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Let $R$ be the solid in $\mathbb{R}^{3}$ bounded by the surfaces $x=0, y=0, z=0$ and $x^{2}+y^{2}+z^{2}=1$. Evaluate
$\iiint_{R} 3\left(x^{3}+x y^{2}+x z^{2}\right) d V$
$\frac{\pi}{4}$
$\frac{\pi}{32}$
$\frac{\pi}{8}$
$\frac{\pi}{16}$
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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

Convert the integral to spherical coordinates because the region of integration is a sphere in the first octant

step 2

The spherical coordinates are given by $x = \rho \sin(\phi) \cos(\theta)$, $y = \rho \sin(\phi) \sin(\theta)$, and $z = \rho \cos(\phi)$

step 3

The Jacobian determinant for the transformation to spherical coordinates is $\rho^2 \sin(\phi)$

step 4

Set up the limits of integration for $\rho$ from 0 to 1, $\phi$ from 0 to $\frac{\pi}{2}$, and $\theta$ from 0 to $\frac{\pi}{2}$ to cover the first octant of the sphere

step 5

The integral becomes $\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} 3(\rho^3 \sin^3(\phi) \cos^3(\theta) + \rho^3 \sin^3(\phi) \cos(\theta) \sin^2(\theta) + \rho^3 \sin^3(\phi) \cos(\theta) \cos^2(\phi)) \rho^2 \sin(\phi) d\rho d\theta d\phi$

step 6

Integrate with respect to $\rho$, then $\theta$, and finally $\phi$

step 7

After performing the integration, the result is $\frac{\pi}{4}$

Answer

$\frac{\pi}{4}$

Key Concept

Spherical Coordinates in Integration

Explanation

The triple integral over a spherical region is often easier to evaluate using spherical coordinates, which simplifies the integration process when the function and limits are symmetrical with respect to the origin.

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