Double Integral Area Calculation

Question

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?

Please help me by pointing me to the right direction

Answer 1

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

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

Answer 2

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?

Answer 3

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.