# Thread: Finding volume under sin(x^2)?

1. ## Finding volume under sin(x^2)?

Hi guys, I'm stuck on this problem:

Let S be the solid obtained by rotating the region bounded by the curves y=sin(x^2) and y=0, with 0<= x <=sqrt(pi), about the y-axis. Use cylindrical shells to find the volume of S.

I get as far as finding the circumference (2pi(x)), the height (sin(x^2)) and then the integral from 0 to sqrt(pi) of 2pi(x)(sin(x^2))

I thought I could use integration by parts but I don't know how to find the integral of sin(x^2). I wolframed it but I got something with the "fresnel function" which I've never heard of. Any advice is appreciated

Edit, I just realized I might just be able to use substitution on that integral :S, will try it.

2. The limits of integration are given by $c = \sin(0^2) = 0$ and $d = \sin((\sqrt(\pi))^2) = 1$.

Therefore, the volume is given by $\displaystyle V = \int_0^1 x\sin(x^2) \, dx$.

Like you said, u-substitution is appropriate. Let $u = x^2$. Then, $du = 2xdx$. The new limits of integration are $u(0) = 0$ and $u(1) = 1$ (coincidentally, the limits didn't change).

$\displaystyle \frac{1}{2} \int_0^1 \sin(u) \, du = \frac{1}{2} [ -\cos(u) ]_0^1 = \frac{1}{2} [ -\cos(1) + \cos(0) ] = \frac{1}{2} [ -\cos(1) + 1 ] = \frac{1 - \cos(1)}{2}$

Edit: I just realized you already edited the topic as solved. Sorry for bumping. :\$