Q. 26 - Calculus

Question

Evaluate the iterated integral :

${\int }_1^3 {\int }_{-1}^5 {\int }_0^3 x{y}^{2}zdzdydx$

Step-by-Step Solution

Answer

$\frac{189}{2}$

Step 1. Given information.

Given integral is :

${\int }_1^3 {\int }_{-1}^5 {\int }_0^3 x{y}^{2}zdzdydx$

We have to evaluate it.

Step 2. Evaluate.

Given ${\int }_1^3 {\int }_{-1}^5 {\int }_0^3 x{y}^{2}zdzdydx$

$\begin{array}{r}={\int }_1^3 {\int }_{-1}^5 \left({\int }_0^3 x{y}^{2}zdz\right)dydx\\ ={\int }_1^3 {\int }_{-1}^5 \left(x{y}^2{\int }_0^3 zdz\right)dydx\\ ={\int }_1^3 {\int }_{-1}^5 \left(x{y}^2{\left(\frac{z^2}{2}\right)}_0^3\right)dydx\\ ={\int }_1^3 {\int }_{-1}^5 \left(x{y}^2\left(\frac{9}{2}\right)\right)dydx\end{array}$

Step 3. Integrate with respect to y.

$\begin{array}{r}={\int }_1^3 \left(\frac{9}{2}x{\int }_{-1}^5 y^{2}dy\right)dx\\ ={\int }_1^3 \left(\frac{9}{2}x{\left(\frac{y^3}{3}\right)}_{-1}^{5}dx\right.\\ ={\int }_1^3 \left(\frac{9}{2}x\left(\frac{5^3}{3}-\frac{(-1{)}^3}{3}\right)dx\right.\\ ={\int }_1^3 \left(\frac{9}{2}x\left(42\right)\right)dx\\ ={\int }_1^2 189xdx\end{array}$

Integrate with respect to x. 

$\begin{array}{r}=189{\int }_0^1 xdx\\ =189{\left(\frac{x^2}{2}\right)}_0^1\\ =189\left(\frac{1}{2}-0\right)\\ =\frac{189}{2}\end{array}$