Question

Evaluate the iterated integral :

${\int }_0^1 {\int }_0^{3-3y} {\int }_0^{6y} \frac{x}{y}dzdxdy$

Short answer

$9$

Step-by-step solution

Step 1. Given information.

Given integral is :

${\int }_0^1 {\int }_0^{3-3y} {\int }_0^{6y} \frac{x}{y}dzdxdy$

We have to evaluate it.

Step 2. Evaluate.

Given ${\int }_0^1 {\int }_0^{3-3y} {\int }_0^{6y} \frac{x}{y}dzdxdy$

$\begin{array}{r}={\int }_0^1 {\int }_0^{3-3y} \left[\left({\int }_0^{6y} \frac{x}{y}dz\right)dxdy\right]\\ ={\int }_0^1 {\int }_0^{3-3y} \left[\left(\frac{x}{y}{\int }_0^{6y} dz\right)dxdy\right]\\ ={\int }_0^1 {\int }_0^{3-3y} \left[\frac{x}{y}(z{)}_0^{6y}\right]dxdy\\ ={\int }_0^1 {\int }_0^{3-3y} \left[\frac{x}{y}\left(6y\right)\right]dxdy\\ ={\int }_0^1 {\int }_0^{3-3y} \left(6x\right)dxdy\end{array}$

Step 3. Integrate with respect to z.

Integrate with respect to z

$$\begin{gathered} ={\int }_0^1 \left[\left({\int }_0^{3-3y} 6xdx\right)dy\right] \\ ={\int }_0^1 \left(6{\left(\frac{x^2}{2}\right)}_0^{3-3y}\right)dy \\ ={\int }_0^1 {\left(3x^2\right)}_0^{3-3y}dy \\ \begin{array}{r}={\int }_0^1 3(3-3y{)}^{2}dy\\ ={\int }_0^1 3\left(9-18y+9y^2\right)dy\\ =27{\int }_0^1 dy-54{\int }_0^1 ydy+27{\int }_0^1 y^{2}dy\\ =27(y{)}_0^1-54{\left(\frac{y^2}{2}\right)}_0^1+27{\left(\frac{y^3}{3}\right)}_0^1\\ =27-27+9\\ =9\end{array} \end{gathered}$$