# Evaluate an integral in polar form

This problem is driving me crazy :

$$\int_{0}^{1}\int_0^{\sqrt{2y-y^2}}{dx.dy}$$

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 !

• What have you done so far to try and solve it. Show some working, people don't appreciate "please do this" – Henry Lee Nov 5 '18 at 19:20
• @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 ! – John adams Nov 5 '18 at 19:22
• Shouldn't the answer be $\frac{\pi}{2}$? – John Douma Nov 5 '18 at 19:43
• @JohnDouma No its pi/4 , please solve it. – John adams Nov 5 '18 at 20:46
• 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. – John Douma Nov 5 '18 at 20:51

## 3 Answers

Hint:

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?

• Can you please help me in deriving the boundaries only ? – John adams Nov 5 '18 at 19:31
• Are the limits will be : for r 0-->1 and for theta 0--> pi/4 ? – John adams Nov 5 '18 at 20:35
• 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 ? – John adams Nov 5 '18 at 21:19
• @Johnadams Yes, your approach and the answer are correct. – 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 $$1$$ $$I=\int_0^{2\pi}\int_0^1 rdrd\theta=\pi$$

• please check your limits for polar coordinates. – 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: $$\int_{-1}^{0}\sqrt{1-u^2} du = \frac{1}{2}(u \sqrt{1-u^2} + arcsin(u))|_{-1}^{0}= \frac{1}{2}arcsin(-1) = \frac{\pi}{4}$$

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