# A double integral into polar coordinates

I have the double integral

$$\int^{10}_0 \int^0_{-\sqrt{10y-y^2}} \sqrt{x^2+y^2} \,dx\,dy$$

And I am asked to evaluate this by changing to polar coordinates.

• When you see $10y-y^2$ it should occur to you to try completing the square: $$1-y-y^2 = 25 - (5-y)^2$$ The graph of $(\,x = -\sqrt{\text{that expression}} \, )$ is a half-circle of radius $5$ centered at $(x,y) = (0,5). \qquad$ – Michael Hardy Jan 2 '18 at 16:30
• see my comment below. Do you think that I'm right? – user504516 Jan 2 '18 at 16:32

Set $x$ equal to the lower bound of your inner integral. This line defines a boundary of the region of integration. Plot it, along with the other boundaries $x=y=0$ and $y=10$ and see if you can express the region more naturally in polar coordinates. Make the necessary transformation and it should become clear how to proceed.

• The upper limit is the line $x=0$ and the lower limit is $x^2+(y-5)^2=5^2$, corresponding to a circle centered in $C(0,5)$. The conversion to polars simplifies the integral, indeed. Geometrically, we have that ${-\sqrt{10y-y^2}}≤x≤0$ (rearranging this leads to $0≤x^2+(y-5)^2≤5^2$) and $0≤y≤10$. So this is a quarter circle of radius $5$ in the second quadrant. So the appropriate bounds will be $0≤r≤5$ and $π/2≤φ≤π$. Αm I correct? – user504516 Jan 2 '18 at 16:13
• @John : I'm accustomed to calling it $\theta$ rather than $\varphi.$ Your bounds for that are right, but I don't know why you have a quarter circle rather than a half circle. Note that $y$ goes all the way up to $10.$ And your bounds for $r$ depend on $\theta.$ See my posted answer. – Michael Hardy Jan 2 '18 at 16:42

Complete the square: $$10y-y^2 = 25 - (5-y)^2,$$ so the graph of $x = -\sqrt{10y-y^2} = - \sqrt{5^2 - (5-y)^2}$ is the left half of the circle $x^2 + (5-y)^2 = 5^2.$

Exercises with polar coordinates will have shown you that $$\tag 1 r = 5\sin\theta$$ is that circle. If you multiply both sides of $(1)$ by $r,$ you get $$r^2 = 5r\sin\theta$$ which becomes $$x^2+y^2 = 5y$$ and by completing the square, then becomes $$x^2 + (y-5)^2 = 5^2.$$

Thus you have $$\int_{\pi/2}^\pi \int_0^{5\sin\theta} r\, dr\,d\theta.$$

• thank you, I am trying to understand the reason for the bounds for $r$ depend on $θ$ – user504516 Jan 2 '18 at 16:46
• Is $$\int_{\pi/2}^\pi \int_0^{5\sin\theta} r^2\, dr\,d\theta?$$ – user504516 Jan 2 '18 at 17:06
• @John : Draw the picture and you'll see it. Draw a ray going from the origin into the second quadrant. See where it intersects the half-circle. The distance from the origin to that intersection point depends on the direction in which the ray points from the origin. Observed also that I explained why $r$ runs from $0$ to $5\sin\theta,$ and that $5\sin\theta$ depends on $\theta. \qquad$ – Michael Hardy Jan 2 '18 at 22:09