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

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

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$\mathrm{ℝ}=\left\{(x,y,z)\mid a_1\le x\le a_2,b_1\le y\le b_2,c_1\le z\le c_2\right\}$

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.