Find the volume of the solid that lies within the hemisphere $x^2 + y^2 + z^2 =6$, z ≤ 0, and outside the cone $z=-\sqrt{x^2+y^2}$. I am not sure what to do for this question, I tried subtracting the area of the sphere from the cone, I also tried spherical but a unsure of what the limits for phi would be.
[Math] Triple Integral In a Sphere Outside of a Cone
calculusintegrationmultivariable-calculus
Best Answer
Due to symmetry, the solid is identical to the one which lies within the hemisphere $x^2 + y^2 + z^2 = 6$, $z \geq 0$ and outside the cone $z = \sqrt{x^2 + y^2}$, the only difference being that one is the mirror image of the other across $xy$-plane. (This is done just to avoid negative signs. This is a personal choice, and you can very well work with either of the solids). We will work with the solid which lies above $xy$-plane.
Notice that this solid is identical to the solid of revolution if we revolve $\Omega = \{(x,y)| \; x,y \geq 0; \; x^2 + y^2 \leq 6; \; x \geq y \}$ around $y$-axis.
Using Disk Method, the volume of this solid of revolution is given by:
$V = \pi \int \limits_{y = 0}^{\sqrt{3}} [(6 - y^2) - y^2] dy = 4 \sqrt{3} \pi$.
Alternatively, using triple integration,
$V = \int \limits_{\phi = 0}^{2 \pi} \; \; \int \limits_{\theta = \pi/4}^{\pi/2} \; \; \int \limits_{\rho = 0}^{\sqrt{6}} % \rho^2 \; \sin{\theta} \; d\rho \; d\theta \; d\phi = % 4 \sqrt{3} \pi$.