Examples

Integration over measure of area

Integration over measure of area

Problem: Compute the integral ${\iint }_{\mathcal{R}}{x}^{2}ydA$ where $\mathcal{R}$ is the region in the figure:

Solution: We rewrite the integral as an iterated integral and then compute. The region can be written as $x\in [1,3]$ and $y\in [1/x,\sqrt{x}]$. So we have:

$$\begin{array}{rl}{\iint }_{\mathcal{R}}{x}^{2}ydA& ={\int }_1^3{\int }_{1/x}^{\sqrt{x}}{x}^{2}ydydx\\ & ={\int }_1^3{\frac{1}{2}{x}^2{y}^2|}_{1/x}^{\sqrt{x}}dx={\int }_1^3\frac{1}{2}{x}^3-\frac{1}{2}dx\\ & ={\frac{1}{8}{x}^4-\frac{1}{2}x|}_1^3=9.\end{array}$$

Area under curve as double integral

Area under curve as double integral

Problem: Find the area under the curve $y=x^2$ for $x\in [0,1]$ using a double integral.

Solution: Compute:

$$\begin{array}{rl}{\iint }_{\mathcal{R}}dydx& ={\int }_0^1{\int }_0^{x^2}dydx\\ & ={\int }_0^{1}y{|}_0^{x^2}dx={\int }_0^1{x}^{2}dx\\ & ={\frac{x^3}{3}|}_0^1=\frac{1}{3}.\end{array}$$

Note that this agrees with the usual method: ${\int }_0^1{x}^{2}dx=\frac{1}{3}$.