# Double Integral Area Calculation

Evaluate $$\iint_A xy\,dx\,dy$$ where $A$ is bounded by $x^2+y^2-2x=0$, $y^2=2x$ and $y=x$.

After I sketched the curves, I found these points,

So by Integrating the above $xy$ by $dy\,dx$ I will get the area enclosed by those curves, right? there are 3 points and I'm confused how should I solve this problem.

if I consider the limits upper $(2,2)$ and lower limits $(0,0)$ is that correct?

• Try converting to polar coordinates. Then $\theta$ will be in $[\frac {\pi}{4},\frac {\pi}{2}]$ and $r$ in $[2\cos\theta, \csc\theta \cot\theta].$ Alternatively, break the region in two, one with $x$ in [0,1], and one with $x$ in $(1,2]$ – Doug M Sep 13 '17 at 18:11
• @DougM thanks for the answer. but we weren't taught about the polar coordinates. I still didn't get your second method. let's say I divided it. what will be the limits then? and I still must use xy dxdy for the both parts? Sorry i'm not very familiar with multiple integrations as I was taught them recently – user2511652 Sep 13 '17 at 18:18

HINT

To solve the leftmost curve for $x$ note it is equivalent to $$(x-1)^2 = 1-y^2$$ and therefore, as you are above the $x$-axis, you get $$x = 1 + \sqrt{1-y^2}.$$

From your picture then, $$\iint_A dA = \int_{y=0}^{y=1} \int_{x=1 + \sqrt{1-y^2}}^{x=y^2/2} dx\,dy + \int_{y=1}^{y=2} \int_{x=1 + \sqrt{1-y^2}}^{x=y} dx\,dy$$

Can you take it from here?

• I understand it now sir. thank you very much for your answer – user2511652 Sep 13 '17 at 18:27
• btw your method gives me 'indeterminate' for the answer when I solved it in using 'Microsoft Mathematics' – user2511652 Sep 14 '17 at 2:20
• @user2511652 not sure about the software nor about how you used it... – gt6989b Sep 14 '17 at 3:07

If you have not yet learned how to transform the coordinate system in double integration...

The region bound by the line the circle and the parabola can be broken into two.

$\displaystyle \int_0^1 \int_{\frac 12y^2}^{\sqrt{1 -y^2}+1} xy\ dx\ dy + \int_1^2 \int_{\frac 12 y^2}^y xy\ dx\ dy$

Alternatively:

$\displaystyle \int_0^1 \int_{\sqrt{2x -x^2}}^{\sqrt {2x}} xy\ dy\ dx + \int_1^2 \int_{x}^{\sqrt {2x}} xy\ dy\ dx$

and in polar coordinates

$\displaystyle \int_{\frac \pi4}^\frac\pi2 \int_{2\cos\theta}^{2\csc\theta\cot\theta} r\sin\theta\cos\theta \ dr\ d\theta$

• thanks for the answer. I will try learn much more as possible. I will also read up on polar coordinates. thanks again – user2511652 Sep 13 '17 at 18:25
• Finding the correct limits is usually the difficult part when performing double (triple) integration. I like to imagine a line that crosses my region (vertical or horizontal whichever appears will be easier) , and then try to find equations that would define the endpoints of that line. – Doug M Sep 13 '17 at 18:27
• Yeah, it just confused me on this problem too. ah, i see. so it does the trick :) thank you – user2511652 Sep 13 '17 at 18:32
• in the second part of your equation, shouldn't the limits to be from to x to squareroot of 2x ? – user2511652 Sep 14 '17 at 9:25
• As pointed out by @user2511652, the limits in the second integral are wrong. So are the limits in the first integral. The integral goes from $\operatorname{max}\{\sqrt{2x-x^2},x\}$ to $\sqrt{2x}$ for $x\in[0,\,2]$. – Tom-Tom Sep 14 '17 at 13:08

If the portion of $A$ where $0\le x\le 1$ is broken into infinite many vertical Riemann rectangles, then the height of each rectangle is $h_1=\sqrt{2x}-\sqrt{2x-x^2}$ and its width is $dx$.

If the portion of $A$ where $1\le x\le 2$ is broken into infinite many vertical Riemann rectangles, then the height of each rectangle is $h_1=\sqrt{2x}-x$ and its width is $dx$.

Thus the total area of $A$ is

\begin{align} \operatorname{area}(A) &= \int_0^1 h_1 \, dx + \int_1^2 h_2 \, dx \\ &= \int_0^1 \left( \sqrt{2x}-\sqrt{2x-x^2} \right) \, dx + \int_1^2 \left( \sqrt{2x}-x \right) \, dx \\ \end{align}

Not that integrating $\sqrt{2x-x^2}$ is going to be easier than other approaches.