#uw/notes #uw/class/math224 
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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 
$$\int_{\alpha}^\beta \int_{a}^b f(r\cos(\theta), r\sin(\theta))r \ dr d\theta$$


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 
$$$$