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I'm trying to find the volume of the region enclosed by the curves above rotated along the $y$-axis. I know that my lower and upper bounds will be $a=0$ and $b=1$. I'm not sure if I should set each curve equal to $x$, and even then I'm not sure how to go about setting up the integral.

I asked a similar question to this yesterday but it was along the $x$-axis. Thank you!

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    $\begingroup$ Draw the two curves. They intersect at $0$ and $\pm1$. You can now rewrite the functions as $x=\sqrt[3]{y}$ and $x=y$ $\endgroup$ – Andrei Oct 1 '17 at 19:08
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hint: Due to symmetry, one can find the volume of the upper part by dishwasher method and double the answer. We have: $V = 2\displaystyle \int_{0}^1 \pi((\sqrt[3]{y})^2 - y^2)dy$. Can you continue?

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  • $\begingroup$ I can definitely continue, but I don't really understand how you got to that point. If I start with this, $$V=\int_0^1 \pi(outer radius-inner radius)dx$$ How do I determine which is the outer and inner? Could I just input an arbitrary number such as one into both curves and compare which is larger? Also, how do I determine if I need to use this washer method? Is it always going to be used when rotating about the y-axis? $\endgroup$ – JustHeavy Oct 1 '17 at 19:12
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    $\begingroup$ pick up a calculus book or google online the topic and read the diswasher method, you can should see my points. $\endgroup$ – DeepSea Oct 1 '17 at 19:14
  • $\begingroup$ I have this as the final volume, ${28\pi\over15}$ Does that seem correct? I tried wolfram and it came up with a different answer, I may have done something wrong in the process. I'll quickly show you what I did: $$= 2\pi \int_0^1 y^{2/3}-y^2dy$$ Now, $$= 2\pi({3\over5}-{1\over3})-0$$ Which equals my final answer after some simplifying: $${28\pi\over15}$$ $\endgroup$ – JustHeavy Oct 1 '17 at 19:27
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    $\begingroup$ Your answer is incorrect. $\dfrac{3}{5}- \dfrac{1}{3} = \dfrac{4}{15}$. Thus the "correct" answer is: $\dfrac{8\pi}{15}$. $\endgroup$ – DeepSea Oct 1 '17 at 19:29
  • $\begingroup$ Hmm. Okay, I'll show you everything I did, maybe you can point out my mistake: $$V=2\int_0^1 \pi ((y^{1/3})^2 - y^2)dy$$ Next, $$= 2\pi \int_0^1 y^{2/3}-y^2dy$$ And, $$2\pi({{y^{5\over3}}\over{5\over3}}-{{y^3}\over3}) ]_0^1$$ So, $$2\pi({3\over5}-{1\over3})-2\pi(0-0)$$ Equals, $${6\pi \over 5} - {2\pi \over 3}$$ $\endgroup$ – JustHeavy Oct 1 '17 at 19:35

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