Nanoman
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?
Delete
Share
This Question is Closed
Nanoman
Best Response
You've already chosen the best response.
0
The main problem I have is that I can't solve for x.
Nanoman
Best Response
You've already chosen the best response.
0
Are you there?
Nanoman
Best Response
You've already chosen the best response.
0
@jim_thompson5910
Callisto
Best Response
You've already chosen the best response.
0
Put y =0
0 = (1/2)sin^2 (x^2)
sin^2 (x^2) = 0
sin (x^2) = 0
x^2 =...???
Nanoman
Best Response
You've already chosen the best response.
0
0?
Callisto
Best Response
You've already chosen the best response.
0
and??
Nanoman
Best Response
You've already chosen the best response.
0
0 is all I can think of.
Callisto
Best Response
You've already chosen the best response.
0
|dw:1358903484725:dw|
Nanoman
Best Response
You've already chosen the best response.
0
Oh, right! Sine hits zero on every multiple of pi
Nanoman
Best Response
You've already chosen the best response.
0
So x = sqrt(pi), 0?
Callisto
Best Response
You've already chosen the best response.
0
For that domain, yes.
Nanoman
Best Response
You've already chosen the best response.
0
Right. The domain was a clue.
Nanoman
Best Response
You've already chosen the best response.
0
I need to use the washer method right?
Callisto
Best Response
You've already chosen the best response.
0
I think we need to use shell method?!
Nanoman
Best Response
You've already chosen the best response.
0
oh, right. just a sec....
Nanoman
Best Response
You've already chosen the best response.
0
x = sqrtpi for the radis, y = (1/2)sin^2 (x^2) for the height?
Callisto
Best Response
You've already chosen the best response.
0
x = radius... y = height
Nanoman
Best Response
You've already chosen the best response.
0
\[2\pi \int\limits_{0}^{\sqrt{\pi}} (\sqrt{\pi}/2) \sin^2 (x^2)\] ?
Callisto
Best Response
You've already chosen the best response.
0
\[V = 2 \pi \int_a^brh(width)\]r = x, h = y, width = dx
Callisto
Best Response
You've already chosen the best response.
0
y is a function of x, you know.
Nanoman
Best Response
You've already chosen the best response.
0
Doesn't x = sqrtpi and y = (1/2)sin^2 (x^2)?