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Q. 41

Expert-verifiedFound in: Page 1055

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.

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

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

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

We are given,

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

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

Using this definition, we get

The given iterated integral ${\int}_{-3}^{3}{\int}_{\sqrt{9-{x}^{2}}}^{\sqrt{9-{x}^{2}}}{\int}_{-\sqrt{9-{x}^{2}-{y}^{2}}}^{\sqrt{9-{x}^{2}-{y}^{2}}}f(x,y,z)dzdydx$ 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.

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