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I am given the equation $\sin x$ and I'm to rotate the interval $(0,\pi)$ around the $x$-axis. I think this is equal to $\int_0^\pi \pi\left(1^2-(\sin x)^2\right)\ dx.$ Am I correct? If not, where did I go wrong? Can you also help me find the indefinite integral of $(\sin x)^2$

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    I appreciate your feedback, sir, but I have indeed studied these answers and I am just trying to get everything straight. Sorry for any annoyance this caused you.2012-12-16

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If the region is between $y=\sin x$ and the $x$-axis, then you should have disks of radius $\sin^2 x$; your integral is for revolving the region between $y=\sin x$ and $y=1$ about the $x$-axis.

To integrate $\sin^2x$ and $\cos^2x$ you use the half-angle formulas:

$\sin^2x=\frac{1-\cos 2x}2\quad\text{and}\quad\cos^2x=\frac{1+\cos 2x}2\;.\tag{1}$

These are both deduced from the double angle formula $\cos 2x=\cos^2x-\sin^2x$ and the Pythagorean identity $\sin^2x+\cos^2x=1$. For example, to get the first formula in $(1)$, start with

$\cos 2x=\cos^2x-\sin^2x=\left(1-\sin^2x\right)-\sin^2x=1-2\sin^2x$

and solve for $\sin^2x$.

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    @Lizi: You’re welcome. Note that if you revolve it about the $y$-axis, you’ll get shells of radius $x$, and the volume will be $2\pi\int_0^{\pi}x\sin x~dx\;,$ which you can integrate by parts.2012-12-15