Find the volume of the solid that lies under the paraboloid $z = 8x^2 + 8y^2$ above the $xy$-plane, and inside the cylinder $x^2 + y^2 = 2x$.

I am trying to figure out the double integral in terms of $r$ and I don't know why I am wrong. This is what I wrote: $$\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}(8r^2)r dr d\theta.$$

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    $\begingroup$ This is correct. Why do you think it is wrong? (BTW, welcome to MSE, but you should start to learn how to write good mathematics in MathJax :)) $\endgroup$ – Ted Shifrin Oct 15 '20 at 18:00

You have everything:

$$V = \displaystyle\int_{-0.5\pi}^{0.5\pi}\int_{0}^{2\cos(\theta)}8r^3drd\theta$$ $$V = \displaystyle\int_{-0.5\pi}^{0.5\pi} 8\left( \frac{r^4}{4}\mid_0^{2\cos(\theta)}\right)d\theta$$ $$V = \displaystyle\int_{-0.5\pi}^{0.5\pi} 32 \cos^4(\theta)d\theta$$

$$V = 32\left( \frac{1}{32}(12\theta+8\sin(2\theta)+\sin(4\theta)) \right)\mid_{-0.5\pi}^{0.5\pi} = 12\pi $$

  • $\begingroup$ Okay, it is asking for it in the form of multiplying by 16 on the outside of the evaluated at [ ] pi/2 to 0 instead of 32. How would I write it that way? $\endgroup$ – Kate C Oct 15 '20 at 19:06
  • $\begingroup$ sorry, I have problems understanding your english as I'm not native speaker. Can you reformulate your question ? $\endgroup$ – Luis Felipe Oct 15 '20 at 21:11

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