Q. 41 - Calculus

Question

Describe the three-dimensional region expressed in each iterated integral in Exercises 35–44.

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

Step-by-Step Solution

Answer

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

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

Step 1. Given Information

We are given, 

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

Step 2. The three dimensional region.

By the definition of triple integral ${\int }_{a_1}^{a_1}{\int }_{b_1}^{b_2}{\int }_{c_1}^{c_2}f(x,y,z)dzdydx$ represent the volume of the solid region $\mathrm{ℝ}=\left\{(x,y,z)\mid a_1\le x\le a_2,b_1\le y\le b_2,c_1\le z\le c_2\right\}$

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 $z=\sqrt{9-x^2-y^2}$ 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.