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I'm having a bit of a hard time with this problem from a Calc 1 Textbook:

Find the volume of the solid obtained by rotating the region bounded by the given curves along the given axis:

$y=x^2$

$x=y^2$

about the x-axis

My Work:

$\pi\displaystyle\int_0^1(\sqrt{x}-x^2)^2dx$

$\pi\displaystyle\int_0^1(x-2\sqrt{x}x^2+x^4)dx$

$\pi(\frac{1}{2}-\frac{4}{7}+\frac{1}{5})$

$\frac{9\pi}{70}$

Book's Answer: $\frac{3\pi}{10}$

What did I do wrong? Thanks.

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What you need to do is first find the volume when $\sqrt x$ is rotated, and then subtract the part when $x^2$ is rotated, i.e., $$V = \pi\int_0^1 (\sqrt x)^2\, dx - \pi\int_0^1 (x^2)^2\, dx = \frac{3\pi}{10}.$$

What you did is rotate $\sqrt x - x$, which will give a different volume.

(Notice that unlike usual integration, this operation of rotating is not linear; i.e. $\pi\int_a^b (f+g)^2\,dx \neq \pi\int_a^b f^2\,dx + \pi\int_a^b g^2\,dx$.)

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  • $\begingroup$ Didn't I subtract $\sqrt{x}$ and $x^2$?. The book said $\frac{3\pi}{10}$ $\endgroup$ – N. Bar Jun 22 '19 at 20:41
  • $\begingroup$ Notice what I said about linearity, you cannot subtract them inside the integral because the integrand is squared. That's like assuming $(f+g)^2 = f^2+g^2$. $\endgroup$ – Luke Collins Jun 22 '19 at 20:42
  • $\begingroup$ Ah, I see. Also, didn't see the edit where you changed the answer. $\endgroup$ – N. Bar Jun 22 '19 at 20:43
  • $\begingroup$ Yes, I copied the answer wrong from Mathematica. $\endgroup$ – Luke Collins Jun 22 '19 at 20:44
  • $\begingroup$ Thanks, this really helped. $\endgroup$ – N. Bar Jun 22 '19 at 20:52

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