1.8 KiB
#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
$$\begin{gather}
0\leq r \leq 1 \ 0 \leq \theta \leq \frac{\pi}{2}
\end{gather}$$
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}$$