Compute the volume under sphere $x^2+y^2+z^2=4$ above $xy$-plane and above cone $z=\sqrt{x^2+y^2}$ using triple integral.
I try to plot as follows:
I try using triple integral
$$\int\limits_{-\sqrt{2}}^{\sqrt{2}} \int\limits_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}} \int\limits_{\sqrt{x^2 +y^2}}^{\sqrt{4-x^2-y^2}} dzdydx$$
But too difficult to solve that integral.
Now, I think to transform to cylindrical coordinates or sphere coordinates but I confused to determine lower bound and upper bound of triple integral. Anyone can explain the ways to solve this problem?
Best Answer
The solid has cylindrical symmetry. For $z \in [0, \sqrt{2}]$, you have a cone; for $z \in [\sqrt{2}, 2]$, you have a section of a sphere. Then
$$V = \left[ \int_0^\sqrt{2} \int_0^{2 \pi} \int_0^z + \int_\sqrt{2}^2 \int_0^{2 \pi} \int_0^\sqrt{4 - z^2} \right] r dr d \theta dz$$
Note that because the solid is axisymmetric, the triple integrals can be reduced to double integrals (with a suitable multiplication by $2 \pi$):
$$V = 2 \pi \left[ \int_0^\sqrt{2} \int_0^z + \int_\sqrt{2}^2 \int_0^\sqrt{4 - z^2} \right] r dr dz$$
Since we know that $\int r dr = \frac{1}{2} r^2 + C$, then we have
$$V = \pi \left[ \int_0^\sqrt{2} z^2 dz + \int_\sqrt{2}^2 (4 - z^2) dz \right]$$
Evaluating the integrals, we obtain
$$V = \pi \left[ \frac{2}{3} \sqrt{2} + 4 (2 - \sqrt{2}) - \frac{8}{3} + \frac{2}{3} \sqrt{2} \right] = \pi \left[ \frac{16}{3} - \frac{8}{3} \sqrt{2} \right]$$