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.

• Remember, for the second part, that, to maximise $V(c),$ you need to solve $V'(c)=0.$ – beep-boop Jul 31 '14 at 22:29
• I see. Unfortunately I don't even know how to find V(c) which is the first part of the problem D: – Panthy Jul 31 '14 at 22:30
• Shouldn't the integral be $(sin(x)-c)^2$ on the inside? – illysial Aug 1 '14 at 0:43
• Possible duplicate: math.stackexchange.com/questions/361000/… – user84413 Aug 1 '14 at 15:25

After several hours of frustration I have finally solved it:

$$V=\pi\int_0^\pi(\sin(x)-c)^2dx$$ $$V=\pi\int_0^\pi((\sin^2(x)+c^2-2c\sin(x))dx$$ $$V=\pi\int_0^\pi\left[\frac12(1-\cos(2x))+c^2-2c\sin(x)\right]dx$$ $$V=\pi\left[c^2x+2c\cos(x)+\frac{x}2-\frac14\sin(2x)\right]_0^\pi$$ $$V=\pi\left(\pi c^2-4c+\frac{\pi}2\right)$$ $$V=\pi^2c^2-4\pi c+\frac{\pi^2}2$$

• I think you want to start out by finding where $y=\sin x$ intersects $y=c$, if you integrate with respect to $x$. – user84413 Aug 1 '14 at 15:26

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 $\frac{dV}{dc}=0$.