Let C be the set of points with x^2 + y^2 at least z^2 and x^2 + y^2 + z^2 at most 4. Find the volume of C using spherical coordinates.

Step-by-step solution

1. 1

   Identify the two surfaces. $x^{2}+y^{2}+z^{2}=4$ is the sphere of radius $2$, and $x^{2}+y^{2}=z^{2}$ is a double cone through the origin whose half-angle is $45^\circ$. The region is the part of the ball outside the cone — the sphere with an ice-cream cone removed from the top and another from the bottom.

2. 2

   Translate both conditions into spherical coordinates. With $x=\rho\sin\phi\cos\theta$, $y=\rho\sin\phi\sin\theta$, $z=\rho\cos\phi$:

   $x^{2}+y^{2}=\rho^{2}\sin^{2}\phi,\qquad z^{2}=\rho^{2}\cos^{2}\phi,$

   so $x^{2}+y^{2}\ge z^{2}$ becomes $\sin^{2}\phi\ge\cos^{2}\phi$, i.e. $\tan^{2}\phi\ge 1$, i.e.

   $\frac{\pi}{4}\le\phi\le\frac{3\pi}{4}.$

   The sphere condition is simply $\rho\le 2$. This is why spherical coordinates are the right choice: both surfaces become coordinate limits.

3. 3

   Set up the integral with the correct Jacobian. The volume element is $dV=\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta$:

   $V=\int_{0}^{2\pi}\int_{\pi/4}^{3\pi/4}\int_{0}^{2}\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta.$

   The limits are all constants, so the triple integral factors into three separate one-dimensional integrals.

4. 4

   Evaluate each factor.

   $\int_{0}^{2}\rho^{2}d\rho=\frac{8}{3},\qquad \int_{\pi/4}^{3\pi/4}\sin\phi\,d\phi=\left[-\cos\phi\right]_{\pi/4}^{3\pi/4}=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\sqrt2,\qquad \int_{0}^{2\pi}d\theta=2\pi.$

5. 5

   Multiply the three results.

   $V=\frac{8}{3}\cdot\sqrt2\cdot 2\pi=\frac{16\sqrt2\,\pi}{3}\approx 23.695.$

6. 6

   Check against the whole ball. The full ball of radius $2$ has volume $\tfrac{4}{3}\pi(8)=\tfrac{32\pi}{3}\approx 33.51$, so the region keeps a fraction $\tfrac{16\sqrt2/3}{32/3}=\tfrac{\sqrt2}{2}\approx 70.7\%$ of it. Removing two $45^\circ$ cones should take a substantial but minority bite, so $70.7\%$ remaining is exactly the right order of magnitude.