#uw/notes #uw/class/math224 
- - -
Instead of being bound by a rectangular region regions in polar coordinates are often bound with 
$$\begin{gather}
	a\leq r\leq b \\ \alpha \leq \theta \leq \beta
\end{gather}$$

![[PolarRegion.excalidraw]]

if a function is defined as $f(x,y)$ and the region it is integrated over is polar the integral will often look similar to 
$$\int_{\alpha}^\beta \int_{a}^b f(r\cos(\theta), r\sin(\theta))r \ dr d\theta$$


> [!example] Example
> Find the volume of the solid below $z=1-x^2-y^2$ and above the first quadrant on the xy plane
> 
> rearranging the equation gives $x^2+y^2=1-z$ which shows that each horizontal slice of the function is a circle centered at $(0,0)$ with a radius of $\sqrt{ 1-z }$
> 
> since the slice of the equation at $z=0$ forms the circle $x^2+y^2=1$ the bounds 
> $$\begin{gather}
> 	0\leq r \leq 1 \\ 0 \leq \theta \leq \frac{\pi}{2}
> \end{gather}$$
> can be used to find the volume of the solid.
> 
> writing out the integral with these bounds gives
> $$\begin{gather} 
>  \int \int_{R} f(x,y) dA \\[2ex] \int_{\alpha}^\beta \int_{a}^b f(r\cos(\theta), r\sin(\theta))r \ dr d\theta \\[2ex]
> 	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-(r\cos \theta)^2-(r\sin \theta)^2)r \ dr d\theta
> \end{gather}$$
> This integral can now be solved similar to other double integrals
> $$\begin{gather}
> 	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-(r\cos \theta)^2-(r\sin \theta)^2)r \ dr d\theta  \\[2ex]
> 	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-r^2\cos^2 \theta-r^2\sin^2 \theta)r \ dr d\theta \\[2ex] 
> 	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-r^2)r \ dr d\theta \\[2ex]
> 	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 r-r^3 \ dr d\theta \\[2ex]
> 	\int_{0}^{\frac{\pi}{2}} \left( \left.\frac{r^2}{2}-\frac{r^4}{4} \right\vert_{0}^1 \right) d\theta \\[2ex] 
> 	\int_{0}^{\frac{\pi}{2}} \frac{1}{4} d\theta \\[2ex] 
> 	V=\frac{\pi}{8}
> \end{gather}$$


When are polar coordinates useful for evaluating double integrals?
- Domain has a simple formula in terms of $(r,\theta)$
- Integrand has a simple formula in terms of $(r, \theta)$

Polar regions with functions as boundaries can also be integrated similar to cartesian ones
$$\int_{\alpha}^\beta \int_{h_{1}(\theta)}^{h_{2}(\theta)} f(r\cos \theta, r\sin \theta)r \ dr d\theta$$