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Find the volume of the solid bounded by $z=1-4(x^2+y^2)$ and the $x-y$ plane.

I have used the cylindrical coordinates to set up the following integral

$$\int_{0}^{2\pi} \int_{0}^{1/2} \int_{0}^{1-4r^2} \; r \; dz \; dr \; d \theta$$

Is this set up correctly?

$$\int_{0}^{2\pi} \int_{0}^{1/2} \int_{0}^{1-4r^2} \; r \; dz \; dr \; d \theta = \int_{0}^{2\pi} \int_{0}^{1/2} r -4r^3 \; dr \; d \theta = \int_{0}^{2\pi} \frac{1}{16} \; d \theta = \frac{\pi}{8} $$

But my textbook lists the answer as $\frac{7 \pi}{2}$

What am I doing wrong ?

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  • $\begingroup$ Just a quicker way to evaluate. We see that there is no $\theta$ anywhere in the 2nd and 3rd integral, nor in the function itself, and so we know $\theta$ will not appear. Hence we can do $\int^{2\pi}_{0}d\theta=2\pi$ first (don't have to wait until the very end). But yes that's a side note, looks like textbook incorrect here $\endgroup$ – K Split X Dec 13 '17 at 21:04
  • $\begingroup$ Your answer is correct, your book is wrong (which book, by the way? And where in that book?) $\endgroup$ – DonAntonio Dec 13 '17 at 21:08
  • $\begingroup$ MJ Strauss GL Bradley KJ Smith Calculus 3rd Edition. Exercise 12.5 Question 23 $\endgroup$ – So Lo Dec 13 '17 at 21:20
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The answer should be $\frac{\pi}{8}$, I believe you just type wrongly in the end.

To prove that the book's answer doesn't make sense, consider the upper unit hemisphere that is centered at the origin which cover this object.

The volume of the hemisphere would be $\frac23 \pi$ which is less than $\frac72 \pi$.

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