# example of Riemann double integral

Let us determine the value of the double integral

 $\displaystyle I\;:=\;\iint_{D}\frac{dx\,dy}{(1\!+\!x^{2}\!+\!y^{2})^{2}}$ (1)

where $D$ is the triangle by the lines  $x=0$,  $y=0$  and  $x\!+\!y=1$.

Since the triangle is defined by the inequalities  $0\leqq x\leqq 1,\;\;0\leqq y\leqq 1\!-\!x$,  one can write

 $\displaystyle I$ $\displaystyle\;=\;\int_{0}^{1}\!\int_{0}^{1-x}\frac{dx\,dy}{(1\!+\!x^{2}\!+\!y% ^{2})^{2}}\;=\;\int_{0}^{1}\frac{dx}{(1\!+\!x^{2})^{2}}\int_{0}^{1-x}\frac{dy}% {\left[1+\left(\frac{y}{\sqrt{1+x^{2}}}\right)^{2}\right]^{2}}$ $\displaystyle\;=\;\int_{0}^{1}\frac{1}{(1\!+\!x^{2})^{2}}\cdot\frac{\sqrt{1\!+% \!x^{2}}}{2}\operatornamewithlimits{\Big{/}}_{\!\!\!y=0}^{\,\quad 1-x}\left(% \arctan\frac{y}{\sqrt{1\!+\!x^{2}}}+\frac{\frac{y}{\sqrt{1+x^{2}}}}{1+\frac{y^% {2}}{1+x^{2}}}\right)\,dx$ $\displaystyle\;=\;\int_{0}^{1}\left(\frac{1}{2}(1\!+\!x^{2})^{-\frac{3}{2}}% \arctan\frac{1\!-\!x}{\sqrt{1\!+\!x^{2}}}+\frac{1\!-\!x}{(1\!-\!x\!+\!x^{2})(1% \!+\!x^{2})}\right)\,dx.$

Some appropriate substitution (http://planetmath.org/ChangeOfVariablesInIntegralOnMathbbRn)

 $x\;:=\;x(u,\,v),\quad y\;:=\;y(u,\,v)$

directly to the form (1) could offer a better is

 $\displaystyle\iint_{D}f(x,\,y)\,dx\,dy\>=\;\iint_{\Delta}\!f(x(u,\,v),\,y(u,\,% v))\left|\frac{\partial(x,\,y)}{\partial(u,\,v)}\right|\,du\,dv.$ (2)

What kind a change of variables would be good?  One idea were to use some “natural substitution”, i.e. such one that would give constant limits (http://planetmath.org/DefiniteIntegral).  For example, the equations

 $x\!+\!y\;:=\;u,\quad\frac{y}{x}\;:=\;v,$

map the triangular domain (http://planetmath.org/Domain2) $D$ to the “rectangle  $\Delta\!:\;\;0\leqq u\leqq 1,\quad 0\leqq v<\infty.$
 $\frac{\partial(x,\,y)}{\partial(u,\,v)}\;=\;\frac{u\!+\!v^{2}}{(v\!+\!1)^{3}}.$

By (2), we have

 $I\;=\;\int_{0}^{1}\!\int_{0}^{\infty}\!\frac{(v\!+\!1)^{4}}{u^{2}\!+\!2v^{2}\!% +\!2v\!+\!1}\!\cdot\!\frac{u\!+\!v^{2}}{(v\!+\!1)^{3}}\,du\,dv\;=\;\int_{0}^{% \infty}\!(v\!+\!1)\,dv\int_{0}^{1}\frac{u\!+\!v^{2}}{u^{2}\!+\!2v^{2}\!+\!2v\!% +\!1}\,du.$

But here after integrating with respect to $u$, one obtains a difficult single integral  .  Thus, when the , the integrand may become awkward.

A second idea would be to try to make the integrand simpler.  For this end, the transition to the polar coordinates

 $x\;:=\;r\cos\varphi,\quad y\;:=\;r\sin\varphi$

in (1) is more suitable.  We have

 $\frac{\partial(x,\,y)}{\partial(r,\,\varphi)}\;=\;\left|\begin{matrix}\cos% \varphi&-r\sin\varphi\\ \sin\varphi&r\cos\varphi\end{matrix}\right|\;\equiv\;r.$

The Pythagorean theorem   gives the equation  $r^{2}\,=\,x^{2}\!+\!y^{2}\,=\,(r\cos\varphi)^{2}+(1-r\cos\varphi)^{2}$,  i.e.

 $r^{2}\cos 2\varphi-2r\cos\varphi+1\;=\;0,$
 $r\;=\;\frac{2\cos\varphi\pm\sqrt{4\cos^{2}\varphi-4\cos 2\varphi}}{2\cos 2% \varphi}\;=\;\frac{\cos\varphi\pm\sin\varphi}{\cos^{2}\varphi-\sin^{2}\varphi};$

this is $\displaystyle\frac{1}{\cos\varphi+\sin\varphi}$, since the “+” alternative can be excluded by choosing e.g.  $\varphi=\frac{\pi}{2}$.  Thus

 $\Delta\!:\quad 0\leqq\varphi\leqq\frac{\pi}{2},\quad 0\leqq r\leqq\frac{1}{% \cos\varphi+\sin\varphi}$

and

 $I\;=\;\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\!\int_{0}^{\frac{1}{\cos\varphi+\sin% \varphi}}\frac{2r\,dr}{(1\!+\!r^{2})^{2}}\,d\varphi\;=\;\frac{1}{2}\int_{0}^{% \frac{\pi}{2}}\frac{d\varphi}{2+\sin 2\varphi}.$

Here, the http://planetmath.org/node/9380Weierstrass substitution  $\tan\varphi\,:=\,t$  easily yields the final result

 $\displaystyle I\;=\;\frac{2\pi\sqrt{3}}{9}.$ (3)
Title example of Riemann double integral ExampleOfRiemannDoubleIntegral 2013-03-22 19:12:22 2013-03-22 19:12:22 pahio (2872) pahio (2872) 11 pahio (2872) Example msc 26A42 msc 28-00 SubstitutionNotation ChangeOfVariablesInIntegralOnMathbbRn ExampleOfRiemannTripleIntegral