# Evaluating double integral.

Find the double integral $$I=\iint_D y dy dx$$ where $$D$$ is area bounded by

$$D= \{(x,y): x^2+y^2\leq 1, x^2+y^2\leq2x, y\leq0 \}$$

First off, function I am evaluating is $$z=y$$ and it's just one plane in $$\mathbb{R}^3$$.

Now, this is the $$D$$ I am looking for:

Now, I should integrate over the intersection of these two circles where $$y<0$$. Obviously, I should use polar coordinates so $$x=r\cos(\theta)\\ y=r\sin(\theta)$$, Jacobian is $$r$$ so it remains now to find the bounds of integration, however I am not quite sure how to do that, since the intersection point of two circles is $$(x,y)=\left(\frac{1}{2}, -\frac{\sqrt3}{2}\right)$$ I suppose that angle should go from $$-\frac{\pi}{6}$$ to $$0$$ but then, I could write out the equations of circles to get bounds for $$r$$, where $$r$$ should go from the blue circle to red circle, which means $$r \in [2\cos\theta, 1]$$ , but I am not quite sure is that correct way to do it because when i calculate it this way I get the positive result $$I=\frac{5}{12}$$, which is not what I expected since I am integrating over an area where $$x$$ is positive and $$y$$ is negative which means that I am actually finding a volume of body that's either in fourth or eighth octant, but, when I take a look at the function I am integrating I see that $$z=y$$ meaning that sign of $$y$$ will determine the sign of $$z$$ so I am finding a volume of a body that's located in eighth octant, so I expected a negative value here. Obviously, something is wrong, it might be $$D$$ or bounds of integration or calculations or ,the worst case probably, my reasoning. Any help is appreciated.

• symmetry with respect to what? I am actually integrating the part where $y\leq 0$, not the whole intersection, just like i stated above. Commented Nov 26, 2017 at 12:28
• Right, i didn't notice. Discarding my comment.
– user65203
Commented Nov 26, 2017 at 12:57

Your integral can be written (using the geometry of the domain), as $$I=\int_{\frac{-\sqrt{3}}{2}}^{0} \int_{1-\sqrt{1-y^2}}^{ \sqrt{1-y^2}} y dx dy = \int_{\frac{-\sqrt{3}}{2}}^{0} y \left(2 \sqrt{1-y^2}-1 \right) dy$$

Edit following OP's comment pointing out that the domain contains only region below x-axis:

Making the substitution $y=\sin\theta$,

$$I=\int_{\frac{-\pi}{3}}^0 \sin\theta (2\cos \theta-1) \cos\theta d\theta = -\frac{5}{24}$$

• i don't understand how you got your upper bound for variable $y$. I explicitly stated that i am looking for the area $y\leq 0$ right bellow the picture, and if you check out $D$ again, you'll see that i did the same thing in the definition of $D$. Shouldn't then upper bound be zero actually? Commented Nov 26, 2017 at 12:25
• @Mathemagical are you sure about the integration limits? $1-\sqrt[1-y ^2]$ is the arc of the circle centered in the origin translated up by 1. $\sqrt[1-y ^2]$ is the arc of the circle centered in the origin in the first quadrant.
– Upax
Commented Nov 26, 2017 at 13:20
• @cdummie: I think you simply did a small mistake. The results is negative and you forget to divide by 2 i.e. I = -5/24
– Upax
Commented Nov 26, 2017 at 13:22
• @cdummie ah. I missed that. Sorry. In that case, the answer, I got on the lower half was -5/24 and on the upper 5/24. Let me edit the answer. Commented Nov 26, 2017 at 13:50
• @cdummie: the result I got is the same of Yves Daoust.
– Upax
Commented Nov 26, 2017 at 13:57

The length of the section of the area at ordinate $y$ is $\sqrt{1-y^2}-(1-\sqrt{1-y^2})$, and it cancels at $y=-\sqrt3/2$.

Then

$$I=\int_{y=-\sqrt3/2}^0 y\,\left(2\sqrt{1-y^2}-1\right)\,dy=\left.-\frac23(1-y^2)^{3/2}-\frac{y^2}2\right|_{y=-\sqrt3/2}^0.$$

The solution I propose is based on the Green's theorem. Let R the interior of the region delimited by D, and let (P(x,y),Q(x,y)) a differentiable vector field. Then $$\int_C P dx + Q dy = \int \int_R (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}) dx dy$$ Now if we assume $P = -y^2/2$ and $Q = 0$ we have $$-\int_C \frac{y^2}{2} dx = \int \int_R y dx dy$$ We can then evaluate the integral over C. To do so we use the polar coordinated. We have two circles, but the only difference is that one is translated along the x-axis with respect to the other. If we consider the circle centered into the origin we have: $$x(\theta)= \cos(\theta)$$ and $$y(\theta)= \sin(\theta)$$ So $$dx= -\sin(\theta) d\theta$$ Then

$$-\int_{C_1} \frac{y^2}{2} dx = \frac{1}{2}\int \sin^{3}(\theta)$$ where $C_1$ is the circle centered into the origin. For the second circle $dx$ is the same so you can find exactly the same expression. Once integrating with with the correct integration limits you get the value -5/24.