Question

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

${\int }_{-3}^3{\int }_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}{\int }_0^{9-y^2-z^2}f(x,y,z)dxdzdy$

Short answer

The three-dimensional region represents the volume of the region is the right by the parabolic $x=9-y^2-z^2$bounded on the left by the yz plane. ,

Step-by-step solution

Step 1. Given Information.

We are given,

${\int }_{-3}^3{\int }_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}{\int }_0^{9-y^2-z^2}f(x,y,z)dxdzdy$

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 triple integral represents the volume of the region is the right by the parabolic $x=9-y^2-z^2$ bounded on the left by the yz plane.

Since from the given integral we observe that

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

Which represent the parabolic region and

$z=\sqrt{9-y^2}$

It represent yz plane.

Thus, the given iterated integral represents the volume of the region is the right by the parabolic $x=9-y^2-z^2$bounded on the left by the yz plane.