# Double integral over a region

Given $f(x,y)=\displaystyle\frac{x^2}{x^2+y^2}$ and $D=\{(x,y) : 0 \leq x \leq 1, x^2 \leq y \leq 2-x^2\}$ i have to solve $\displaystyle\int\displaystyle\int_Df(x,y)dA$.

Here's my try:

(1) Changing variables

$x = \sqrt{v-u}$, $y= v+u$.

(1.1) Since $0 \leq x \leq 1$, then $0 \leq v-u \leq 1 \rightarrow u \leq v \leq 1+u$

(1.2) Since $x^2 \leq y \leq 2-x^2$, then $v-u \leq v+u \leq 2-v+u \rightarrow -u \leq u \rightarrow 0\leq u$ and $v \leq 2-v \rightarrow v \leq 1$

(1.3) It seems that now i should integrate over $S = \{(u,v) : 0\leq u \leq v \leq 1 \}$ (the upper triangle in $[0,1]\times[0,1]$ ?), so i may as well put $S = \{(u,v) : 0 \leq v \leq 1, 0 \leq u \leq v \}$.

(2) Alright, what do i need to calculate the integral?

(2.1) First, i should calculate the Jacobian

$\displaystyle\frac{\partial(x,y)}{\partial(u,v)} = \left| \begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \end{array} \right| = \left| \begin{array}{cc} -\frac{1}{2\sqrt{v-u}} & \frac{1}{2\sqrt{v-u}} \\ 1 & 1 \\ \end{array} \right| = -\frac{1}{\sqrt{v-u}}$

(2.2) Then i have to solve $\displaystyle\int_0^1\displaystyle\int_0^v \frac{v-u}{(v-u)+(v^2+2uv+u^2)}\bigg(-\frac{1}{\sqrt{v-u}} \bigg)dvdu$ $=-\displaystyle\int_0^1\displaystyle\int_0^v \frac{\sqrt{v-u}}{v^2+v(1+2u) + (u^2-u) }dvdu$

(2.3) Well, here i'm stuck.I've been thinking about taking $z = \sqrt{v-u}$ and then $dz = \displaystyle\frac{1}{2\sqrt{v-u}}dv$ wich means $dv = 2zdz$, this would lead to an integral of the form $2\displaystyle\int\displaystyle\int \frac{z^2}{z^2+z(1+4u)+ 4u^2 }dvdu = 2\displaystyle\int\displaystyle\int \frac{z^2}{(z+(\frac{1}{2}+2u))^2-2u }dvdu$ and if i put $w = z+(\frac{1}{2}+2u)$ i'll have $2\displaystyle\int\displaystyle\int \frac{(w-\frac{1}{2}-2u)^2}{w^2-2u }dwdu$ but it seems that the last one will lead to some ugly shaped solution and i would have a hard time getting the final answer. What would be the best way to solve this?

• Are you supposed to use that change of variables or you did by your own idea? I mean is that a given hint of your text or not? Commented Jul 10, 2013 at 5:37
• It was a given hint.
– Cure
Commented Jul 10, 2013 at 5:40
• For what it's worth, the answer should be $1+\frac{\pi }{16}-\frac{\sqrt{7}}{8} \log \left(8+3 \sqrt{7}\right)$. Commented Jan 8, 2014 at 21:01
• @heropup: which is equal to $1+\frac\pi{16}+\frac{\sqrt7}{8}\log(8-3\sqrt7)$
– robjohn
Commented Jan 9, 2014 at 15:30

\begin{align} &\int_0^1\int_{x^2}^{2-x^2}\frac{x^2}{x^2+y^2}\,\mathrm{d}y\,\mathrm{d}x\tag{1}\\ &=\int_0^1\int_{x}^{\frac2x-x}\frac{x}{1+y^2}\,\mathrm{d}y\,\mathrm{d}x\tag{2}\\ &=\int_0^1x\left(\arctan\left(\frac2x-x\right)-\arctan(x)\right)\,\mathrm{d}x\tag{3}\\ &=\int_0^1x\arctan\left(\frac{2(1-x^2)}{x(3-x^2)}\right)\,\mathrm{d}x\tag{4}\\ &=\int_0^1\frac{x^2}{2}\frac{\frac{2(3+x^4)}{x^2(3-x^2)^2}}{1+\left(\frac{2(1-x^2)}{x(3-x^2)}\right)^2}\,\mathrm{d}x\tag{5}\\ &=\int_0^1\frac{x^2(3+x^4)}{x^2(3-x^2)^2+4(1-x^2)^2}\,\mathrm{d}x\tag{6}\\ &=\int_0^1\left(1-\frac1{2(1+x^2)}+\frac{5x^2-4}{2(x^4-3x^2+4)}\right)\,\mathrm{d}x\tag{7}\\ &=1-\frac\pi8+\int_0^1\frac14\left(\frac{\sqrt7x-2}{x^2-\sqrt7x+2}-\frac{\sqrt7x+2}{x^2+\sqrt7x+2}\right)\,\mathrm{d}x\tag{8}\\ &=1-\frac\pi8+\int_0^1\frac{\sqrt7}{8}\left(\frac{2x-\sqrt7}{x^2-\sqrt7x+2}-\frac{2x+\sqrt7}{x^2+\sqrt7x+2}\right)\,\mathrm{d}x\\ &+\int_0^1\frac32\left(\frac1{(2x-\sqrt7)^2+1}+\frac1{(2x+\sqrt7)^2+1}\right)\,\mathrm{d}x\tag{9}\\ &=1-\frac\pi8+\frac{\sqrt7}{8}\log\left(\frac{3-\sqrt7}{3+\sqrt7}\right)\\ &+\frac34(\arctan(2-\sqrt7)+\arctan(2+\sqrt7))\tag{10}\\ &=1+\frac\pi{16}+\frac{\sqrt7}{8}\log(8-3\sqrt7)\tag{11} \end{align} Justification:
$\ \;(1)$: get limits from the problem
$\ \;(2)$: change variables $y\mapsto xy$
$\ \;(3)$: integrate in $y$
$\ \;(4)$: combine arctans
$\ \;(5)$: integrate by parts
$\ \;(6)$: algebra
$\ \;(7)$: partial fractions
$\ \;(8)$: integrate and partial fractions
$\ \;(9)$: separate into easily integrable pieces
$(10)$: integrate
$(11)$: simplify: $\quad\frac{3-\sqrt7}{3+\sqrt7}=8-3\sqrt7\quad$ and $\quad\arctan(2-\sqrt7)+\arctan(2+\sqrt7)=\frac\pi4$

• Isn't the change of variables in $(2)$ $y\mapsto \frac{y}{x}$ instead of $y\mapsto xy$? Commented Jan 9, 2014 at 13:12
• @AméricoTavares: I am replacing all occurrences of $y$ by $xy$. The limits have to undergo the opposite transform to counter.
– robjohn
Commented Jan 9, 2014 at 14:22
• French town on the Cote d'Azure ;) Commented Jan 9, 2014 at 15:05
• @DanielFischer: Cannes? :-) Thanks
– robjohn
Commented Jan 9, 2014 at 15:27
• Sorry, I was confused by a different notation, the u-substitution $u=y/x$. Commented Jan 9, 2014 at 17:39

$$I = \int_0^1\int_{x^2}^{2-x^2} \frac{x^2}{x^2+y^2}\mathrm{d}y\,\mathrm{d}x = \frac{\pi}{16}+ 1 + \frac{\sqrt{7}}{8}\log\left(8-3\sqrt{7}\right)$$

I am sorry but I can not help you with your substitution. My guess is that the absolute value of the integral diverges, hence such a substitution is not legal. -

However the integral can be evaluated straight forward, with a bit of an hassle.

Note that the integrand has quite nice symmetry around the origo, however the domain is not centered about origo so this can not be used. Now the drawing gives the integral in the header, and the innermost integral can be calculated as follows $$\int_{x^2}^{2-x^2} \frac{x^2}{x^2+y^2}\mathrm{d}y = \left[ x \arctan\left(\frac{y}{x}\right) \right]_{x^2}^{2-x^2} = -x \arctan x - x \arctan \left( \frac{2-x^2}{x}\right)$$ We can now write the double integral as an sum of two integrals $$\iint_D \frac{x^2}{x^2+y^2} \mathrm{d}A = \int_1^0 x \arctan x\,\mathrm{d}x + \int_1^0 x \arctan \left( \frac{2-x^2}{x}\right)\,\mathrm{d}x$$ For simplicity label integrals $J$ and $K$, the indefinite integral of $J$ can be evaluated by parts $$\int x \arctan x\,\mathrm{d}x = \left( \frac{x}{2}+\frac{1}2\right)\arctan x - \int \frac{x/2+1/2}{x^2+1}$$ where $u = \arctan x$ and $v = (x^2-1)/2$ were choosen. The last integral is trivial (why?) to show, so the definite integral becomes $$J = \int_1^0 x \arctan x\,\mathrm{d}x = \frac{1}{2} - \frac{\pi}{4}$$ The last integral can be dealt with in a similar fashion. See that $$K = \int_1^0 x \arctan \left( \frac{2-x^2}{x}\right)\,\mathrm{d}x = \frac{\pi}{8} + \frac{1}{2} \int_0^1 \frac{x^2(x^2+2)}{4-3x^2+x^4}\,\mathrm{d}x$$ Where again integration by parts was used this time with $v = x^2$. This is unfortunately where my bag of clever tricks run out, by some clever manipulation note that $$\frac{x^2(x^2+2)}{4-3x^2+x^4} = 1 - \frac{4-5x^2}{4-3x^2+x^4} = 1 - \frac{1}{2} \left( \frac{x-\sqrt{7}}{x^2-\sqrt{7}x+2} + \frac{x+\sqrt{7}}{x^2+\sqrt{7}x+2} \right)$$ Where the clever factorization $$-3x^2+4+x^4 = (x^2+\sqrt{7}x+2)(x^2-\sqrt{7}x+2)$$ was used, along with partial fractions. The first integral is trivial over $1$ and the last integral can be shown to be $$\int \frac{2 \pm \sqrt{7}}{x^2\pm\sqrt{7}x+2} = \pm \frac{\sqrt{7}}{4} \log\left( x^2 \pm \sqrt{7}x+2\right) - \frac{3}{2} \arctan\left(2x \pm \sqrt{7}\right)$$ This is again simply splitting the integrand once again into two simpler integrals. Putting all of this together the last integral can be expressed as \begin{align*} & = \frac{1}{2} \int_0^1 \frac{x^2(x^2+2)}{4-3x^2+x^4}\,\mathrm{d}x \\ & = \frac{1}{2} - \frac{1}{4} \int_0^1 \frac{x-\sqrt{7}}{x^2-\sqrt{7}x+2} + \frac{x+\sqrt{7}}{x^2+\sqrt{7}x+2} \\ & = \frac{1}{2} - \frac{\sqrt{7}}{8}\log(3+\sqrt{7}) + \frac{3}{4} \arctan(\sqrt{7}+2) \\ & \phantom{ = \frac{1}{2}} + \frac{\sqrt{7}}{8}\log(3-\sqrt{7}) - \frac{3}{4} \arctan(\sqrt{7}-2) \end{align*} Which can be simplified down to $$\frac{1}{2} + \frac{3\pi}{16} - \frac{\sqrt{7}}{4} \log(8 + 3\sqrt{7})$$ By combining the arctan and the logarithmic terms. Finaly the integral can be written as \begin{align*} \iint_D \frac{x^2}{x^2+y^2} \mathrm{d}A & = \int_1^0 x \arctan x\,\mathrm{d}x + \int_1^0 x \arctan \left( \frac{2-x^2}{x}\right)\,\mathrm{d}x \\ & = \left( \frac{1}{2} - \frac{\pi}{4} \right) + \left( \frac{\pi}{8} + \frac{1}{2} + \frac{3\pi}{16} - \frac{\sqrt{7}}{4} \log(8 + 3\sqrt{7})\right)\\ & = \frac{\pi}{16}+ 1 + \frac{\sqrt{7}}{8}\log\left(8-3\sqrt{7}\right) \end{align*} Which is what one wanted to show. Where it was used that $1/(8+3\sqrt{7})=8-3\sqrt{7}$.

The might be some slight typing errors in the above calculations and most of the grunt work is left for you to show. But this is at least one outline of a general solution.

• This was essentially my solution, but I did not post it because it did not use a bivariate transformation as requested. Commented Jan 8, 2014 at 23:27
• I'm sorry, how do you get the "clever factorization" and how do you know how to manipulate the expression? Be patient, I'm still learning :-) Commented Jan 9, 2014 at 15:18
• @twinprime: since the $x^3$ coefficient of $x^4-3x^2+4$ is $0$ and the $x^0$ coefficient is $4$, we know that $$x^4-3x^2+4=(x^2+ax+b)(x^2-ax+4/b)$$ The rest is pretty straightforward.
– robjohn
Commented Jan 9, 2014 at 20:53
• @robjohn Thanks a lot! Commented Jan 10, 2014 at 10:16

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\begin{align} &\int_{D}\fermi\pars{x,y}\,\dd x\,\dd y = \int_{0}^{1}\dd x\,x^{2} \int_{-\infty}^{\infty}{\Theta\pars{y - x^{2}}\Theta\pars{2 - x^{2} - y} \over x^{2} + y^{2}}\,\dd y = \int_{0}^{1}\dd x\,x^{2}\int_{x^{2}}^{2 - x^{2}}{\dd y \over x^{2} + y^{2}} \\[3mm]&= \int_{0}^{1}\dd x\,x\int_{x}^{2/x - x}{\dd y \over 1 + y^{2}} = \int_{0}^{1}x\bracks{\arctan\pars{{2 \over x} - x} - \arctan\pars{x}}\,\dd x \\[3mm]&= \left.\half\,x^{2}\bracks{\arctan\pars{{2 \over x} - x} - \arctan\pars{x}} \right\vert_{0}^{1} - \int_{0}^{1}\half\,x^{2}\bracks{% {-2/x^{2} - 1 \over \pars{2/x - x}^{2} + 1} - {1 \over x^{2} + 1}}\,\dd x \\[3mm]&= \half\int_{0}^{1}\bracks{% {x^{2}\pars{2 + x^{2}} \over \pars{2 - x^{2}}^{2} + x^{2}} + {x^{2} \over x^{2} + 1}} \,\dd x = \half\int_{0}^{1}\bracks{% {x^{4} + 2x^{2} \over x^{4} - 3x^{2} + 4} + 1 - {1 \over x^{2} + 1}} \,\dd x \\[3mm]&= \half\int_{0}^{1}\bracks{% 1 + {5x^{2} - 4\over x^{4} - 3x^{2} + 4} + 1 - {1 \over x^{2} + 1}} \,\dd x = \color{#0000ff}{\large 1 - {\pi \over 8} + \half\underbrace{\int_{0}^{1}{5x^{2} - 4 \over x^{4} - 3x^{2} + 4}\,\dd x} _{\ds{\approx\ -0.653199}}} \end{align}

The last integral can be easily calculated by using a 'partial fraction technique'.

• What does $\Theta$ mean?
– Cure
Commented Jan 13, 2014 at 13:04
• @Dante $\large\Theta:{\mathbb R}\verb*\*\left\lbrace 0\right\rbrace \to {\mathbb R}$ such that $\large\Theta\left(x\right) = 0$ when $\large x < 0$ and $\large = 1$ when $\large x > 0$. It's the Heaviside Step function. See --- > en.wikipedia.org/wiki/Heaviside_step_function Commented Jan 14, 2014 at 6:23

The result differs from what expected by heropup but I cannot see the mistake.

The domain is symmetric with respect to $y=1$.

$$\iint_{D} f(x,y) dxdy= 2\int_{0}^{1}\int_{x^2}^{1}\frac{1}{1+\frac{y^2}{x^2}}dydx$$

$$= 2\int_{0}^{1} [t \arctan \frac{y}{t}]_{t^2}^{1} dt=2\int_{0}^{1} t \arctan \frac{1}{t} dt -2\int_{0}^{1} t \arctan t \ dt$$

$$2 \int_{0}^{1} t \arctan \frac{1}{t} dt= [t^2 \arctan \frac{1}{t} +t -\arctan t]_{0}^{1}= \frac{\pi}{4} - \lim_{t \rightarrow 0^+} t^2 \arctan \frac{1}{t}+ 1 -\frac{\pi}{4}= 1$$

$$2\int_{0}^{1} t \arctan t \ dt= [(1+t^2) \arctan t- t]_{0}^{1}=\frac{\pi}{2}- 1$$

• The integrand isn't symmetric in $y$ about the line $y = 1$. Thus you can't split and multiply by $2$. Commented Jan 8, 2014 at 23:02