# Sketch the region $R$ and evaluate the double integral $\iint 2y\;\mathrm dA$

Let $R$ be the region in the first quadrant bounded above by the circle $(x-1)^2 + y^2 = 1$ and below by the line $y = x$ . Sketch the region $R$ and evaluate the double integral $\iint 2y \;\mathrm dA$ .

• Please stop copying and pasting your homework assignments here. This is not a free homework service. – Ron Gordon Jul 10 '13 at 16:44
• Could you please show us what you have tried? I'm sure we would be happy to help if you did ... It's just that we need to know that you're interested in learning from solving the problem. If you show that initiative, then people are always willing to help! – Amitesh Datta Jul 10 '13 at 16:47

We will do it using rectangular coordinates, though polar is tempting in any problem that involves circles.

The sketch is an essential part of the work. (I do not think I could do the calculation of the integral without a picture.)

We have a circle centre $(1,0)$ and radius $1$, and the familiar line $y=x$. These two curves meet. It will be useful to know where. Put $y=x$ in the equation $(x-1)^2+y^2=1$. Expand. We quickly get $2x^2-2x=0$. So the curves meet at $(0,0)$ and $(1,1)$.

Shall we integrate first with respect to $x$ or with respect to $y$? We take a quick think about the alternatives. I think $y$ first will be best.

If we integrate first with respect to $y$, then $y$ will travel from $x$ to the top of the circle. From $y^2=1-(x-1)^2$, we find that the top half of the circle has equation $y=\sqrt{1-(x-1)^2}=\sqrt{2x-x^2}$. After we have integrated with respect to $y$, we integrate with respect to $x$, with $x$ travelling from $0$ to $1$. So we want $$\int_0^1 \left(\int_x^{\sqrt{2x-x^2}} 2y\,dy \right)\,dx.$$ The integrations are easy. And even the square roots disappear!

Remark: If we choose to integrate first with respect to $x$, note that the left half of the circle has equation $x=1-\sqrt{1-y^2}$. So if we integrate first with respect to $x$, then $y$ travels from $1-\sqrt{1-y^2}$ to $y$, and then $y$ travels from $0$ to $1$. The integration with respect to $x$ is easy, we get $2yx$. The substitution now yields a slightly messy function of $y$, but doable.

First you need to find the region: you can see it here. The minor segment is your required region and bounded by $$x=0$$$$x=1$$$$y=x$$$$y=\sqrt{1-(x-1)^2}=\sqrt{2x-x^2}$$ so all you need to do is changing $dA$ into $dydx$ and put those upper and lower limit on the integral.

$$\iint 2y\;\mathrm dA=\int^1_0\int^{\sqrt{2x-x^2}}_x 2y\;\mathrm dy\mathrm dx=\int^1_02x-2x^2\:\mathrm dx=\frac13$$