Chapter 7.3, Problem 77E

a.

To determine

To find: The value of $c$ that minimizes the volume of the solid, and the minimum volume.

Answer to Problem 77E

The value of $c$ that minimizes the volume is $c=\frac{2}{\pi }$ and the minimum volume is $\frac{{\pi }^2}{2}-4$ .

Explanation of Solution

Given information:

The arc $y=\sin x$ , $0\le x\le \pi$ is revolved about the line $y=c$ , $0\le c\le 1$ to generate a solid.

Calculation:

When the arc $y=\sin x$ intersects with the line $y=c$ , then $c=\sin x$ .

The solution of the equation $c=\sin x$ in the interval $0\le x\le \pi$ when $0<c<1$ are $x={\sin }^{-1}\left(c\right)$ and $x=\pi -{\sin }^{-1}\left(c\right)$ .

Then volume of the solid can be given by $V={\displaystyle \underset{0}{\overset{{\sin }^{-1}\left(c\right)}{\int }}\pi {\left(c-\sin x\right)}^{2}dx}+{\displaystyle \underset{{\sin }^{-1}\left(c\right)}{\overset{\pi -{\sin }^{-1}\left(c\right)}{\int }}\pi {\left(\sin x-c\right)}^{2}dx}+{\displaystyle \underset{\pi -{\sin }^{-1}\left(c\right)}{\overset{\pi }{\int }}\pi {\left(c-\sin x\right)}^{2}dx}$ .

Since ${\left(c-\sin x\right)}^2=c^2+{\sin }^{2}x-2c\sin x$ and ${\left(\sin x-c\right)}^2=c^2+{\sin }^{2}x-2c\sin x$ , the integrands of all the integrals are same. So, the volume can be given as $V={\displaystyle \underset{0}{\overset{\pi }{\int }}\pi \left(c^2+{\sin }^{2}x-2c\sin x\right)dx}$ .

Evaluate the integral.

  $\begin{array}{c}V={\displaystyle \underset{0}{\overset{\pi }{\int }}\pi \left(c^2+{\sin }^{2}x-2c\sin x\right)dx}\\ =\pi {\left[c^{2}x+\frac{x}{2}-\frac{\sin \left(2x\right)}{4}+2c\cos x\right]}_0^{\pi }\\ =\pi \left[\left(c^2\pi +\frac{\pi }{2}-\frac{\sin \left(2\pi \right)}{4}+2c\cos \pi \right)-\left(c^{2}0+\frac{0}{2}-\frac{\sin \left(0\right)}{4}+2c\cos 0\right)\right]\\ =\pi \left[\left(c^2\pi +\frac{\pi }{2}-0+2c\left(-1\right)\right)-\left(0+0-0+2c\right)\right]\\ =\pi \left(c^2\pi +\frac{\pi }{2}-4c\right)\\ =c^2{\pi }^2+\frac{{\pi }^2}{2}-4c\pi \end{array}$

Now, find the value of $c$ that minimizes the volume $V\left(c\right)=c^2{\pi }^2+\frac{{\pi }^2}{2}-4c\pi$ .

Since $V^{\prime }\left(c\right)=2c{\pi }^2-4\pi$ , the equation $V^{\prime }\left(c\right)=0$ gives, $c=\frac{2}{\pi }$ .

Observe that $V^{″}\left(c\right)=2{\pi }^2$ which is positive. So, the volume is minimum at $c=\frac{2}{\pi }$ .

Substitute $c=\frac{2}{\pi }$ in $V\left(c\right)=c^2{\pi }^2+\frac{{\pi }^2}{2}-4c\pi$ to find the minimum volume.

  $\begin{array}{c}V\left(\frac{2}{\pi }\right)={\left(\frac{2}{\pi }\right)}^2{\pi }^2+\frac{{\pi }^2}{2}-4\left(\frac{2}{\pi }\right)\pi \\ =4+\frac{{\pi }^2}{2}-8\\ =\frac{{\pi }^2}{2}-4\end{array}$

Conclusion:

The value of $c$ that minimizes the volume is $c=\frac{2}{\pi }$ and the minimum volume is $\frac{{\pi }^2}{2}-4$ .

b.

To determine

To find: The value of $c$ in $\left[0,1\right]$ that maximizes the volume of the solid.

Answer to Problem 77E

  $c=0$

Explanation of Solution

Given information:

The arc $y=\sin x$ , $0\le x\le \pi$ is revolved about the line $y=c$ , $0\le c\le 1$ to generate a solid.

Calculation:

From part (a), the minimum value occurs when $c=\frac{2}{\pi }$ in the interval $0<c<1$ . Now, find the value of the volume function at $c=0$ and $c=1$ .

At $c=0$ , the volume is given by $V={\displaystyle \underset{0}{\overset{\pi }{\int }}\pi {\left(\sin x\right)}^{2}dx}$ .

  $\begin{array}{c}V={\displaystyle \underset{0}{\overset{\pi }{\int }}\pi {\left(\sin x\right)}^{2}dx}\\ =\pi {\left[\frac{x}{2}-\frac{\sin \left(2x\right)}{4}\right]}_0^{\pi }\\ =\frac{{\pi }^2}{2}\end{array}$

At $c=1$ , the line $y=c$ intersects $y=\sin x$ at $x=\frac{\pi }{2}$ .

So, the volume is $V=2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\pi {\left(1-\sin x\right)}^{2}dx}$ .

  $\begin{array}{c}V=2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\pi {\left(1-\sin x\right)}^{2}dx}\\ =2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\pi \left(1+{\sin }^{2}x-2\sin x\right)dx}\\ =2\pi {\left[x+\frac{x}{2}-\frac{\sin \left(2x\right)}{4}+2\cos x\right]}_0^{\frac{\pi }{2}}\\ =2\pi \left[\left(\frac{\pi }{2}+\frac{\pi }{4}-\frac{\sin \left(\pi \right)}{4}+2\cos \left(\frac{\pi }{2}\right)\right)-\left(0+0-\frac{\sin \left(0\right)}{4}+2\cos \left(0\right)\right)\right]\\ =2\pi \left[\left(\frac{\pi }{2}+\frac{\pi }{4}\right)-2\right]\\ =\frac{3{\pi }^2}{2}-4\pi \end{array}$

So, the maximum volume occurs at $c=0$ .

Conclusion:

The value of $c$ that maximizes the volume of the solid is $c=0$ .

c.

To determine

To graph: The solid’s volume as a function of $c$ , first for $0\le c\le 1$ and then larger domain, and determine what happens to the volume as $c$ moves away from $\left[0,1\right]$ , give reason for the answer.

Explanation of Solution

Given information:

The arc $y=\sin x$ , $0\le x\le \pi$ is revolved about the line $y=c$ , $0\le c\le 1$ to generate a solid.

Calculation:

From part (a), the volume of the solid is given by the function $V\left(c\right)=c^2{\pi }^2+\frac{{\pi }^2}{2}-4c\pi$ .

Graph this function for $0\le c\le 1$ as shown below.

  

Now, graph the function for a larger domain of $c$ .

  

It can be observed that the volume of the solid keeps increasing as $c$ moves away from $\left[0,1\right]$ .

The answer makes sense as $c$ moves away from $\left[0,1\right]$ , the distance between the arc $y=\sin x$ and the line $y=c$ increasing which increases the radius of each tiny cross-section which results in greater volume of the solid.