Q. 44 - Calculus

Question

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

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

Step-by-Step Solution

Answer

The three-dimensional region represents the volume of the region is the right by the parabolic $(y-3{)}^2=x^2+z^2$ bounded on the left by the xz plane. ,

Step 1. Given Information.

We are given, 

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

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 ${\int }_{-3}^3{\int }_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}{\int }_0^{3-\sqrt{x^2+z^2}}f(x,y,z)dydxdz$ represents the volume of the region is the right by the parabolic $(y-3{)}^2=x^2+z^2$ bounded on the left by the xz plane.

Since from the given iterated integral we observe that $y=3-\sqrt{x^2+z^2}$

$$\begin{gathered} \sqrt{x^2+z^2}=3-y \\ \Rightarrow (3-y{)}^2=x^2+z^2 \end{gathered}$$

It represents the hyperbolic plane in xz

Thus the given iterated integral represents the volume of the region is the right by the parabolic $(y-3)^{2}=x^{2}+z^{2}$ bounded on the left by the $x z$ plane.