Finding an Iterated Integral Using Polar Coordinates

Find the iterated integral \(\int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}} }(x^{2}+y^{2})\,{\it dy}\,{\it dx}\).

Solution The region \(R\) is enclosed by the graph of \(y=\sqrt{ 4-x^{2}}\), (a portion of a circle of radius 2), \(0\leq x\leq 2\), and the positive \(x\)- and \(y\)-axes. See Figure 27. In polar coordinates, this region \(R\) is given by \(0\leq r\leq 2\) and \(0\leq \theta \leq \dfrac{\pi }{2} \). Then \[ \begin{eqnarray*} \int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}}}(x^{2}+y^{2})\,{\it dy}\,{\it dx}&=&\displaystyle\iint\limits_{\kern-3ptR}(x^{2}+y^{2}) \, {\it dA}\\[4pt] &=&\displaystyle\iint\limits_{\kern-3ptR}r^{2}r\,{\it dr}\,d\theta\qquad{\color{#0066A7}{\hbox{$x^2 + y^2=r^2;\quad {dA}=r\,{dr}\,d\theta$}}}\\[4pt] &=&\int_{0}^{\pi /2}\int_{0}^{2}r^{3}\,dr\,d\theta =\int_{0}^{\pi /2}\left[\dfrac{r^{4}}{4}\right] _{0}^{2}\,d\theta\\[4pt] &=&\int_{0}^{\pi /2}4\,d\theta = \big[ 4\theta \big] _{0}^{\pi /2}=2\pi \end{eqnarray*} \]