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 $

  • $\begingroup$ Have you tried a diagram? Can you see how your integrand's sign varies? $\endgroup$ – J.G. Mar 16 at 18:57
  • $\begingroup$ You need not to integrate to find the area. $\endgroup$ – user Mar 16 at 19:05
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    $\begingroup$ @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 $\endgroup$ – Ninad Munshi Mar 16 at 19:08
  • $\begingroup$ @J.G. I did draw the circles .. I still cannot see what I am missing can you give me a hint? $\endgroup$ – Adamrk Mar 16 at 19:10
  • $\begingroup$ @NinadMunshi exactly , according to the book I should use integrals here for practice $\endgroup$ – Adamrk Mar 16 at 19:11

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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 $\sin\theta = \cos (\frac{\pi}{2} - \theta)$.

  • $\begingroup$ 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 $ $\endgroup$ – Adamrk Mar 16 at 19:18
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    $\begingroup$ OK see my edit. $\endgroup$ – Math Lover Mar 16 at 19:25

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