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

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 ?

• 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 – K Split X Dec 13 '17 at 21:04
• Your answer is correct, your book is wrong (which book, by the way? And where in that book?) – DonAntonio Dec 13 '17 at 21:08
• MJ Strauss GL Bradley KJ Smith Calculus 3rd Edition. Exercise 12.5 Question 23 – So Lo Dec 13 '17 at 21:20

The answer should be $\frac{\pi}{8}$, I believe you just type wrongly in the end.
The volume of the hemisphere would be $\frac23 \pi$ which is less than $\frac72 \pi$.