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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 have typeset your question into latex. Please verify that I have edited it correctly. –  Neal Dec 15 '12 at 23:19
    
Yep, that is right. –  Lizi Dec 15 '12 at 23:23
    
This is not the first, not the second, but the third question in quick succession on washers by you. This is not good form. Plus, you'd benefit more from studying the answers you do have rather than asking new questions about the same concept. –  JohnD Dec 15 '12 at 23:46
    
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. –  Lizi Dec 16 '12 at 0:43
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1 Answer 1

up vote 1 down vote accepted

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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So there is no need for washers? –  Lizi Dec 15 '12 at 23:29
    
@Lizi: Not if the region is the one between $y=\sin x$ and the $x$-axis. Draw the picture: is there any hole in the middle of the disk? –  Brian M. Scott Dec 15 '12 at 23:33
    
I see it now, I was rotating it around the y-axis and so I was pretty confused. Thank you. –  Lizi Dec 15 '12 at 23:34
    
@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. –  Brian M. Scott Dec 15 '12 at 23:36
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