Evaluate the area of given region using double integral technique

Exercise:

Evaluate $\iint\limits_{D}{\left( 4xy-{{y}^{3}} \right)dA}$ where $D$ is region bounded by two curves are $\sqrt{x}\,\,\And \,\,{{x}^{3}}$

Solution: First we need to sketch the given region

We have $\sqrt{x}={{x}^{3}}\Leftrightarrow x={{x}^{5}}\Leftrightarrow x\left( 1-{{x}^{4}} \right)=0\Leftrightarrow x=0\,or\,{{x}^{4}}=1\Leftrightarrow x\in \left\{ 0,1 \right\}$( since $x\ge 0$ )

So $D=\left\{ \left( x,y \right)\in {{\mathbb{R}}^{2}}\,:\,\,\,0\le x\le 1\,\,\And \,\,\,{{x}^{3}}\le y\le \sqrt{x} \right\}\,$

So $\iint\limits_{D}{f\left( x,y \right)dA=\int_{0}^{1}{\int_{{{x}^{3}}}^{\sqrt{x}}{f\left( x,y \right)dydx}}}$

$=\int_{0}^{1}{\left( \int_{{{x}^{3}}}^{\sqrt{x}}{\left( 4xy-{{y}^{3}} \right)dy} \right)dx}$

$=\int_{0}^{1}{\left( 2x{{y}^{2}}-\frac{{{y}^{4}}}{4} \right)_{{{x}^{3}}}^{\sqrt{x}}dx}$

$=\int_{0}^{1}{\left( 2{{x}^{2}}-\frac{{{x}^{2}}}{4}-2{{x}^{7}}+\frac{{{x}^{12}}}{4} \right)dx}=\left( \frac{2}{3}{{x}^{3}}-\frac{1}{12}{{x}^{3}}-\frac{2}{8}{{x}^{8}}+\frac{{{x}^{13}}}{52} \right)_{0}^{1}=\frac{2}{3}-\frac{1}{12}-\frac{1}{4}+\frac{1}{52}=\frac{55}{156}$