Calculus - Value of $\int\int _A xy dxdy$ between two circles using double integration (no polar coordinates)

Question

This is my first post on this forum, so I'm sorry in advance if I come to the wrong section or something ...

I am currently stuck on an exercise of an exam given in my math college. The exercise is the following:

Statement: Let A be the set of points in the plane bounded by the two circles of equation: $x^2+y^2 = 1$ and $(x-1)^2 + (y-1)^2 = 1$.

Exercise: Draw A, then calculate the integral $\int\int _A xy dxdy$.

However, I don't know whether to switch to polar coordinates (which would make it easier to calculate I presume), as it is given as a hint:

Hint: "Start with an integration with respect to y. In the integration with respect to x, make a change of variable $x = 1 - sin(t)$"

I tried to do this by posing: $x^2+y^2 =1\Leftrightarrow x^2 = 1 - y^2$, which I tried to inject into (x-1)^2 + (y-1)^2 = 1, which after transformation, resulted in x+y = 2, which is not great because I can at most get $\sqrt{1-x^2}+y = 2$, using what was done few lines before.

So here are my questions:

• How do I integrate without polar coordinates, more precisely how do I choose my bounds?
• I don't see how to choose my bounds if I switch to polar coordinates. I suspect that $\theta \in \left[0;\frac{\pi}{2} \right]$, but it stops here at most.

Thanks in advance for your help!

Here's the two circles

Answer 1

The double integral in usual, cartesian coordinates is

$$\int_0^1\int_{1-\sqrt{1-(x-1)^2}}^{\sqrt{1-x^2}}dy\,dx=\int_0^1\left[\sqrt{1-x^2}+\sqrt{1-(x-1)^2}-1\right] dx$$

For the first integral resulting on the rightmost one above substitute $\;x=\sin t\;$, and for the second one substitute $\;x-1=\sin t\;$ . The last one equals $1$, of course. You thus get

$$\int_0^{\pi/2}\cos^2t\,dt+\int_{-\pi/2}^0 \cos^2t\,dt-1=\frac\pi2-1$$

Pretty simple, no need of polar coordinates

Answer 2

Given circles are $C1: x^2 + y^2 = 1$ and $C2: (x-1)^2 + (y-1)^2 = 1$

At intersection $x+y = 1 \implies 1-x = \sqrt{1-x^2} \implies x = 0, 1$. That leads to $y = 1, 0$. So intersection points are $(0, 1)$ and $(1, 0)$.

For $C2, x = 1 \pm \sqrt{1 - (y-1)^2} = 1 \pm \sqrt{2y-y^2}$. As we are between $0 \leq x \leq 1, x = 1 - \sqrt{2y-y^2}$. So if we integrate wrt $dx$ first and then $dy$, the integral to find area of intersection can be written as,

$\displaystyle \int_0^1 \int_{1-\sqrt{2y-y^2}}^{\sqrt{1-y^2}} xy \ dx \ dy = \frac{\pi}{4} - \frac{2}{3}$

To evaluate it in polar coordinates, please use the hint given. In the diagram if you shift the origin to $O' (0, 1)$, it is easy to see that $C1$ forms between $- \pi \leq \theta \leq 0$ with equation $r = - 2 \sin\theta$ and $C2$ forms between $- \pi/2 \leq \theta \leq \pi/2$ with equation $r = 2 \cos\theta$.

To get to the above equations, you can plug in,

$x = r \cos\theta, y = 1 + r \sin\theta$ in $x^2+y^2 = 1$ and $(x-1)^2+(y-1)^2 = 1$

At intersection of both circles, $2 \cos \theta = - 2 \sin\theta \implies \theta = - \frac{\pi}{4}$. As you can see, between $- \frac{\pi}{2} \leq \theta \leq -\frac{\pi}{4}$, we are bound by $C2$ and between $- \frac{\pi}{4} \leq \theta \leq 0$, we are bound by $C1$.

So the integral is,

$\displaystyle \int_{-\pi/2}^{-\pi/4} \int_0^{2\cos\theta} r^2 \cos\theta \ (1 + r \sin\theta) \ dr \ d\theta$ +
$\displaystyle \int_{-\pi/4}^{0} \int_0^{-2\sin\theta} r^2 \cos\theta \ (1 + r \sin\theta) \ dr \ d\theta$