I have to evaluate $\displaystyle\iint_Df(x,y)\;dxdy$ for $f(x,y) = \sqrt{4x^2-y^2}$ with $D = \{(x,y)\in\mathbb{R}^2: 0\leq x \leq 1, 0\leq y \leq x\}$.
It seems that i can't solve for $\displaystyle\int_0^1 \displaystyle\int_0 ^x\sqrt{4x^2-y^2} dydx$ but working with iterated integrals I could solve for $\displaystyle\int_0^1 \displaystyle\int_{\color{red}{1}} ^{\color{red}{y}}\sqrt{4x^2-y^2} dxdy$.
A small sketch of what i did until now:
$\displaystyle\int_{\color{red}{1}} ^{\color{red}{y}}\sqrt{4x^2-y^2} dx = $
$\displaystyle\int_{\color{red}{1}} ^{\color{red}{y}}\displaystyle\frac{4x^2}{\sqrt{4x^2-y^2}}dx – \displaystyle\int_{\color{red}{1}} ^{\color{red}{y}}\displaystyle\frac{y^2}{\sqrt{4x^2-y^2}}dx = $
$
4\left(\displaystyle\frac{x^2}{\sqrt{4x^2-y^2}} -\displaystyle\frac{1}{4}\displaystyle\int \sqrt{4x^2-y^2}dx\right)-y^2\displaystyle\frac{y}{2|y|}\arcsin\left(\displaystyle\frac{2}{y}x \right) $
$\implies \displaystyle\int_{\color{red}{1}} ^{\color{red}{y}}\sqrt{4x^2-y^2} dx = $
$\displaystyle\frac{1}{2}\left(\displaystyle\frac{4x^2}{\sqrt{4x^2-y^2}}-y^2\displaystyle\frac{y}{2|y|}\arcsin\left(\displaystyle\frac{2}{y}x \right) \right)$.
[wrong]
And it seems that after I evaluate the last one i get $\displaystyle\frac{y^2}{\sqrt{3y^2}} = \displaystyle\frac{y^2}{\sqrt{3}|y|}$
Then if I solve the integral of the expresion above for x I get $\displaystyle\frac{1}{2\sqrt{3}}$ if $y>0$ and $\displaystyle\frac{-1}{2\sqrt{3}}$ if $y<0$.[/wrong]
I'm almost sure i made a mistake somewhere. Can someone find any errors?
Best Answer
I am posting this about finding the proper limits not to solve the definite integrals. We have $$D = \{(x,y)\in\mathbb{R}^2: 0\leq x \leq 1, 0\leq y \leq x\}$$ and it seems you want to change the order of the integrals. OK! As you might already do, the following plot makes the region of $D$ clear.
Changing the order of integrals makes $y$ to be the outer variable and $x$ to be the inner one. As you see $D$, I made a line parallel to $x$ -axe intersecting the region. Now if we read the points (follow the arrows), the line enters in with $x=y$ and enters out with $x=1$, so we have $$x|_{y}^1,~~y|_{0}^1$$ It seems switching the limits for $y$ is needed in your second integrals.