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