Question

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

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

Short answer

The three-dimensional region is given by the equation,

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

It represent the represents the volume of the cylinder of radius $3$and height $3$.

Step-by-step solution

Step 1. Given Information.

We are given,

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

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

Given triple integral ${\int }_{-3}^3{\int }_{\sqrt{9-x^2}}^{\sqrt{9-x^2}}{\int }_{-3}^{3}f(x,y,z)dzdydx$represents the volume of the cylinder of radius $3$and height $3$.

Since from the given iterated integral we observe that

role="math" localid="1650352416990" $$\begin{gathered} y=\sqrt{9-x^2} \\ \Rightarrow x^2+y^2=9 \\ {x}^2+y^2=3^2 \end{gathered}$$

Which represent the circle of radius $3$ and center at origin.

The top of the cylinder is a circle.

Hence it represent the represents the volume of the cylinder of radius $3$and height $3$.