# Find the volume of the solid where the solid is bounded below the sphere x^2 + y^2 + z^2 = 4 and...

## Question:

Find the volume of the solid where the solid is bounded below the sphere {eq}x^2 + y^2 + z^2 = 4 {/eq} and above the cone {eq}z = \sqrt {x^2 + y^2} {/eq} in the first octant.

## Spherical Coordinates:

Recall that we can find the volume of a solid in three-dimensional space by setting up and evaluating a triple integral. In the situation described above, we have a solid that is bounded by a sphere and a cone. Note that these two surfaces are easy to describe using spherical coordinates, so we will find it convenient to use spherical coordinates here.

{eq}x=\rho \cos \theta \sin \phi {/eq}

{eq}y = \rho \sin \theta \sin \phi {/eq}

{eq}z = \rho \cos \phi {/eq}

{eq}\rho^2 = x^2+y^2+z^2 {/eq}

{eq}\begin{align*} \rho^2 \sin^2 \phi &= x^2 + y^2 \end{align*} {/eq}

{eq}dV = \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta {/eq}

## Answer and Explanation:

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View this answerFirst, observe that we are in the first octant and so we have {eq}\theta \in [0, \frac{\pi}{2} ] {/eq}. Now, we also know that the other angle is...

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Chapter 13 / Lesson 10Learn how to convert between Cartesian, cylindrical and spherical coordinates. Discover the utility of representing points in cylindrical and spherical coordinates.