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I am trying to change the order of integration of the following double integral $$\int _0^{2a} dx\int _{\sqrt{2ax-x^2}}^{\sqrt{4ax}} f(x,y) \ dy .$$ (a>0)

I've sketched the domain but am struggling to change it from type 1 domain to type 2 domain. Any help?

enter image description here

I've attached an image of the geometry of the domain and I've shaded it. (Well the image shows the case when a=1.)

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    I haven't encountered the terms "Type 1 domain" and "Type 2 domain"... would you mind defining these. Also, perhaps sharing a photo of your sketch would help us evaluate your work. – 2017-02-21

3 Answers 3

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Let $y=\sqrt{4ax}$ ($0\le x\le 2a$) and then $x=\frac{y^2}{4a}$. Let $y=\sqrt{2ax-x^2}$ ($0\le x\le 2a$) and then $$ x=a\pm\sqrt{a^2-y^2}. $$ Thus the area is divided into three parts:

(1). $\frac{y^2}{4a}\le x\le a-\sqrt{a^2-y^2}, 0\le y\le a$;

(2). $a+\sqrt{a^2-y^2}\le x\le 2a, 0\le y\le a$;

(3). $\frac{y^2}{4a}\le x\le 2a, a\le y\le 2\sqrt 2a$.

So \begin{eqnarray} &&\int _0^{2a} dx\int _{\sqrt{2ax-x^2}}^{\sqrt{4ax}} f(x,y) \ dy\\ &=&\int _0^{a} dy\int _{\frac{y^2}{4a}}^{a-\sqrt{a^2-y^2}} f(x,y) \ dx+\int _0^{a} dy\int^{2a}_{a+\sqrt{a^2-y^2}} f(x,y)dx+\int_a^{2\sqrt2a} dy\int^{2a}_{\frac{y^2}{4a}} f(x,y) \ dx. \end{eqnarray}

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    Is there no way of combining his into a single integral. This is what I got buy just the way the question is worded would seem like you can. But it doesn't look like you can. Guess this is what the question meant if not. – 2017-02-22
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    @Ben, I doubt it is possible. – 2017-02-22
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    Did you make a mistake in your very last integral. Should the upper y limit be $2\sqrt 2 a $ not $2a$? – 2017-02-22
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    @Ben, Yes, you are right. – 2017-02-22
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I hope it can help you

enter image description here

$1)\,\,\,\,\,\,\,\,\,\,\,\,\int_{y=a}^{y=\sqrt{8}a} (\int_{x=\frac{y^2}{4a}}^{x=2a}f(x,y)\, dx)\,dy$

$\color{lime}{2)}\,\,\,\,\int_{y=0}^{y=a} (\int_{x=\frac{y^2}{4a}}^{x=a-\sqrt{a^2-y^2}}f(x,y)\, dx)\,dy$

$\color{#f8f}{3)}\,\,\,\,\,\,\,\,\int_{y=0}^{y=a} (\int_{x=a+\sqrt{a^2-y^2}}^{x=2a}f(x,y)\, dx)\,dy$

$\int _0^{2a} dx\int _{\sqrt{2ax-x^2}}^{\sqrt{4ax}} f(x,y) \ dy$

$=\int_{y=a}^{y=\sqrt{8}a} (\int_{x=\frac{y^2}{4a}}^{x=2a}f(x,y)\, dx)\,dy+\int_{y=0}^{y=a} (\int_{x=\frac{y^2}{4a}}^{x=a-\sqrt{a^2-y^2}}f(x,y)\, dx)\,dy+\int_{y=0}^{y=a} (\int_{x=a+\sqrt{a^2-y^2}}^{x=2a}f(x,y)\, dx)\,dy$

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Here is what the domain looks like with $a=1$:

enter image description here

You cannot switch the order of integration: the bounds are different when for example $y=1/2$ and $y=2$.

What is the $f(x,y)$ explicitly?

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    I don't have $f(x,y)$ explicitly - only know that it is continuous. I have a question asking me to sketch the domain and to use the sketch to change the order of integration - I'm aware that you can't just switch the order. – 2017-02-21