anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
The main problem I have is that I can't solve for x.
anonymous
  • anonymous
Are you there?
anonymous
  • anonymous
@jim_thompson5910

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More answers

Callisto
  • Callisto
Put y =0 0 = (1/2)sin^2 (x^2) sin^2 (x^2) = 0 sin (x^2) = 0 x^2 =...???
anonymous
  • anonymous
0?
Callisto
  • Callisto
and??
anonymous
  • anonymous
0 is all I can think of.
Callisto
  • Callisto
|dw:1358903484725:dw|
anonymous
  • anonymous
Oh, right! Sine hits zero on every multiple of pi
anonymous
  • anonymous
So x = sqrt(pi), 0?
Callisto
  • Callisto
For that domain, yes.
anonymous
  • anonymous
Right. The domain was a clue.
anonymous
  • anonymous
I need to use the washer method right?
Callisto
  • Callisto
I think we need to use shell method?!
anonymous
  • anonymous
oh, right. just a sec....
anonymous
  • anonymous
x = sqrtpi for the radis, y = (1/2)sin^2 (x^2) for the height?
Callisto
  • Callisto
x = radius... y = height
anonymous
  • anonymous
\[2\pi \int\limits_{0}^{\sqrt{\pi}} (\sqrt{\pi}/2) \sin^2 (x^2)\] ?
Callisto
  • Callisto
\[V = 2 \pi \int_a^brh(width)\]r = x, h = y, width = dx
Callisto
  • Callisto
y is a function of x, you know.
anonymous
  • anonymous
Doesn't x = sqrtpi and y = (1/2)sin^2 (x^2)?

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