis of rotation: \(y=6\)

method: both

Final Answer

\( V = 36\pi/5 \)

Problem Statement

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

axis of rotation: \(y=6\)

method: both

Hint

plots built with GeoGebra

Solution

1. Set up the plots with labels - See the hint for the plots and enclosed area. Although we could have put everything on one plot, we decided to do two separate plots for clarity. Check with your instructor to see what they require. It may help to use colour to separate each technique.

2. Choose The Integrals - The two integrals we need are \(V=\pi\int_a^b{R^2-r^2~dx}\) for the washer-disc method and \(V=2\pi\int_c^d{ph~dy}\) for the cylinder-shell method.

3. Determine p, h, r and R - By now, you should be able to know how to determine the distances, since you worked the practice problems on the washer-disc and cylinder-shell pages. So we just present the answers here.

washer-disc method

\(r=5\)

\(R=6-x^2\)

cylinder-shell method

\(p=6-y\)

\(h=\sqrt{y}\)

4. Set Up the Integrals

washer-disc method

\(\displaystyle{ V = \pi\int_0^1{ (6-x^2)^2-5^2 ~ dx } }\)

cylinder-shell method

\(\displaystyle{ V = 2\pi\int_0^1{ (6-y)\sqrt{y} ~dy } }\)

Note: The limits of integration for these two integrals are the same. However, that will not usually occur.

5. Evaluate the Integrals

washer-disc method

\(\displaystyle{ V = \pi\int_0^1{ (6-x^2)^2-5^2 ~ dx } }\)

\(\displaystyle{ V = \pi\int_0^1{ 11-12x^2+x^4 ~ dx } }\)

\(\displaystyle{ V = \pi \left[ 11x-\frac{12x^3}{3}+\frac{x^5}{5} \right]_0^1 }\)

\(\displaystyle{ V = \pi \left[ 11-4+\frac{1}{5} \right] = \frac{36\pi}{5} }\)

cylinder-shell method

\(\displaystyle{ V = 2\pi\int_0^1{ (6-y)\sqrt{y} ~dy } }\)

\(\displaystyle{ V = 2\pi\int_0^1{ 6\sqrt{y}-y^{3/2} ~dy } }\)

\(\displaystyle{ V = 2\pi \left[ \frac{6y^{3/2}}{3/2} - \frac{y^{5/2}}{5/2} \right]_0^1 }\)

\(\displaystyle{ V = 2\pi \left[ 4 - \frac{2}{5} \right] = \frac{36\pi}{5} }\)

Final Answer

\( V = 36\pi/5 \)

close solution

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

axis of rotation: \(x=2\)

method: both

Problem Statement

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

axis of rotation: \(x=2\)

method: both

Hint

plots built with GeoGebra

Problem Statement

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

axis of rotation: \(x=2\)

method: both

Final Answer

\( V = 13\pi/6 \)

Problem Statement

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

axis of rotation: \(x=2\)

method: both

Hint

plots built with GeoGebra

Solution

1. Set up the plots with labels - See the hint for the plots and enclosed area. Although we could have put everything on one plot, we decided to do two separate plots for clarity. Check with your instructor to see what they require. It may help to use colour to separate each technique.

2. Choose The Integrals - The two integrals we need are \(V=\pi\int_c^d{R^2-r^2~dy}\) for the washer-disc method and \(V=2\pi\int_a^b{ph~dx}\) for the cylinder-shell method.

3. Determine p, h, r and R - By now, you should be able to know how to determine the distances, since you worked the practice problems on the washer-disc and cylinder-shell pages. So we just present the answers here.

washer-disc method

\(r=2-\sqrt{y}\)

\(R=2\)

cylinder-shell method

\(p=2-x\)

\(h=1-x^2\)

4. Set Up the Integrals

washer-disc method

\(\displaystyle{ V = \pi\int_0^1{ 2^2 - (2-\sqrt{y})^2 ~ dy } }\)

cylinder-shell method

\(\displaystyle{ V = 2\pi\int_0^1{ (2-x)(1-x^2) ~dx } }\)

Note: The limits of integration for these two integrals are the same. However, that will not usually occur.

5. Evaluate the Integrals

washer-disc method

\(\displaystyle{ V = \pi\int_0^1{ 2^2 - (2-\sqrt{y})^2 ~ dy } }\)

\(\displaystyle{ V = \pi\int_0^1{ 4 - 4 + 4\sqrt{y} - y ~ dy } }\)

\(\displaystyle{ V = \pi \left[ \frac{4y^{3/2}}{3/2} - \frac{y^2}{2} \right]_0^1 }\)

\(\displaystyle{ V = \pi \left[ \frac{8}{3} - \frac{1}{2} \right] = \frac{13\pi}{6} }\)

cylinder-shell method

\(\displaystyle{ V = 2\pi\int_0^1{ (2-x)(1-x^2) ~dx } }\)

\(\displaystyle{ V = 2\pi\int_0^1{ 2-2x^2-x+x^3 ~dx } }\)

\(\displaystyle{ V = 2\pi \left[ 2x-\frac{2x^3}{3}-\frac{x^2}{2}+\frac{x^4}{4} \right]_0^1 }\)

\(\displaystyle{ V = 2\pi \left[ 2 - \frac{2}{3}-\frac{1}{2}+\frac{1}{4} \right] = \frac{13\pi}{6} }\)

Final Answer

\( V = 13\pi/6 \)

close solution

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

axis of rotation: x-axis

method: both

Problem Statement

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

axis of rotation: x-axis

method: both

Hint

plots built with GeoGebra

Problem Statement

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

axis of rotation: x-axis

method: both

Final Answer

\( V = 8\pi/27 \)

Problem Statement

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

axis of rotation: x-axis

method: both

Hint

plots built with GeoGebra

Solution

1. Set up the plots with labels - See the hint for the plots and enclosed area. Although we could have put everything on one plot, we decided to do two separate plots for clarity. Check with your instructor to see what they require. It may help to use colour to separate each technique.

2. Choose The Integrals - The two integrals we need are \(V=\pi\int_a^b{R^2-r^2~dx}\) for the washer-disc method and \(V=2\pi\int_c^d{ph~dy}\) for the cylinder-shell method.

3. Determine p, h, r and R - By now, you should be able to determine the distances, since you worked the practice problems on the washer-disc and cylinder-shell pages. So we just present the answers here.

washer-disc method

\(r=1/(x+1) \)

\(R=1-x/3\)

cylinder-shell method

\(p=y\)

\(h=(3-3y)-(1/y-1)=4-3y-1/y\)

4. Set Up the Integrals

washer-disc method

\(\displaystyle{ V = \pi\int_0^2{ \left(1-\frac{x}{3} \right)^2 - \left( \frac{1}{x+1} \right)^2 ~dx } }\)

cylinder-shell method

\(\displaystyle{ V = 2\pi\int_{1/3}^1{ y(4-3y-1/y) ~dy } }\)

5. Evaluate the Integrals

washer-disc method

\(\displaystyle{ V = \pi\int_0^2{ \left(1-\frac{x}{3} \right)^2 - \left( \frac{1}{x+1} \right)^2 ~dx } }\)

\(\displaystyle{ V = \pi\int_0^2{ 1-\frac{2x}{3}+\frac{x^2}{9} - \frac{1}{(x+1)^2} ~dx } }\)

\(\displaystyle{ V = \pi \left[ x-\frac{2x^2}{6}+\frac{x^3}{27} \right]_0^2 - \pi \int_0^2{ \frac{dx}{(x+1)^2} } }\)

In the last line, the remaining integral is easily evaluated using integration by substitution where \(u=x+1\). It evaluates to \(2/3\).

\(\displaystyle{ V = \pi \left[ 2-\frac{4}{3}+\frac{8}{27}-\frac{2}{3} \right] = \frac{8\pi}{27} }\)

cylinder-shell method

\(\displaystyle{ V = 2\pi\int_{1/3}^1{ y(4-3y-1/y) ~dy } }\)

\(\displaystyle{ V = 2\pi \int_{1/3}^1{ 4y-3y^2-1 ~dy } }\)

\(\displaystyle{ V = 2\pi \left[ \frac{4y^2}{2}-\frac{3y^3}{3} -y \right]_{1/3}^1 }\)

\(\displaystyle{ V = 2\pi \left[ 2-1-1 \right] -2\pi \left[ \frac{2}{9} - \frac{1}{27} - \frac{1}{3} \right] = \frac{8\pi}{27} }\)

Final Answer

\( V = 8\pi/27 \)

close solution

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

axis of rotation: x-axis

method: both

Problem Statement

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

axis of rotation: x-axis

method: both

Hint

plots built with GeoGebra

Problem Statement

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

axis of rotation: x-axis

method: both

Final Answer

\( V = 4\pi/15 \)

Problem Statement

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

axis of rotation: x-axis

method: both

Hint

plots built with GeoGebra

Solution

1. Set up the plots with labels - See the hint for the plots and enclosed area. Although we could have put everything on one plot, we decided to do two separate plots for clarity. Check with your instructor to see what they require. It may help to use colour to separate each technique.

2. Choose The Integrals - The two integrals we need are \(V=\pi\int_a^b{R^2-r^2~dx}\) for the washer-disc method and \(V=2\pi\int_c^d{ph~dy}\) for the cylinder-shell method.

3. Determine p, h, r and R - By now, you should be able to know how to determine the distances, since you worked the practice problems on the washer-disc and cylinder-shell pages. So we just present the answers here.

washer-disc method

\(r=x \)

\(R=x^{1/3}\)

cylinder-shell method

\(p=y\)

\(h=y-y^3\)

4. Set Up the Integrals

washer-disc method

\(\displaystyle{ V = \pi\int_0^1{ x^{2/3}-x^2 ~dx } }\)

cylinder-shell method

\(\displaystyle{ V = 2\pi\int_0^1{ y(y-y^3) ~dy } }\)

Note: The limits of integration for these two integrals are the same. However, that will not usually occur.

5. Evaluate the Integrals

washer-disc method

\(\displaystyle{ V = \pi\int_0^1{ x^{2/3}-x^2 ~dx } }\)

\(\displaystyle{ V = \pi \left[ \frac{x^{5/3}}{5/3} -\frac{x^3}{3} \right]_0^1 }\)

\(\displaystyle{ V = \pi \left[ \frac{3}{5} -\frac{1}{3} \right] = \frac{4\pi}{15} }\)

cylinder-shell method

\(\displaystyle{ V = 2\pi\int_0^1{ y(y-y^3) ~dy } }\)

\(\displaystyle{ V = 2\pi\int_0^1{ y^2-y^4 ~dy } }\)

\(\displaystyle{ V = 2\pi \left[ \frac{y^3}{3} - \frac{y^5}{5} \right]_0^1 }\)

\(\displaystyle{ V = 2\pi \left[ \frac{1}{3} - \frac{1}{5} \right] = \frac{4\pi}{15} }\)

Final Answer

\( V = 4\pi/15 \)

close solution

area: \( y=\sqrt{\ln(x^2)}, y=\sqrt{\ln x}, y=1 \)

axis of rotation: x-axis

method: both

Problem Statement

area: \( y=\sqrt{\ln(x^2)}, y=\sqrt{\ln x}, y=1 \)

axis of rotation: x-axis

method: both

Hint

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

plots built with GeoGebra

Problem Statement

area: \( y=\sqrt{\ln(x^2)}, y=\sqrt{\ln x}, y=1 \)

axis of rotation: x-axis

method: both

Final Answer

\( V = \pi[1+e-2e^{1/2}] \)

Problem Statement

area: \( y=\sqrt{\ln(x^2)}, y=\sqrt{\ln x}, y=1 \)

axis of rotation: x-axis

method: both

Hint

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

plots built with GeoGebra

Solution

1. Set up the plots with labels - See the hint for the plots and enclosed area. Although we could have put everything on one plot, we decided to do two separate plots for clarity. Check with your instructor to see what they require. It may help to use colour to separate each technique.

2. Choose The Integrals - The two integrals we need are \(V=\pi\int_a^b{R^2-r^2~dx}\) for the washer-disc method and \(V=2\pi\int_c^d{ph~dy}\) for the cylinder-shell method.

3. Determine p, h, r and R and set up integrals - From the plots, we can tell that cylinder-shell method will be easier since only one integral is required. But we were asked to set up both, so let's start with the easy one. This is probably what you would do on an exam.

cylinder-shell method

Our integral is \(V=2\pi\int_c^d{ph~dy}\). Since this is in terms of y, we will probably need our functions solved for x. So let's go ahead and do that.
\( y=\sqrt{\ln(x^2)} \to y^2 = \ln(x^2)\)
This is a valid operation since x and y are both positive.
\( y^2 = \ln(x^2) \to e^{y^2} = e^{\ln(x^2)} = x^2 \to x = e^{y^2/2}\)
Similarly for the second equation, we have
\(y=\sqrt{\ln x} \to y^2 = \ln x \to x=e^{y^2}\)
So for our two curves, we have

upper curve

\(y=\sqrt{\ln(x^2)}\)

\(x=e^{y^2/2}\)

lower curve

\(y=\sqrt{\ln(x)}\)

\(x=e^{y^2}\)

Of course, these equations do not hold for all x and y, just in the first quadrant, which our area is in. So we are okay.
So, we determined that the equation we need is \(V=2\pi\int_c^d{ph~dy}\). The equation for p is pretty easy, it is just y. The equation for h is also pretty easy if you remember how to get the length of a line between two curves in the y direction (from the area between two curves page). It is just the equation of the line on the right minus the equation of the line on the left. In this case, we have \(h=e^{y^2} - e^{y^2/2}\).
These give the resulting integral of \(V=2\pi\int_0^1{y(e^{y^2} - e^{y^2/2})~dy}\)

washer-disc method

We decided to do this method second, since finding the volume is more difficult in that we need two integrals. At the point \(x=e^{1/2}\), the top end of the rectangle jumps from the curve \(y=\sqrt{\ln(x^2)}\) to \(y=1\). So we are going to have one integral on the interval \([1,e^{1/2}]\) and a second integral on the interval \([e^{1/2},e]\). Let's work left to right, so we we will start with the interval \([1,e^{1/2}]\). The distance r is from the axis of rotation, the x-axis, to the lower plot. So this distance is \(r=\sqrt{\ln x}\). The distance R is also pretty easy. It goes from the x-axis to the upper plot, which is \(R=\sqrt{\ln(x^2)}\). These results give our first integral as \(V_1=\pi \int_1^{e^{1/2}}{ \ln(x^2)-\ln(x) ~dx}\).
For the second interval/integral, we have the distance r the same as the previous interval and for the distance R, we have \(R=1\) no matter where the example rectangle is. So our second integral is \(V_2=\pi\int_{e^{1/2}}^e{1-\ln(x)~dx}\)
Our total volume we have \(V=V_1+V_2 = \pi \int_1^{e^{1/2}}{ \ln(x^2)-\ln(x) ~dx} + \pi\int_{e^{1/2}}^e{1-\ln(x)~dx}\).
At this point, we may be tempted to try to combine these integrals somehow but it is better to leave them separate for now, for two reasons. First, it may not be possible to combine them and, in most cases, we can't. Secondly, trying to do so introduces the possibility of making a mistake. So we will leave them as they are.

4. Evaluate the Integrals
Let's look at the washer-disc method integrals first. Notice that in both integrals we have \(\int{\ln(x)~dx}\). This integral requires the technique integration by parts, which you have probably not had yet. So, we gave you the answer in the hint. If you want to see how to do this integral, see the integral by parts page.

washer-disc method

\(\displaystyle{ V = \pi \int_1^{e^{1/2}}{ \ln(x^2)-\ln(x) ~dx} + \pi\int_{e^{1/2}}^e{1-\ln(x)~dx} }\)

Since \(\ln(x^2) = 2\ln 2\), the integrand of the first integral is \(2\ln(x)-\ln(x)=\ln(x)\)

\(\displaystyle{ V = \pi \int_1^{e^{1/2}}{ \ln(x) ~dx} + \pi\int_{e^{1/2}}^e{1-\ln(x)~dx} }\)

\(\displaystyle{ V = \pi \left[ x\ln x - x \right]_1^{e^{1/2}} + \pi\left[ x-(x\ln x -x) \right]_{e^{1/2}}^e }\)

\(\displaystyle{ V = \pi[ 1+e-2e^{1/2}]}\)

cylinder-shell method

\(\displaystyle{ V=2\pi\int_0^1{y(e^{y^2} - e^{y^2/2})~dy} }\)

Use integration by substitution.

\(u=y^2 \to du=2y~dy \)

Converting the limits of integration from y's to u's, we have

\(y=1 \to u=1\) and \(y=0 \to u=0\)

\(\displaystyle{ V=2\pi\int_0^1{ e^u - e^{u/2} (du/2) } }\)

\(\displaystyle{ V= \pi\left[ e^u - 2e^{u/2} \right]_0^1 }\)

\(\displaystyle{ V = \pi[ 1+e-2e^{1/2}]}\)

Final Answer

\( V = \pi[1+e-2e^{1/2}] \)

close solution

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

axis of rotation: x-axis

method: both

Problem Statement

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

axis of rotation: x-axis

method: both

Hint

plot built with GeoGebra

Problem Statement

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

axis of rotation: x-axis

method: both

Final Answer

\( V = \pi/9 \)

Problem Statement

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

axis of rotation: x-axis

method: both

Hint

plot built with GeoGebra

Solution

1. Set up the plot with labels - See the hint for the plot and enclosed area. There is no way to easily solve the equation for x, so we cannot do the cylinder-shell method.

2. Choose The Integral - The integral we need is \(V=\pi\int_a^b{R^2-r^2~dx}\) for the washer-disc method.

3. Determine r and R and set up integral - Since the axis of rotation runs right along one edge of the area, \(r=0\). The distance R is just \(y=x-x^4 \to R=x-x^4\). So our integral is \(V=\pi\int_0^1{ (x-x^4)^2 ~dx}\)

4. Evaluate the Integral

washer-disc method

\(\displaystyle{ V=\pi\int_0^1{ (x-x^4)^2 ~dx} }\)

\(\displaystyle{ V=\pi\int_0^1{ x^2-2x^5+x^8 ~dx} }\)

\(\displaystyle{ V=\pi \left[ \frac{x^3}{3}-\frac{2x^6}{6} +\frac{x^9}{9} \right]_0^1 }\)

\(\displaystyle{ V=\pi \left[ \frac{1}{3}-\frac{1}{3} +\frac{1}{9} \right] = \frac{\pi}{9} }\)

Final Answer

\( V = \pi/9 \)

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