Use a double integral in polar coordinates to find the area of the region inside the circle $x^2 + y^2 = 4$ but outside the circle $(x − 1)^2 + y^2 = 1.$


Area 1



$$\int_0^{2\pi} \int_0^2 r \,dr\,d\theta$$

Area 2

$$(x-1)^2+y^2= 1$$


$$\int_0^{2\pi} \int_0^{2\cosθ} r \,dr\,d\theta$$

$$\text{Area} = \text{Area} 1 - \text{Area} 2 $$

Is the set up ok? Im not sure if the region of integration is correct.

  • 1
    $\begingroup$ Where you wrote things like $x^2$+$y^2$=1, I changed it to $x^2+y^2=1$. That is proper MathJax usage. $\endgroup$ Apr 23, 2018 at 14:37

1 Answer 1

  • For the first area this is correct.
  • For the second one you obtain: $$\int_0^{2 \pi} \int_0^{2 \cos(\theta)} r \,dr \,d\theta=\int_0^{2 \pi} 2 \cos^2(\theta) \,d \theta=2 \pi$$ but the correct result is $\pi$ as it is a circle of radius $1$.

The problem is that by taking $\theta \in (0,2\pi)$ you go through the circle two times.

In my opinion the best way to avoid the problem is to write: $$\text{Area}_2 =\int_0^{2 \pi} \int_0^\infty \chi_{0 \leq r \leq 2 \cos(\theta)} r \,dr \,d\theta$$ so you obtain: $$\text{Area}_2 =\int_0^\pi \int_0^{2\cos(2 \theta} 1 r \,dr \,d\theta$$ which give $\pi$ as expected.


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