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Practice Problems: Rotational Volumes

To determine the volume of rotational symmetries there are two main protocols to follow: the Washer/Disk Method or the Shell Method. The washer method is a generalization of the disk method. Begin by sketching a 2D model, and DO NOT forget to include the cross-section. Most students are unable to solve these problems because they forgot to include this part. The cross-section is important for various reasons. Not only does it quickly help you determine which method to use, it also helps to measure heights and radii. 

To compute the radius, always measure FROM (not TO but FROM) the axis of the rotation. 

  • Washer/Disk Method
    • Cross-Sections are taken perpendicular to the Axis of Rotation
  • (Cylindrical) Shell Method
    • Cross-Sections are taken parallel to the Axis of Rotation

Virtual Lessons

Need some additional help understanding how to apply this integration technique? Click Here to visit the virtual lesson section.

Practice Problems

Determine the volume for the solid by integrating congruent cross-sections.

  1. \(\displaystyle \begin{align}y &= x^{-1} \\ y& = 0\\ 1\leq &x\leq 4\end{align}\)
  2. \(\displaystyle \begin{align}y&= \texttt{ln}(x) \\y &= 1\\ 0\leq &x\leq \texttt{e}\end{align} \)
  3. \(\displaystyle \begin{align}y &= 2x \\y &= x+1 \\ 0\leq &x\leq 1 \end{align} \)
  4. \(\displaystyle \begin{align} f(x) &= x^{2/3} \\ g(x) &= x^{3/2} \\  0\leq &x \leq 1 \end{align} \)
  1. \(\displaystyle \begin{align}f(x) &= \texttt{arcsin}(x) \\ 0\leq &y\leq 1 \end{align}\)
  2. \(\displaystyle \begin{align} r^2 &= (x – a)^2 + y^2  \\ V &= 2{\pi}^2 ar^2 \end{align}\)
  3. \(\displaystyle \begin{align}f(x) &= 4 – x^2 \\ g(x) &= 2- x \\  0\leq &x\leq 2 \end{align}\)
  4. \(\displaystyle \begin{align} f(x) &= x^3 \\ g(x) &= x \\ 0\leq &x \leq 1 \end{align}\)
  1. \(\displaystyle \begin{align}f(x) &= (x-2)^3 \\ g(x) &= x- 2 \\ 1\leq x\leq 2 \end{align} \)
  2. \(\displaystyle 2\pi \int^1_0 (2x-x^2) dx \)
  3. \(\displaystyle 2\pi \int^1_0 (y+2)\left(\sqrt{1-y^2}\right) dy \)

Suggestive Solution Guide

Sketch the indicated region bounded by the given graphs and compute the volume that is generated by rotating the region about the x-axis:

 

Sketeched area to be rotated around the x-axis. Given how we already have the endoints and area under the curve, the only thing left is to select the appropiate method of integration. Given how we will rotate the area around the x-axis and the differential segment is perpendicular to the axis of rotation without any airgap, we choose the disk mehtod. Let the radius defined by the length of the segment be the function itself \(\left( y = \frac{1}{x} \right)\)

\(
\begin{align*}
& V = \pi \int^4_1 \left( \frac{1}{x} \right)^2 dx = \pi \int^4_1 \frac{1}{x^2} dx = \pi \left[ \frac{-1}{x} \right] \Bigg\rvert^4_1 = \pi \left[ \left(\frac{-1}{4}\right) – \left(\frac{-1}{1}\right)\right] = \pi \left[ \left(\frac{-1}{4}\right) + 1 \right] \\[5pt]
& \color{green}{V = \frac{3}{4}\pi}
\end{align*}
\)

Sketch the indicated region bounded by the given graphs and compute the volume that is generated by rotating the region about the x-axis:

Sketched area to be rotated around the x-axis. Because we already know that the area is going to be rotated around the x-axis, then the graphic can’t go beyond said rotational axis. This is why we have to split the integration area into two different integrals (where the changes are the endpoints and the radius. First, let’s calulate the volume of the pink-shaded area using the disk method:

\(
\begin{align*}
& V = \pi \int^1_0 (1)^2 dx = \pi[x]\bigg\rvert^1_0 = \pi[ 1 -0] = \pi
\end{align*}
\)

Now, for the second area (shaded in blue), use the washer method because of the airgap between the axis of rotation and the area to be rotated. Let \(R = 1, r = \texttt{ln}(x)\), and the integration limits be \(1 \le x \le \texttt{e}\). Using this data yields:

\(
\begin{align*}
& V = \pi \int^{\texttt{e}}_1 \left[ (1)^2 – \left(\texttt{ln}(x)\right)^2 \right] dx
\end{align*}
\)

In order to solve for the second integrand, use IBP. Let \(u=\left(\texttt{ln}(x)\right)^2\) and \(dv = dx\), so \(du = \frac{2\texttt{ln}(x)}{x}dx\) and \(v = x\). Applying those substitutions yields:

\(
\begin{align*}
\int \left(\texttt{ln}(x)\right)^2 dx = x\texttt{ln}^2(x) \:- 2\int \frac{\texttt{ln}(x)}{x}(x) dx = x\texttt{ln}^2(x) \:- 2\int \texttt{Ln}(x)dx
\end{align*}
\)

Apply IBP once again. Let \(u=\texttt{ln}(x)\) and \(dv = dx\), so \(du = \frac{1}{x}dx\) and \(v = x\). Applying those substitutions yields:

\(
\begin{align*}
x\texttt{ln}^2(x) \:- 2\int \texttt{ln}(x)dx
& = x\texttt{ln}^2(x) \:- 2\left[x\texttt{ln}(x) \:- \int \frac{1}{x}(x)dx\right] \\[5pt]
& = x\texttt{ln}^2(x) \:- 2x\texttt{ln}(x) \:+ 2\int dx \\[5pt]
& = x\texttt{ln}^2(x) \:- 2x\texttt{ln}(x) \:+ 2x \\[5pt]
\end{align*}
\)

Use the previous result to find the final volume:

\(
\begin{align*}
& V = \pi \int^{\texttt{e}}_1 \left[ (1)^2 – \left(\texttt{ln}(x)\right)^2 \right] dx \\[5pt]
& V = \pi \left[ x \:- \left( x\texttt{ln}^2(x) \:- 2x\texttt{Ln}(x) \:+ 2x \right) \right] \Bigg\rvert^{\texttt{e}}_1
= \pi \left[ x \:- x\texttt{ln}^2(x) \:+ 2x\texttt{ln}(x) \:- 2x \right] \Bigg\rvert^{\texttt{e}}_1 \\[5pt]
& V = \pi \left[ \left(\texttt{e} \:- \texttt{e}\texttt{ln}^2(\texttt{e}) \:+ 2\texttt{e}\texttt{ln}(\texttt{e}) \:- 2\texttt{e}\right) \:- \left(1 \:- \texttt{ln}^2(1) \:+ 2\texttt{ln}(1) \:- 2\right) \right] \\[5pt]
& V = \pi [(\texttt{e} \:- \texttt{e} \:+ 2\texttt{e} \:- 2\texttt{e}) \:- (1 \:- 0 + 0 – 2)] \\[5pt]
& V = \pi[2 \:- 1] = \pi
\end{align*}
\)

Finally, combine the results of both definite integrals to obtain the final answer:

\(
\begin{align*}
& V = \pi \int^1_0 (1)^2 dx \:+ \pi \int^{\texttt{e}}_1 \left[ (1)^2 – \left(\texttt{ln}(x)\right)^2 \right] dx = \pi + \pi\\[5pt]
& \color{green}{V = 2\pi}
\end{align*}
\)

Compute the volume generated by rotating the indicated bounded region about the x-axis:

Sketched area to be rotated around the x-axis. Given the existence of a gap between the rotational area and the axis of rotation, we use the washer method. Let R = x + 1 and r = 2x. Applying this data into the integrand yields:

\(
\begin{align*}
& V = \pi \int^1_0 \left[(x+1)^2\ – (2x)^2 \right] dx = \pi \int^1_0 (-3x^2 + 2x + 1) dx = \pi \left[-x^3 \:+ x^2 \:+ x\right] \Bigg\rvert^1_0 \\[5pt]
& V = \pi\left[ \left(-(1)^3 \:+ 1^2 \:+ 1\right) \:- \left(-\frac{0^3}{3} \:+ 0^2 \:+ 0\right) \right] \\[5pt]
& \color{green}{V = \pi}
\end{align*}
\)

Compute the colume generated by rotating the region bounded by the graphs of the given equations:

  • (a) About the x-axis

Sketched area to rotate around the axis. Given the airgap between the shaded area and the axis of rotation, and that the chosen differential is perpendicular to the axis, use the washer method. Let \(R = x^{\frac{2}{3}}\) and \(r = x^{\frac{3}{2}}\). Using this data in the integral yields:

\(
\begin{align*}
& V = \pi \int^1_0 \left[ \left(x^{\frac{2}{3}}\right)^2 \:- \left(x^{\frac{3}{2}}\right)^2 \right] dx = \left[ \frac{3}{7}x^{\frac{7}{3}} \:- \frac{1}{4}x^4 \right] \Bigg\rvert^1_0 = \left[ \frac{3}{7}(1)^{\frac{7}{3}} \:- \frac{1}{4}(1)^4 \right] \:- \left[ \frac{3}{7}(0)^{\frac{7}{3}} \:- \frac{1}{4}(0)^4 \right] \\[5pt]
& \color{green}{V = \frac{5}{28}\pi}
\end{align*}
\)

 

  • (b) About x=-2

Sketched area and updated axis of rotation. Given how we can’t change the orientation of the differential (mainly because both ends have to touch both functions), we choose to use the shell method. For that, we need to identify the radius and the height. Let \(R = x^{\frac{2}{3}} – x^{\frac{3}{2}}\) and \(H = x + 2\). Using this values in the integral yields:

\(
\begin{align*}
& V = 2\pi \int^1_0 (x+2) \left(x^{\frac{2}{3}} – x^{\frac{3}{2}}\right) dx = 2\pi \int^1_0 \left( x^{\frac{5}{3}} \:- x^{\frac{5}{2}} \:+ 2x^{\frac{2}{3}} \:- 2x^{\frac{3}{2}} \right)dx = 2\pi\left[ \frac{3}{8}x^{\frac{8}{3}} \:- \frac{2}{7}x^{\frac{7}{2}} \:+ (2)\frac{3}{5}x^{\frac{5}{3}} \:- (2)\frac{2}{5}x^{\frac{5}{2}} \right] \Bigg\rvert^1_0 \\[5pt]
& V = 2\pi \left( \left[ \frac{3}{8}(1)^{\frac{8}{3}} \:- \frac{2}{7}(1)^{\frac{7}{2}} \:+ \frac{6}{5}(1)^{\frac{5}{3}} \:- \frac{4}{5}(1)^{\frac{5}{2}} \right] \:- \left[ \frac{3}{8}(0)^{\frac{8}{3}} \:- \frac{2}{7}(0)^{\frac{7}{2}} \:+ \frac{6}{5}(0)^{\frac{5}{3}} \:- \frac{4}{5}(0)^{\frac{5}{2}} \right] \right) \\[5pt]
& V = 2\pi \left[ \frac{3}{8} \:- \frac{2}{7} \:+ \frac{6}{5} \:- \frac{4}{5} \right] = 2\pi \left[\frac{137}{280}\right] \\[5pt]
& \color{green}{V = \frac{137}{140}\pi}
\end{align*}
\)

Compute the volume generated by rotating the region bounded by the graphs of the given equations:

  • (a) About the y-axis

Sketched area to be rotated around the y-axis.
Given the parallel orientation of the differential line regarding the axis of rotation, use the disk method to compute the revolution solid. Calculate the main f(x) function in terms of y: \(f(y) = sin(y)\). The limits of integration are \(0 \le y \le 1\):

\(
\begin{align*}
& V = \pi \int^1_0 \texttt{sin}^2(y) dy = \frac{\pi}{2} \int^1_0 (1 \:- \texttt{cos}(2y))dy \\[5pt]
& V = \frac{\pi}{2} \left[ y \:- \frac{1}{2}\texttt{sin}(2y) \right] \Bigg\rvert^1_0 = \frac{\pi}{2} \left[ \left((1) \:- \frac{1}{2}\texttt{sin}(2(1))\right) \:- \left((0) \:- \frac{1}{2}\texttt{sin}(2(0))\right) \right] \\[5pt]
& V = \frac{1}{2} \left[ 1 \:- \frac{1}{2}\texttt{sin}(2) + \frac{1}{2}\texttt{sin}(0) \right] = \frac{1}{2} \left[ 1 \:- \texttt{sin}(2) \right] \\[5pt]
& \color{green}{V = \frac{1}{2} \:- \frac{1}{2}\texttt{sin}(2)}
\end{align*}
\)

 

  • (b) About x=2

Sketched area to be rotated around \(x = 2\) as an axis of revolution.
The main problem remains almost the same as the last one, but this time we have an airgap in between the rotational area instead.
Given the orientation of the differential line (as it is perpendicular to the axis of rotation), the best method to use in this case is the washer method.

\(
\begin{align*}
& R = 2; \Rightarrow R^2 = 4 \\[5pt]
& r = 2 \:- \texttt{sin}(y); \Rightarrow r^2 = (2 \:- \texttt{sin}(y))^2 = 4 \:- 4\texttt{sin}(y) + \texttt{sin}^2(y) \\[15pt]
& V = \pi \int^1_0 (R^2 \:- r^2) dy = \pi \int^1_0 \left[ 4 \:- (4 \:- 4\texttt{sin}(y) + \texttt{sin}^2(y)) \right] dy \\[5pt]
& V = \pi \int^1_0 \left( 4\texttt{sin}(y) \:- \texttt{sin}^2(y) \right) dy = \pi \left[ 4\int^1_0 \texttt{sin}(y)dy \:- \int^1_0 \texttt{sin}^2(y)dy \right] \\[5pt]
& V = \pi \left[ -4\texttt{cos}(y) \:- \frac{1}{2}\int^1_0 (1 \:- \texttt{cos}(2y))dy \right] = \pi \left[ -4\texttt{cos}(y) \:- \frac{1}{2}y + \frac{1}{4}\texttt{sin}(2y) \right] \Bigg\rvert^1_0 \\[5pt]
& V = \pi \left( \left[ -4\texttt{cos}(1) \:- \frac{1}{2}(1) + \frac{1}{4}\texttt{sin}(2(1)) \right] \:- \left[ -4\texttt{cos}(0) \:- \frac{1}{2}(0) + \frac{1}{4}\texttt{sin}(2(0)) \right] \right) \\[5pt]
& V = \pi \left[ -4\texttt{cos}(1) \:- \frac{1}{2} + \frac{1}{4}\texttt{sin}(2) + 4 \right] \\[5pt]
& \color{green}{V = \pi \left[ \frac{1}{4}\texttt{sin}(2) \:- 4\texttt{cos}(1) + \frac{7}{2} \right]}
\end{align*}
\)

 

Prove that by rotating the circle \((x-a)^2 + y^2 = r^2\) about the y-axis generates a torus whose volume is \(V = 2\pi^2ar^2\):

Sketched area to be rotated around the y-axis. For the sake of an easier integration process, let everything be with respect to y. The initial equation of the circle would be:

\(
\begin{align*}
(x-a)^2 + y^2 = r^2 \\[5pt]
(x-a)^2 = r^2 – y^2 \\[5pt]
x-a = \sqrt{r^2-y^2} \\[5pt]
x = \sqrt{r^2 – y^2} + a
\end{align*}
\)

 

Given the function is a circle, the integration limits with regards to y are -r and r. We will use the washer method in order to calculate the volume of the shape:

\(
\begin{align*}
& R = \sqrt{r^2 – y^2} + a; \Rightarrow R^2 = \left(a+\sqrt{r^2 – y^2}\right)^2 = a^2 + 2a\sqrt{r^2 – y^2} + r^2-y^2 \\[5pt]
& r = -\sqrt{r^2 – y^2} + a; \Rightarrow R^2 = \left(a-\sqrt{r^2 – y^2}\right)^2 = a^2 – 2a\sqrt{r^2 – y^2} + r^2-y^2 \\[5pt]
& V = \pi \int^r_{-r} \left(R^2 – r^2\right) dy = \pi \int^r_{-r} \left(\left[a^2 + 2a\sqrt{r^2 – y^2} + r^2-y^2\right] – \left[ a^2 – 2a\sqrt{r^2 – y^2} + r^2-y^2 \right] \right) dy \\[5pt]
& V = \pi \int^r_{-r} \left(a^2 + 2a\sqrt{r^2 – y^2} + r^2 \:- y^2 \:-  a^2 + 2a\sqrt{r^2 – y^2} \:- r^2 \:+ y^2 \right) dy = \pi \int^r_{-r} 4a\sqrt{r^2-y^2} dy \\[5pt]
& V = 4a\pi \int^r_{-r} \sqrt{r^2 \:- y^2} dy = 8a\pi \int^r_0 \sqrt{r^2 \:- y^2} dy
\end{align*}
\)

 

Apply trigonometric substitution in order to solve the final integral. Let \(y = r\texttt{sin}(\theta)\), so \(dy = r\texttt{cos}(\theta)d\theta\). Applying this subsitution in the radical yields:

\(
\begin{align*}
& \sqrt{r^2 \:- y^2} = \sqrt{r^2 \:- (r\texttt{sin}(\theta))^2} = \sqrt{r^2 \:- r^2\texttt{sin}^2(\theta)} = \sqrt{r^2 (1 – \texttt{sin}^2(\theta))} = \sqrt{r^2\texttt{cos}^2(\theta)} = r\texttt{cos}(\theta) \\[5pt]
& V = 8a\pi \int^r_0 \sqrt{r^2 \:- y^2} = 8a\pi \int^r_0 \left[r\texttt{cos}(\theta)\right] \left[r\texttt{cos}(\theta)d\theta\right] = 8ar^2\pi \int^r_0 \texttt{cos}^2(\theta)d\theta = 8ar^2\pi \left(\frac{1}{2}\right)\int^r_0 \left(1 + \texttt{cos}(2\theta)\right) d\theta \\[5pt]
& V = 4ar^2\pi \left[ \theta \:+ \frac{1}{2}\texttt{sin}(2\theta) \right] \Bigg\rvert^r_0 = 4ar^2\pi \left[ \theta \:+ \texttt{sin}(\theta)\texttt{cos}(\theta) \right] \Bigg\rvert^r_0 \\[5pt]
\end{align*}
\)

 

Substitute back to the original variable by using the \(\texttt{sin}(\theta) = \frac{y}{r}\) substitution done earlier:

\(
\begin{align*}
& V = 4ar^2\pi \left[ \theta \:+ \texttt{sin}(\theta)\texttt{cos}(\theta) \right] \Bigg\rvert^r_0 = 4ar^2 \pi \left[ \texttt{Arcsin}\left( \frac{y}{r} \right) \:+ \left(\frac{y}{r}\right)\left(\frac{\sqrt{r^2-y^2}}{r}\right)\right] \Bigg\rvert^r_0 \\[5pt]
& V = 4ar^2 \pi \left[ \texttt{Arcsin}\left( \frac{y}{r} \right) \:+ \frac{y\sqrt{r^2-y^2}}{r^2} \right] \Bigg\rvert^r_0 \\[5pt]
& V = 4ar^2 \pi \left[ \left( \texttt{Arcsin}\left( \frac{r}{r} \right) \:+ \frac{r\sqrt{r^2-r^2}}{r^2}\right) \:- \left( \texttt{Arcsin}\left( \frac{0}{r} \right) \:+ \frac{0\sqrt{r^2-0^2}}{r^2}\right) \right] = 4ar^2 \pi \left[ \texttt{Arcsin}(1) \:+ 0 \:- \texttt{Arcsin}(0) \:- 0 \right] \\[5pt]
& V = 4ar^2\pi \left[ \frac{\pi}{2} \:- 0 \right] = 4ar^2\pi\left[ \frac{\pi}{2}\right] \\[5pt]
& \color{green}{V = 2ar^2\pi^2}
\end{align*}
\)

Compute the volume generated by rotating the region bounded by the graphs of the given equation about the y-axis:

Sketched area to be rotated. Given how the differential is located parallel to the axis of rotation, use the shell method. Let \(R = x \) and \(H = 4 \:- x^2 \:- (2 \:- x) = 2 + x \:- x^2\). Using the aforementioned values into the integrand yields:

\(
\begin{align*}
& V = 2\pi \int^2_0 x(2 + x \:- x^2) dx = 2\pi \int^2_0 (2x + x^2 \:- x^3)dx = 2\pi \left[ \frac{1}{2}2x^2 + \frac{1}{3}x^3 \:- \frac{1}{4}x^4 \right] \Bigg\rvert^2_0 = 2\pi \left[ x^2 + \frac{1}{3}x^3 \:- \frac{1}{4}x^4 \right] \Bigg\rvert^2_0 \\[5pt]
& V = 2\pi \left( \left[ (2)^2 + \frac{1}{3}(2)^3 \:- \frac{1}{4}(2)^4 \right] \:- \left[ (0)^2 + \frac{1}{3}(0)^3 \:- \frac{1}{4}(0)^4 \right] \right) = 2\pi \left[4 + \frac{8}{3} \:- 4\right] = 2\pi \left[\frac{8}{3}\right] \\[5pt]
& \color{green}{V = \frac{16}{3}\pi}
\end{align*}
\)

Compute the volume generated by rotating the region bounded by the graphs of \(f(x)=x^3, g(x)=x \le 0 \le 1 \):

  • About x = 1

Use the shell method because of the orientation of the differential line. Let \(R = 1-x \) and \(x \:- x^3\). Updating these values in the integral yields:

\(
\begin{align*}
& V = \pi \int^1_0 (1-x)(x \:- x^3)dx = \pi \int^1_0 (x^4 \:- x^3 \:- x^2 + x)dx \\[5pt]
& V = \pi \left[ \frac{1}{5}x^5 \:- \frac{1}{4}x^4 \:- \frac{1}{3}x^3 + \frac{1}{2}x^2 \right] \Bigg\rvert^1_0 \\[5pt]
& V = \pi \left( \left[ \frac{1}{5}(1)^5 \:- \frac{1}{4}(1)^4 \:- \frac{1}{3}(1)^3 + \frac{1}{2}(1)^2 \right] \:- \left[ \frac{1}{5}(0)^5 \:- \frac{1}{4}(0)^4 \:- \frac{1}{3}(0)^3 + \frac{1}{2}(0)^2 \right] \right) \\[5pt]
& V = \pi \left[ \frac{1}{5} \:- \frac{1}{4} \:- \frac{1}{3} + \frac{1}{2} \right] \\[5pt]
& \color{green}{V = \frac{7}{60}\pi}
\end{align*}
\)

 

  • About x = -2

Once again, use the shell method. The problem is very similar to the last one. Let \(R = x+2 \) and \(x \:- x^3\). Updating these values in the integral yields:

\(
\begin{align*}
& V = \pi \int^1_0 (x+2)(x \:- x^3)dx = \pi \int^1_0 (-x^4 \:- 2x^3 + x^2 + 2x)dx \\[5pt]
& V = \pi \left[ -\frac{1}{5}x^5 \:- \frac{2}{4}x^4 + \frac{1}{3}x^3 + \frac{2}{2}x^2 \right] \Bigg\rvert^1_0  = \pi \left[ -\frac{1}{5}x^5 \:- \frac{1}{2}x^4 + \frac{1}{3}x^3 + x^2 \right] \Bigg\rvert^1_0 \\[5pt]
& V = \pi \left( \left[ -\frac{1}{5}(1)^5 \:- \frac{1}{2}(1)^4 + \frac{1}{3}(1)^3 + (1)^2 \right] \:- \left[ -\frac{1}{5}(0)^5 \:- \frac{1}{2}(0)^4 + \frac{1}{3}(0)^3 + (0)^2 \right] \right) \\[5pt]
& V = \pi \left[ -\frac{1}{5} \:- \frac{1}{2} + \frac{1}{3} + 1 \right] \\[5pt]
& \color{green}{V = \frac{19}{30}\pi}
\end{align*}
\)

 

  • About y = 2

Given the orientation of the differential line with respect to the axis of rotation, use the washer method:

\(
\begin{align*}
& R = 2 \:- x^3; \Rightarrow R^2 = (2 \:- x^3)^2 = 4 \:- 4x^3 + x^6 \\[5pt]
& r = 2 \:- x; \Rightarrow r^2 = (2 \:- x)^2 = 4 \:- 4x + x^2 \\[10pt]
& V = \pi \int^1_0 (R^2 \:- r^2)dx = \pi \int^1_0 (4 \:- 4x^3 + x^6 \:- (4 \:- 4x + x^2)) dx = \pi \int^1_0 (x^6 \:- 4x^3 \:- x^2 + 4x)dx \\[5pt]
& V = \pi \left[ \frac{1}{7}x^7 \:- \frac{4}{4}x^4 \:- \frac{1}{3}x^3 + \frac{4}{2}x^2 \right] \Bigg\rvert^1_0 = \pi \left[ \frac{1}{7}x^7 \:- x^4 \:- \frac{1}{3}x^3 + 2x^2 \right] \Bigg\rvert^1_0 \\[5pt]
& V = \pi \left( \left[ \frac{1}{7}(1)^7 \:- (1)^4 \:- \frac{1}{3}(1)^3 + 2(1)^2 \right] \:- \left[ \frac{1}{7}(0)^7 \:- (0)^4 \:- \frac{1}{3}(0)^3 + 2(0)^2 \right] \right) \\[5pt]
& V = \pi \left[ \frac{1}{7} \:- 1 \:- \frac{1}{3} + 2 \right] \\[5pt]
& \color{green}{V = \frac{17}{21}\pi}
\end{align*}
\)

 

  • About y = -2

Once again, apply the washer method because of the orientation of the differential line:

\(
\begin{align*}
& R = x + 2; \Rightarrow R^2 = (x + 2)^2 = x^2 + 4x + 4 \\[5pt]
& r = x^3 + 2; \Rightarrow r^2 = (x^3 + 2)^2 = x^6 + 4x^3 + 4 \\[10pt]
& V = \pi \int^1_0 (R^2 \:- r^2)dx = \pi \int^1_0 (x^2 + 4x + 4 \:- (x^6 + 4x^3 + 4))dx = \pi \int^1_0 (-x^6 \:- 4x^3 + x^2 + 4x) dx \\[5pt]
& V = \pi \left[ -\frac{1}{7}x^7 \:- \frac{4}{4}x^4 + \frac{1}{3}x^3 + \frac{4}{2}x^2 \right] \Bigg\rvert^1_0 = \pi \left[ -\frac{1}{7}x^7 \:- x^4 + \frac{1}{3}x^3 + 2x^2 \right] \Bigg\rvert^1_0 \\[5pt]
& V = \pi \left( \left[ -\frac{1}{7}(1)^7 \:- (1)^4 + \frac{1}{3}(1)^3 + 2(1)^2 \right] \:- \left[ -\frac{1}{7}(0)^7 \:- (0)^4 + \frac{1}{3}(0)^3 + 2(0)^2 \right] \right) \\[5pt]
& V = \pi \left[ -\frac{1}{7} \:- 1 + \frac{1}{3} + 2 \right] \\[5pt]
& \color{green}{V =\frac{25}{21}\pi}
\end{align*}
\)

 

Compute the volume generated by rotating the region bounded by the graphs of the given equations:

  • (a) About x=-1

First, sketch all of the given functions and identify a common area among them:

Finally, choose the proper method of volumetric intergation. Notice how the differencial line is parallel to the axis of rotation. Because of this, use the shell method: Let \(R = x + 1\) and \(H = (x \:- 2)^3 \:- (x \:- 2)= x^3 \:- 6x^2 + 11x \:- 6 \)

\(
\begin{align*}
& V = 2\pi \int^2_1 [(x+1)(x^3 \:- 6x^2 + 11x \:- 6)]dx = 2\pi \int^1_0 [x^4 \:- 5x^3 + 5x^2 + 5x \:- 6]dx \\[5pt]
& V = 2\pi \left[ \frac{1}{5}x^5 \:- \frac{5}{4}x^4 + \frac{5}{3}x^3 + \frac{5}{2}x^2 \:- 6x \right] \Bigg\rvert^2_1 \\[5pt]
& V = 2\pi \left[ \left( \frac{1}{5}(2)^5 \:- \frac{5}{4}(2)^4 + \frac{5}{3}(2)^3 + \frac{5}{2}(2)^2 \:- 6(2) \right) \:- \left( \frac{1}{5}(1)^5 \:- \frac{5}{4}(1)^4 + \frac{5}{3}(1)^3 + \frac{5}{2}(1)^2 \:- 6(1)\right) \right] \\[5pt]
& V = 2\pi \left[ \left( \frac{32}{5} \:- 20 + \frac{40}{3} + 10 \:- 12 \right) \:- \left( \frac{1}{5} \:- \frac{5}{4} + \frac{5}{3} + \frac{5}{2} \:- 6 \right) \right] = 2\pi \left[-\frac{34}{15} -\left(-\frac{173}{60}\right)\right] \\[5pt]
& \color{green}{V = \frac{37}{30}\pi}
\end{align*}
\)

 

  • (b) About x=2

First, sketch all of the given functions and identify a common area among them:

The problem remains mostly unchanged, the only modification being the radius \(R = 2 \:- x\). Updating the previous integral with this change yields:

\(
\begin{align*}
& V = 2\pi \int^2_1 (2 \:- x)(x^3 \:- 6x^2 + 11x \:- 6) dx = 2\pi \int^1_0 (-x^4 + 8x^3 \:- 23x^2 + 28x \:- 12)dx \\[5pt]
& V = 2\pi \left[ -\frac{1}{5}x^5 + \frac{8}{4}x^4 \:- \frac{23}{3}x^3 +\frac{28}{2}x^2 \:- 12x \right] \Bigg\rvert^2_1 = 2\pi \left[ -\frac{1}{5}x^5 + 2x^4 \:- \frac{23}{3}x^3 + 14x^2 \:- 12x \right] \Bigg\rvert^2_1 \\[5pt]
& V = 2\pi \left( \left[ -\frac{1}{5}(2)^5 + 2(2)^4 \:- \frac{23}{3}(2)^3 + 14(2)^2 \:- 12(2) \right] \:- \left[ -\frac{1}{5}(1)^5 + 2(1)^4 \:- \frac{23}{3}(1)^3 + 14(1)^2 \:- 12(1) \right] \right) \\[5pt]
& V = 2\pi \left[ \left(-\frac{32}{5} + 32 \:- \frac{184}{3} + 56 \:- 24 \right) \:- \left(-\frac{1}{5} + 2 \:- \frac{23}{3} + 14 \:- 12\right) \right] = 2\pi\left[-\frac{56}{15}-\left(-\frac{58}{15}\right)\right] \\[5pt]
& \color{green}{V = \frac{4}{15}\pi}
\end{align*}
\)

 

  • (c) About y =-1

First, sketch all of the given functions and identify a common area among them:

Now, as we can’t change the orientation of the differential line, change the method to washer (because of the perperndicular position between the differential and the rotation axis):

\(
\begin{align*}
& R = (x \:- 2)^3 + 1; \Rightarrow R^2 = ((x \:- 2)^3 + 1)^2 = x^6 \:- 12x^5 + 60x^4 \:-158x^3 + 228x^2 \:- 168x + 49 \\[5pt]
& r = (x \:- 2) + 1; \Rightarrow r = (x \:- 1)^2 = x^2 \:- 2x + 1 \\[10pt]
& R^2 \:- r^2 = x^6 \:- 12x^5 + 60x^4 \:- 158x^3 + 228x^2 \:- 168x + 49 \:- ( x^2 \:- 2x + 1) \\[5pt]
& = x^6 \:- 12x^5 + 60x^4 \:-158x^3 + 227x^2 \:- 166x+ 48 \\\\[5pt]
& V = \pi \int^2_1 \left( x^6 \:- 12x^5 + 60x^4 \:-158x^3 + 227x^2 \:- 166x+ 48 \right)dx \\[5pt]
& V = \pi\left[ \frac{1}{7}x^7 \:- \frac{12}{6}x^6 + \frac{60}{5}x^5 \:- \frac{158}{4}x^4 + \frac{227}{3}x^3 \:- \frac{166}{2}x^2 + 48x \right] \Bigg\rvert^2_1 \\[5pt]
& V = \pi\left[ \frac{1}{7}x^7 \:- 2x^6 + 12x^5 \:- \frac{79}{2}x^4 + \frac{227}{3}x^3 \:- 83x^2 + 48x \right] \Bigg\rvert^2_1 \\[5pt]
& V = \pi\left( \left[ \frac{1}{7}(2)^7 \:- 2(2)^6 + 12(2)^5 \:- \frac{79}{2}(2)^4 + \frac{227}{3}(2)^3 \:- 83(2)^2 + 48(2) \right] \:- \left[ \frac{1}{7}(1)^7 \:- 2(1)^6 + 12(1)^5 \:- \frac{79}{2}(1)^4 + \frac{227}{3}(1)^3 \:- 83(1)^2 + 48(1) \right] \right) \\[5pt]
& V = \pi \left[ \left( \frac{128}{7} \:- 128 + 384 \:- 632 + \frac{1816}{3} \:- 332 + 96 \right) \:- \left( \frac{1}{7} \:- 2 + 12 \:- \frac{79}{2} + \frac{227}{3} \:- 83 + 48 \right) \right] = \pi \left[ \left(\frac{244}{21}\right) \:- \left(\frac{475}{12} \right) \right] \\[5pt]
& \color{green}{V = \frac{13}{42}\pi}
\end{align*}
\)

 

  • (d) About y = 1

First, sketch all of the given functions and identify a common area among them:

The problem remains mostly unchanged (we are still using the washer method), so the only modification being both radiuses:

\(
\begin{align*}
& R = 1 \:- (x \:- 2); \Rightarrow r = (3 \:- x)^2 = 9 \:- 6x + x^2 \\[10pt]
& r = 1 \:- (x \:- 2)^3; \Rightarrow R^2 = (1 \:- (x \:- 2)^3)^2 = x^6 \:- 12x^5 + 60x^4 \:- 162x^3 + 252x^2 \:- 216x + 81 \\[5pt]
& R^2 \:- r^2 = ( 9 \:- 6x + x^2) \:- (x^6 \:- 12x^5 + 60x^4 \:- 162x^3 + 252x^2 \:- 216x + 81) \\[5pt]
& = -x^6 + 12x^5 \:- 60x^4 + 162x^3 \:- 251x^2 + 210x \:- 72 \\\\[5pt]
& V = \pi \int^2_1 \left( -x^6 + 12x^5 \:- 60x^4 + 162x^3 \:- 251x^2 + 210x \:- 72 \right)dx \\[5pt]
& V = \pi\left[ -\frac{1}{7}x^7 + \frac{12}{6}x^6 \:- \frac{60}{5}x^5 + \frac{162}{4}x^4 \:- \frac{251}{3}x^3 + \frac{210}{2}x^2 \:- 72x \right] \Bigg\rvert^2_1 \\[5pt]
& V = \pi\left[ -\frac{1}{7}x^7 + 2x^6 \:- 12x^5 + \frac{81}{2}x^4 \:- \frac{251}{3}x^3 + 105x^2 \:- 72x \right] \Bigg\rvert^2_1 \\[5pt]
& V = \pi\left( \left[ -\frac{1}{7}(2)^7 + 2(2)^6 \:- 12(2)^5 + \frac{81}{2}(2)^4 \:- \frac{251}{3}(2)^3 + 105(2)^2 \:- 72(2) \right] \:- \left[ -\frac{1}{7}(1)^7 + 2(1)^6 \:- 12(1)^5 + \frac{81}{2}(1)^4 \:- \frac{251}{3}(1)^3 + 105(1)^2 \:- 72(1) \right] \right) \\[5pt]
& V = \pi \left( \left[ -\frac{128}{7} + 128 \:- 384 + 648 \:- \frac{2008}{3} + 420 \:- 144 \right] \:- \left[ -\frac{1}{7} + 2 \:- 12 + \frac{81}{2} \:- \frac{251}{3} + 105 \:- 72 \right] \right) \\[5pt]
& V = \pi\left( \left[ -\frac{412}{21} \right] \:- \left[ -\frac{853}{42} \right] \right) \\[5pt]
& \color{green}{V = \frac{29}{42}\pi}
\end{align*}
\)

 

Describe the solid whose volume is being represented by the given integral:

\(
\begin{align*}
2\pi \int^1_0 (2x-x^2) dx = 2\pi \int^1_0 x(2-x) dx
\end{align*}
\)

 

First, we start with the original given integral. Then, we factor the term x out of the integrand. Since we have \(2\pi\) at the beginning of the integral, we can safely assume that this is a shell method integral. The radius would be the factored \(x\), the height would be \(2-x\), and the limits of integration would be \(0 \le x \le 1\). The graph is as follows:

Describe the solid whose volume is being represented by the given integral:

\(
\begin{align*}
2\pi \int^1_0 (y+2)\left(\sqrt{1-y^2}\right) dy
\end{align*}
\)

 

Given that the integral starts with \(2\pi\), we can safely assume this integral describes a solid using the shell method. The radius would be \(y + 2\) (meaning that the integration area is far by 2 units from the axis of rotation, let it be on the first or on the second quadrant) and the height would be described by the expression \(\sqrt{1 \:- y^2}\). Upon closer analysis, we notice that the height is the positive half of a circle of radius 1 (hence the limits of integration \(0 \le x \le 1\), yet these describe the half of a semicircle, or a quarter of a full circle). This means that this integral describes the volume of a quarter of a torus of rafius 1 with an offset of 2 units from the axis of rotation. This is a graphical representation of these explanation:

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