Easy way to integrate over region in first quadrant between circles

Question

The integral in question is $$ \iint\limits_D x \ dA$$ where $D$ is the region in the first quadrant lying between circles $x^2+y^2=4$ and $x^2+y^2=2x$.

Changing to polar coordinates I get $$\int_0^{\pi/2} \!\!\int_{2\cos{\theta}}^2 r^2\cos\theta \ dr \ d\theta = \frac{8}{3} \int_0^{\pi/2} \cos\theta-\cos^4\theta \ d\theta$$

Is there any way I can calculate this result without integrating $\cos^4\theta$? I wonder if there are any geometric solutions that avoid polar coordinates completely.

Edit: rectangular coordinates, too easy...

Answer 1

Let $D_1$ be the region described by $x^2+y^2 \le 4,\, x\ge 0,\, y \ge 0$ and $D_2$ the region $(x-1)^2+y^2 \le 1, \, y \ge 0$. \begin{gather*} \iint_{D}x\, dA = \iint_{D_1}x\, dA -\iint_{D_2}x\, dA =\\[2ex] \int_{0}^{2}\left(\int_{0}^{\sqrt{4-y^2}}x\, dx\right)\, dy - \int_{0}^{1}\left(\int_{1-\sqrt{1-y^2}}^{1+\sqrt{1-y^2}}x\, dx\right)\, dy = \\[2ex] \dfrac{1}{2}\int_{0}^{2}(4-y^2)\, dy - \int_{0}^{1}2\sqrt{1-y^2}\, dy = \dfrac{8}{3}-\dfrac{\pi}{2}. \end{gather*} where the last integral can be interpreted as the area of a half unit circle.

Answer 2

With the fundamental identity and the double angle sine formula:

$\cos^4\theta=\cos^2\theta(1-\sin^2\theta)=\cos^2\theta-\frac{1}{2}\sin^2(2\theta)$

Further you might use the double angle cosine formula:

$\cos^2\theta=\frac{1}{2}(1+\cos(2\theta))$

$\sin^2(2\theta)=\frac{1}{2}(1-\sin(4\theta))$

... and you have reduced your integral to simple sine and cosine

Answer 3

The integral $$ I = \displaystyle\int_0^{\frac \pi 2} \cos^4 \theta = \displaystyle\int_0^{\frac \pi 2} \cos^4 \left(\frac \pi 2 - \theta\right) = \displaystyle\int_0^{\frac \pi 2} \sin^4 \theta $$ Note then that $$ 2I = \displaystyle\int_0^{\frac \pi 2} \cos^4 \theta + \sin^4 \theta $$ Now, a little bit of insight, and we see that: $$ \cos^4 \theta + \sin^4 \theta = \frac{\cos 4\theta + 3}4 $$ So it turns into: $$ 2I = \displaystyle\int_0^{\frac \pi 2} \frac{\cos 4\theta + 3}4 $$ Surely this is an easier integral to solve. Upon request, I can derive the identity above.