I am asked to find the volume below $z = x^2 - y^2$ and above $x^2+y^2 \leq 1$ in the $xy$-plane. My attempt goes as follows:
Since $|x| \geq |y|$ for z to be positive, that means I am looking at the areas $\theta \in [-\pi/4, \pi/4]$ and $\theta \in [3\pi/4, 5\pi/4]$. Since the surface z is symmetric I only need to look at one of these, for example $\theta \in [0, \pi/4]$. Using polar coordinates, I get $\int_0^{\pi/4}\int_0^1 r^3\cos(2\theta) drd\theta = \frac{1}{4}\int_0^{\pi/4}\cos(2\theta)d\theta = \frac{1}{8}$. Adding the contributions from all quadrants then gives me $V = \frac{1}{2}$.
Unfortunately this is wrong: the correct answer should be $V = \frac{1}{8}$, which is what I got from just one quadrant. Can anyone help me see what's wrong?
$\endgroup$ 41 Answer
$\begingroup$Your calculations are correct. Doing the integral in cartesian coordinates (not the best idea in this case!): $$ \frac V4=\int_0^{\sqrt2/2}\int_y^{\sqrt{1-y^2}}\int_0^{x^2-y^2}1\,dz\,dx\,dy = \frac18 $$
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