Volume of revolving $ y = \sin(x) $ about a line $ y = c $ Consider the surface formed by revolving $y=\sin(x)$ about the line $y=c$ from some $0\le{c}\le{1}$ along the interval $0\le{x}\le{\pi}$.

Set up and evaluate an integral to calculate the volume $V(c)$ as a function of $c$.
(My attempt) 
$$
\begin{align}
V &= \pi\int_0^\pi[(\sin(x))^2-c^2]dx \\
  &= \pi\int_0^\pi[(\sin^2(x))-c^2]dx \\
  &= \pi\int_0^\pi\left[\frac12(1-\cos(2x))-c^2\right]dx \\
  &= \pi\left[\frac12(x-\sin(x)(\cos(x))-\frac{c^2x}{2}\right]_0^\pi \\
  &= \pi\left[\left(\frac12(\pi-\sin(\pi)\cos(\pi)-\frac{\pi c^2}{2}\right)-\left(\frac12(0-\sin(0)\cos(0)-0\right)\right]
\end{align}
$$
So far... is this correct?
The second part of the question:
What value of c maximises the volume $V(c)$?
^ no idea on that one. help appreciated.
 A: After several hours of frustration I have finally solved it:
$$V=\pi\int_0^\pi(\sin(x)-c)^2dx$$
$$V=\pi\int_0^\pi((\sin^2(x)+c^2-2c\sin(x))dx$$
$$V=\pi\int_0^\pi\left[\frac12(1-\cos(2x))+c^2-2c\sin(x)\right]dx$$
$$V=\pi\left[c^2x+2c\cos(x)+\frac{x}2-\frac14\sin(2x)\right]_0^\pi$$
$$V=\pi\left(\pi c^2-4c+\frac{\pi}2\right)$$
$$V=\pi^2c^2-4\pi c+\frac{\pi^2}2$$
A: The first answer that appeared here has the correct answer, but did not explain why your solution was incorrect. Let's pose the problem in terms of Pappus's $(2^{nd})$ Centroid Theorem: the volume of a planar area of revolution is the product of the area A and the length of the path traced by its centroid $R$, i.e., $2πR$. The bottom line is that the volume is given simply by $V=2πRA$. 
In the present problem, the centroid, $R$ is in the vertical direction and relative to the (horizontal) $x$-axis. Before going there, let's simplify the problem by setting the horizontal axis at $y=c$, i.e., $t=y-c=\sin x-c$. Then,
$$R=\frac{\int_0^{\pi}\int_0^t tdtdx}{\int_0^{\pi}\int_0^t dtdx}=\frac{\frac{1}{2}{}\int_0^{\pi} t^2dt}{A}$$
Now we can write the volume as
$$V=\pi\int_0^{\pi}(\sin x-c)^2dx=\pi\left(\pi c^2-4c+\frac{\pi}{2}\right)$$
So you see that your error was in the centroid term where you had $(\sin^2 x-c^2)$. In essence, you were confusing the centroid of a region between two curves with the centroid relative to a displaced axis.
Finally, to get the value of $c$ for the optimal volume, set $\frac{dV}{dc}=0$.
