Evaluate the integral.

{eq}\iiint_{E} (x^2 + y^2) \, \mathrm{d}V, {/eq} where {eq}E {/eq} is the unit ball.

## Question:

Evaluate the integral.

{eq}\iiint_{E} (x^2 + y^2) \, \mathrm{d}V, {/eq} where {eq}E {/eq} is the unit ball.

## Triple Integral Using Spherical Coordinates

When the region in cartesian coordinates is either a sphere, a spherical sector or some quadric surfaces, it is practical to transform the region into spherical coordinates so that it will look like this

{eq}\displaystyle R = \left \{ (\rho, \phi, \theta) | a_1 \leq \rho \leq a_2, b_1 \leq \phi \leq b_2, c_1 \leq \theta \leq c_2 \right \} {/eq}

## Answer and Explanation: 1

Since {eq}\displaystyle E {/eq} is the unit ball, we can define the region as

{eq}\displaystyle E = \left \{ (\rho, \phi, \theta) | 0 \leq \rho \leq 1, 0 \leq \phi \leq \pi, 0 \leq \theta \leq 2\pi \right \} \quad (1) {/eq}

Now, using equation (1) and the following change of variables for spherical coordinates

{eq}\displaystyle \displaystyle \begin{align*} x &= \rho \sin {\phi} \cos {\theta} \quad (2) \\ y &= \rho \sin {\phi} \sin {\theta} \quad (3) \\ z &= \rho \cos {\phi} \quad (4) \end{align*} {/eq}

The triple integral is computed as

{eq}\displaystyle \iiint_{E} (x^2 + y^2) \, dV = \int_{0}^{\pi}{\int_{0}^{2\pi}{\int_{0}^{1}(( \rho \sin {\phi} \cos {\theta})^2 + (\rho \sin {\phi} \sin {\theta} )^2)(\rho^2 \cos {\phi}d\rho d\theta d\phi)}} \\ \\ \displaystyle = \int_{0}^{\pi} \int_{0}^{2\pi} \int_{0}^{1} \left (\rho^4 \cos {\phi} \left (\sin^2 {\phi} \cos^2 {\theta} + \sin^2 {\phi} \sin^2 {\theta} \right ) \right )d\rho d\theta d\phi \\ \\ \displaystyle = \int_{0}^{\pi} \int_{0}^{2\pi} \int_{0}^{1} \left (\rho^4 \sin^2 {\phi} \cos {\phi} \left (\cos^2 {\theta} + \sin^2 {\theta} \right ) \right )d\rho d\theta d\phi \\ \\ \displaystyle = \int_{0}^{\pi} \int_{0}^{2\pi} \int_{0}^{1} \left (\rho^4 \sin^2 {\phi} \cos {\phi} \right )d\rho d\theta d\phi \\ \\ \displaystyle = \int_{0}^{\pi} \int_{0}^{2\pi} \left [\frac{\rho^5 \sin^2 {\phi} \cos {\phi}}{5} \right ]_{0}^{1} d\theta d\phi \\ \\ \displaystyle = \int_{0}^{\pi} \int_{0}^{2\pi} \left [\frac{\sin^2 {\phi} \cos {\phi}}{5} \right ] d\theta d\phi \\ \\ \displaystyle = \int_{0}^{\pi} \left [\frac{\theta \sin^2 {\phi} \cos {\phi}}{5} \right ]_{0}^{2\pi} d\phi \\ \\ \displaystyle = \int_{0}^{\pi} \left [\frac{2\pi \sin^2 {\phi} \cos {\phi}}{5} \right ] d\phi \\ \\ \displaystyle = \left [\frac{2\pi \sin^3 {\phi}}{15} \right ]_{0}^{\pi} \\ \\ \displaystyle \boxed{\iiint_{E} (x^2 + y^2) \, dV = 0} {/eq}

#### Learn more about this topic:

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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.