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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?

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  • $\begingroup$ Shouldn't it be $\pi\int_0^\sqrt{\pi} sin^2(x^2)dx$, instead? $\endgroup$
    – Guest 86
    Commented Mar 3, 2014 at 23:49
  • $\begingroup$ @Guest86 Well, actually no. You are probably planning on using the disk method. However, notice that the disk method requires the differential volume element to be perpendicular to the axis of rotation. Then, for this case, drawing out the graph shows that you'll need two integrals, instead of just one. So, you use the shell method. The integral is wrong, it should actually be $2\pi\int^\sqrt{\pi}_0x\sin(x^2)dx$. $\endgroup$
    – user122283
    Commented Mar 3, 2014 at 23:56
  • $\begingroup$ Oh right the revolution happens around the y-axis, thank you. $\endgroup$
    – Guest 86
    Commented Mar 4, 2014 at 0:01

1 Answer 1

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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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  • $\begingroup$ can you explain your 3rd arrow, why does the dx change to dx^2 $\endgroup$
    – Physics Rocks
    Commented Mar 3, 2014 at 23:51
  • $\begingroup$ So you have the term $2xdx$. What is $\frac{d(x^2)}{dx}$? It's just $2x$. So, "multiplying out" the $dx$ on both sides gives $d(x^2)=2xdx$. Use this and substitute $2xdx$ for $x^2$ in the integrand. Then use integration by parts, and you get your answer! $\endgroup$
    – user122283
    Commented Mar 3, 2014 at 23:53

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