This problem is driving me crazy :


Someone please solve this problem in details by sketching the required boundary and how did he calculate it. The final answer is $\pi/4$ Thanks in advance

Edit : I am still learning polar coordinates , so i have failed to convert this double integral boundary to polar form !

  • 2
    $\begingroup$ What have you done so far to try and solve it. Show some working, people don't appreciate "please do this" $\endgroup$ – Henry Lee Nov 5 '18 at 19:20
  • $\begingroup$ @HenryLee I have tried to solve it but i have failed to convert the boundaries to polar form . This is because i am not yet good in polar coordinates ! $\endgroup$ – John adams Nov 5 '18 at 19:22
  • $\begingroup$ Shouldn't the answer be $\frac{\pi}{2}$? $\endgroup$ – John Douma Nov 5 '18 at 19:43
  • $\begingroup$ @JohnDouma No its pi/4 , please solve it. $\endgroup$ – John adams Nov 5 '18 at 20:46
  • $\begingroup$ Yes. I see that. Look at Key Flex's answer. You have a unit circle centered at $(0,1)$. You are being asked to find the area of the bottom right quarter of it. $\endgroup$ – John Douma Nov 5 '18 at 20:51


Use the fact that $$x=r\cos\theta\\y=r\sin\theta\\r^2=x^2+y^2$$

You have $\int_0^1\int_0^{\sqrt{2y-y^2}}dxdy$

From above we can say that $y=0,y=1$ and $x=0$, $x=\sqrt{2y-y^2}\implies x^2+(y-1)^2=1$, so it is a circle of radius $1$

Now can you continue from here?

  • $\begingroup$ Can you please help me in deriving the boundaries only ? $\endgroup$ – John adams Nov 5 '18 at 19:31
  • $\begingroup$ Are the limits will be : for r 0-->1 and for theta 0--> pi/4 ? $\endgroup$ – John adams Nov 5 '18 at 20:35
  • $\begingroup$ Here is my attempt . My limits are as follows theta from 0 to Pi/2 and r from 0 to 2sin(theta) and the integrand will be rdrd(theta). The answer of this double integration will give me the area of right half of the circle , then multiplying this result by 1/2 will give me the answer . So this attempt is correct ? My answer was pi/4 , but i don't know if its luck or it is the right approach ? $\endgroup$ – John adams Nov 5 '18 at 21:19
  • $\begingroup$ @Johnadams Yes, your approach and the answer are correct. $\endgroup$ – Key Flex Nov 6 '18 at 0:06

note: $$\int_0^1\int_0^{\sqrt{2y-y^2}}dxdy=\int_0^1\int_0^{\sqrt{1-(1-y)^2}}dxdy$$ so if $$x=\sqrt{1-(1-y)^2}$$ $$x^2=1-(1-y)^2$$ $$x^2+(y-1)^2=1$$ so it is a circle with centre $(0,1)$ and a radius of $1$ $$I=\int_0^{2\pi}\int_0^1 rdrd\theta=\pi$$

  • $\begingroup$ please check your limits for polar coordinates. $\endgroup$ – John adams Nov 5 '18 at 20:35

Try the following: $\int_0^1 \int_{0}^{\sqrt{2y-y^2}} dx dy = \int_0^1 x|_{0}^{\sqrt{2y-y^2}} =\int_0^1 \sqrt{2y-y^2} dy = \int_0^1 \sqrt{1-(y-1)^2} dy$

Now Substitute $u:=y-1$, which leads you to $\int_{-1}^{0}\sqrt{1-u^2} du$

This is a standard integral, which is computed multiple times here in the forum (and on the whole internet), so we get: $\int_{-1}^{0}\sqrt{1-u^2} du = \frac{1}{2}(u \sqrt{1-u^2} + arcsin(u))|_{-1}^{0}= \frac{1}{2}arcsin(-1) = \frac{\pi}{4}$

  • $\begingroup$ please solve it using polar coordinates. $\endgroup$ – John adams Nov 5 '18 at 20:44

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