5.3 Double integrals in polar coordinates  (Page 5/7)

$D$ is the region of the disk of radius $2$ centered at the origin that lies in the first quadrant.

$D$ is the region between the circles of radius $4$ and radius $5$ centered at the origin that lies in the second quadrant.

$D$ is the region bounded by the $y$ -axis and $x=\sqrt{1-y^2}.$

$D$ is the region bounded by the $x$ -axis and $y=\sqrt{2-x^2}.$

$D=\left\{\left(x,y\right)|x^2+y^2\le 4x\right\}$

$D=\left\{\left(x,y\right)|x^2+y^2\le 4y\right\}$

In the following graph, the region $D$ is situated below $y=x$ and is bounded by $x=1,x=5,$ and $y=0.$

In the following graph, the region $D$ is bounded by $y=x$ and $y=x^2.$

$f\left(x,y\right)=x^2+y^2,D=\left\{\left(r,\theta \right)|3\le r\le 5,0\le \theta \le 2\pi \right\}$

$f\left(x,y\right)=x+y,D=\left\{\left(r,\theta \right)|3\le r\le 5,0\le \theta \le 2\pi \right\}$

$f\left(x,y\right)=x^2+xy,D=\left\{\left(r,\theta \right)|1\le r\le 2,\pi \le \theta \le 2\pi \right\}$

$f\left(x,y\right)=x^4+y^4,D=\left\{\left(r,\theta \right)|1\le r\le 2,\frac{3\pi }{2}\le \theta \le 2\pi \right\}$

$f\left(x,y\right)=\sqrt[3]{x^2+y^2},$ where $D=\left\{\left(r,\theta \right)|0\le r\le 1,\frac{\pi }{2}\le \theta \le \pi \right\}.$

$f\left(x,y\right)=x^4+2x^2{y}^2+y^4,$ where $D=\left\{\left(r,\theta \right)|3\le r\le 4,\frac{\pi }{3}\le \theta \le \frac{2\pi }{3}\right\}.$

$f(x,y)=\text{sin}\left(\text{arctan}\frac{y}{x}\right),$ where $D=\left\{(r,\theta )|1\le r\le 2,\frac{\pi }{6}\le \theta \le \frac{\pi }{3}\right\}$

$f(x,y)=\text{arctan}\left(\frac{y}{x}\right),$ where $D=\left\{(r,\theta )|2\le r\le 3,\frac{\pi }{4}\le \theta \le \frac{\pi }{3}\right\}$

${\displaystyle \underset{D}{\iint }{e}^{x^2+y^2}\left[1+2\text{arctan}\left(\frac{y}{x}\right)\right]}dA\text{,}D=\left\{\left(r,\theta \right)|1\le r\le 2,\frac{\pi }{6}\le \theta \le \frac{\pi }{3}\right\}$

${\displaystyle \underset{D}{\iint }\left(e^{x^2+y^2}+x^4+2x^2{y}^2+y^4\right)}\text{arctan}\left(\frac{y}{x}\right)dA\text{,}D=\left\{\left(r,\theta \right)|1\le r\le 2,\frac{\pi }{4}\le \theta \le \frac{\pi }{3}\right\}$