# Determining the area between two polar curves using a double integral

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 $(x − 1)^2 + y^2 = 1.$

Attempt:

Area 1

$$x^2+y^2=4$$

$$r=2$$

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

Area 2

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

$$r=2\cosθ$$

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

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

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

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

• 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.