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What is the volume of a solid obtained by rotating the region between the graph of y = (1/2)sin^2 (x^2) and the x-axis for 0 ≤x≤sqrtpi about the y-axis?

Mathematics
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The main problem I have is that I can't solve for x.
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Other answers:

Put y =0 0 = (1/2)sin^2 (x^2) sin^2 (x^2) = 0 sin (x^2) = 0 x^2 =...???
0?
and??
0 is all I can think of.
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Oh, right! Sine hits zero on every multiple of pi
So x = sqrt(pi), 0?
For that domain, yes.
Right. The domain was a clue.
I need to use the washer method right?
I think we need to use shell method?!
oh, right. just a sec....
x = sqrtpi for the radis, y = (1/2)sin^2 (x^2) for the height?
x = radius... y = height
\[2\pi \int\limits_{0}^{\sqrt{\pi}} (\sqrt{\pi}/2) \sin^2 (x^2)\] ?
\[V = 2 \pi \int_a^brh(width)\]r = x, h = y, width = dx
y is a function of x, you know.
Doesn't x = sqrtpi and y = (1/2)sin^2 (x^2)?

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