[11.5] - Double Integrals in Polar Coordinates

  1. Integrate $f(x,y)=x^2+y^2$ over the domain $D=\{(x,y):1\leq (x^2+y^2)\leq 4 \}$ using polar coordinates.
  2. Sketch the region of integration and evaluate by changing to polar coordinates. $$ \int_0^3\int_{x=0}^{\sqrt{9-y^2}}\sqrt{x^2+y^2}\,dx\,dy$$
  3. Find the shaded area of the region inside the curve $r=\theta$, $0\leq \theta \leq 3\pi$. Hint: Make sure you're not double-counting any areas.
  4. Rewrite $\int_{-2}^2\int_{y=0}^{\sqrt{4-x^2}}e^{-x^2-y^2}\,dy\,dx$ as a polar integral, and evaluate it.

Answers

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  1. $$\iint r^2\,dA= \int_{\theta=0}^{2\pi}\int_{r=1}^2 r^2\,r\,dr\,d\theta=\int_{\theta=0}^{2\pi}\left( \left.\frac{r^4}{4}\right|_1^2\right)\,d\theta= \frac{15\pi}{2}$$
  2. $$\int_0^{\pi/2}\int_0^3 r(r\,dr\,d\theta)=\frac{9\pi}{2}$$
  3. The equation of the black line is $r(\theta)=\theta$. $$\iint\,dA=\int_{\theta=\pi}^{3\pi}\int_{r=0}^\theta r\,dr\,d\theta =\frac{13\pi^3}3 $$
    • By hand, you should evaluate the $\int dr$ integral first to bring down the $\theta$ dependent limit before you integrate wrt $\theta$.
    • To avoid double-counting areas, the lower limit on $\theta$ should be $\pi$ (not 0, even though the black line starts at $\theta=0$).
  4. $$\int_{\theta=0}^\pi\int_{r=0}^2 e^{-r^2}r\,dr\,d\theta = \frac\pi2 \left(1-\frac1{e^4}\right)$$