$\begingroup$The region between $y = sin(x^2)$ and the x-axis for $0 \le x \le \sqrt{\pi}$ is revolved around the y-axis. Find the volume of the resulting solid.
I can get all the way to the integral: $ \pi\int_0^\sqrt{\pi} sin(x^2)dx$ but do not know how to proceed. Is this correct so far?
$\endgroup$31 Answer
$\begingroup$By the shell method, $$dV=2\pi x y dx\\ \implies dV=2\pi x\sin(x^2)dx\\ \implies V=2\pi\int^\sqrt{\pi}_{0}x\sin(x^2)dx\\ \implies V=\pi\int^\sqrt{\pi}_{0}\sin(x^2)d(x^2)\\ \implies V=-\pi\cos(x^2)|^\sqrt{\pi}_0\\ \implies V=-\pi\cos(\pi)+\pi\cos(0)=\pi+\pi=2\pi$$
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