# Evaluate the triple integral over E of z dV, where E is the region above z = sqrt(x^2 + y^2) and...

## Question:

Evaluate the triple integral {eq}\iiint_{E} z \, \mathrm{d}V, {/eq} where {eq}E {/eq} is the region above {eq}z = \sqrt{x^2 + y^2} {/eq} and below {eq}x^2 + y^2 + z^2 = 9 {/eq}.

## Cylindrical Coordinates:

One thing is for sure here: we do not want to proceed using rectangular coordinates. Some form of polar coordinates are going to be the way to go. Since we can easily write these surfaces in cylindrical coordinates, let's use them. Recall

{eq}x = r \cos \theta {/eq}

{eq}y = r \sin \theta {/eq}

{eq}z = z {/eq}

{eq}r^2 = x^2+y^2 {/eq}

{eq}\theta = \tan^{-1} \frac{y}{x} {/eq}

{eq}dV = r \ dz \ dr \ d\theta {/eq}

## Answer and Explanation: 1

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View this answerThe upper bound is the cone {eq}z = \sqrt{ x^2 + y^2} = r {/eq}, and the lower bound is the sphere

{eq}\begin{align*} x^2 + y^2 + z^2 &= 9 \\ z^2...

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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.

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