Answer
a. $V(\frac{2}{\pi})=\frac{\pi^2}{2}-4$
b. $V(0)=\frac{\pi^2}{2}$
c. See graph and explanations.
Work Step by Step
a. Step 1. Using the figure given in the Exercise, with disk approximation, we have
$dV=\pi (\Delta y)^2\ dx=\pi (sin(x)-c)^2\ dx=\pi (sin^2x-2c\ sin(x)+c^2)dx=\pi (\frac{1-cos(2x)}{2}-2c\ sin(x)+c^2)dx=\pi (c^2+\frac{1}{2}-\frac{1}{2}cos(2x)-2c\ sin(x))dx$.
Step 2. Thus the volume can be found as
$V=\int_0^{\pi}\pi (c^2+\frac{1}{2}-\frac{1}{2}cos(2x)-2c\ sin(x))dx= \pi((c^2+\frac{1}{2})x-\frac{1}{4}sin(2x)+2c\ cos(x))|_0^{\pi}=\pi[((c^2+\frac{1}{2})(\pi)-\frac{1}{4}sin(2\pi)+2c\ cos(\pi))-((c^2+\frac{1}{2})(0)-\frac{1}{4}sin(0)+2c\ cos(0))]=\pi(c^2\pi+\frac{\pi}{2}-4c)$
Step 3. Letting $V'=0$, we have $2\pi c-4=0$ and $c=\frac{2}{\pi}$. Calculating the volume at all critical values $c=0, \frac{2}{\pi}, 1$, we have
$V(0)=\frac{\pi^2}{2}\approx4.9, V(\frac{2}{\pi})=\frac{\pi^2}{2}-4\approx0.9, V(1)=\frac{3\pi^2}{2}-4\pi\approx2.2$. Thus the minimum volume is
$V(\frac{2}{\pi})=\frac{\pi^2}{2}-4\approx0.9$
b. Based on the above results, the maximum volume is
$V(0)=\frac{\pi^2}{2}\approx4.9$
c. See graph, where the red curve is for $c\in[0,1]$ and the blue curve (shifted a little to see both curves) is for large range of $c$ values. It can be seen that $c=\frac{2}{\pi}$ gives the absolute minimum volume. This is because when $c$ moves away from the original region, the sine function will be revolving around a further away axis and generating a large volume.
