# Shared area of $2$ circles (polar integral)

Given $$x^2+y^2=4y$$ and $$x^2+y^2=4x$$ find the shared area of the 2 circles

What I tried was :

so first I transformed them to polar coordinates I got $$r=4cos\theta$$ for $$x^2+y^2=4x$$ and $$r=4sin\theta$$ for $$x^2+y^2=4y$$

After that I did $$4cos\theta=4sin\theta$$ then $$\theta=\frac{\pi}{4}$$ I am not sure but according to this I believe that $$\frac{\pi}{4}≤\theta≤\frac{\pi}{2}$$ lastly I did the integral

$$\int_\frac{\pi}{4}^\frac{\pi}{2}\frac{1}{2}((4cos\theta)^2-(4sin\theta)^2) d\theta$$ after integrating I got $$-4$$ which is a wrong answer.

what am I doing wrong? thanks for any help and tips! Edit: I forgot to mention that I havent studied double integrals yet , I learned polar system and this formula $$\int_\alpha^\beta\frac{1}{2}(r(\theta))^2 d\theta$$

• Have you tried a diagram? Can you see how your integrand's sign varies? – J.G. Mar 16 at 18:57
• You need not to integrate to find the area. – user Mar 16 at 19:05
• @user I mean sure, but the point of the exercise seems to be to use the tools they've learned in a section to practice setting up and doing problems leading up to more complicated examples like cardioids – Ninad Munshi Mar 16 at 19:08
• @J.G. I did draw the circles .. I still cannot see what I am missing can you give me a hint? – Adamrk Mar 16 at 19:10
• @NinadMunshi exactly , according to the book I should use integrals here for practice – Adamrk Mar 16 at 19:11

Please note that for $$0 \leq \theta \leq \frac{\pi}{4}$$, you are bound by circle $$4 \sin\theta$$ and for $$\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$$, you are bound by circle $$4 \cos\theta$$.
So the area is given by, $$A = \displaystyle 2 \int_0^{\pi/4} \int_0^{4\sin\theta} r \ dr \ d\theta$$
or $$\ \displaystyle 2 \int_0^{\pi/4} \frac{1}{2}(4\sin\theta)^2 \ d\theta$$
You can see that the area between $$0 \leq \theta \leq \frac{\pi}{4}$$ and between $$\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$$ are same. So we multiply by $$2$$ or use the fact that $$\sin\theta = \cos (\frac{\pi}{2} - \theta)$$.
• thank you for the help , but my mistake I forgot to mention I cannot use double integrals as I haven't gotten there yet. what I learned so far is polar system and this formula $\int_\alpha^\beta\frac{1}{2}(r(\theta))^2 d\theta$ – Adamrk Mar 16 at 19:18