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Expert-verified Found in: Page 1055 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Describe the three-dimensional region expressed in each iterated integral in Exercises 35–44.${\int }_{-3}^{3}{\int }_{-\sqrt{9-{x}^{2}}}^{\sqrt{9-{x}^{2}}}{\int }_{-\sqrt{9-{x}^{2}-{y}^{2}}}^{\sqrt{9-{x}^{2}-{y}^{2}}}f\left(x,y,z\right)dzdydx$

The three-dimensional region is given by the planer equation,

${x}^{2}+{y}^{2}+{z}^{2}={3}^{2}$

See the step by step solution

## Step 1. Given Information

We are given,

${\int }_{-3}^{3}{\int }_{-\sqrt{9-{x}^{2}}}^{\sqrt{9-{x}^{2}}}{\int }_{-\sqrt{9-{x}^{2}-{y}^{2}}}^{\sqrt{9-{x}^{2}-{y}^{2}}}f\left(x,y,z\right)dzdydx$

## Step 2. The three dimensional region.

By the definition of triple integral ${\int }_{{a}_{1}}^{{a}_{1}}{\int }_{{b}_{1}}^{{b}_{2}}{\int }_{{c}_{1}}^{{c}_{2}}f\left(x,y,z\right)dzdydx$ represent the volume of the solid region $\mathrm{ℝ}=\left\{\left(x,y,z\right)\mid {a}_{1}\le x\le {a}_{2},{b}_{1}\le y\le {b}_{2},{c}_{1}\le z\le {c}_{2}\right\}$

Using this definition, we get

The given iterated integral ${\int }_{-3}^{3}{\int }_{\sqrt{9-{x}^{2}}}^{\sqrt{9-{x}^{2}}}{\int }_{-\sqrt{9-{x}^{2}-{y}^{2}}}^{\sqrt{9-{x}^{2}-{y}^{2}}}f\left(x,y,z\right)dzdydx$ represents the volume of the sphere with radius $3$ and centered at origin.

Since from the integral limits we observe that localid="1650347847843" $z=\sqrt{9-{x}^{2}-{y}^{2}}$ to $z=\sqrt{9-{x}^{2}-{y}^{2}}$

Squaring on both sides,

${z}^{2}=9-{x}^{2}-{y}^{2}$

Simplifying and rearranging, we get

${x}^{2}+{y}^{2}+{z}^{2}={3}^{2}$

This represents the equation of sphere with radius $3$ and centered at origin

Hence the given iterated integral represents the volume of the sphere with radius $3$ and centered at origin. ### Want to see more solutions like these? 