How do you find the volume of the solid that lies within the sphere
#x^2+y^2+z^2 =25#, above the xy plane, and outside the cone?

Question

Luke Alvarez · Accepted Answer

The cone ? which one ?

Paisley Johnson · Answer

undefined To find the volume of the solid that lies within the sphere $x^2 + y^2 + z^2 = 25$, above the xy-plane, and outside the cone, we need to set up the integral using spherical coordinates.

1. First, we express the equations of the sphere and the cone in spherical coordinates.

Sphere: $x^2 + y^2 + z^2 = 25$ becomes $r^2 = 25$ in spherical coordinates.
   Cone: The cone is defined by $z = \sqrt{x^2 + y^2}$ in Cartesian coordinates. In spherical coordinates, this becomes $z = r \cos(\phi)$.

2. We need to determine the limits of integration for $r$, $\theta$, and $\phi$.

- For $r$: Since we want the region outside the cone but inside the sphere, the limits for $r$ are from the cone's radius to the sphere's radius. That is, from $\sqrt{x^2 + y^2}$ to $5$.

- For $\theta$: We want to sweep through the full angle around the z-axis, so the limits for $\theta$ are from $0$ to $2\pi$.

- For $\phi$: We're above the xy-plane, so $\phi$ ranges from $\frac{\pi}{2}$ to $\pi$.

3. The volume element in spherical coordinates is $dV = r^2 \sin(\phi) \, dr \, d\theta \, d\phi$.

4. Now, we set up the integral for the volume:
   $$ V = \iiint dV = \int_{\frac{\pi}{2}}^{\pi} \int_{0}^{2\pi} \int_{\sqrt{x^2 + y^2}}^{5} r^2 \sin(\phi) \, dr \, d\theta \, d\phi $$

5. Solve the integral.

$$ V = \int_{\frac{\pi}{2}}^{\pi} \int_{0}^{2\pi} \left[ \frac{r^3}{3} \right]_{\sqrt{x^2 + y^2}}^{5} \sin(\phi) \, d\theta \, d\phi $$
   $$ = \int_{\frac{\pi}{2}}^{\pi} \int_{0}^{2\pi} \left( \frac{125}{3} - \frac{r^3}{3} \right) \sin(\phi) \, d\theta \, d\phi $$
   $$ = \int_{\frac{\pi}{2}}^{\pi} \left[ 125\theta - \frac{\theta}{3} \right]_{0}^{2\pi} \, d\phi $$
   $$ = \int_{\frac{\pi}{2}}^{\pi} \left( 250\pi - \frac{2\pi}{3} \right) \, d\phi $$
   $$ = \left[ 250\pi\phi - \frac{2\pi\phi}{3} \right]_{\frac{\pi}{2}}^{\pi} $$
   $$ = 125\pi - \frac{5\pi}{3} $$
   $$ = \frac{370\pi}{3} $$

Therefore, the volume of the solid is $\frac{370\pi}{3}$.

How do you find the volume of the solid that lies within the sphere #x^2+y^2+z^2 =25#, above the xy plane, and outside the cone?