Calculate $\iint\frac{dxdy}{(1+x^2+y^2)^2}$ over a triangle

Question

Calculate
$$\iint\frac{dxdy}{(1+x^2+y^2)^2}$$ over the triangle $(0,0)$, $(2,0)$, $(1,\sqrt{3})$.

So I tried changing to polar coordinates and I know that the angle is between $0$ and $\frac{\pi}{3}$ but I couldn't figure how to set the radius because it depends on the angle.

Answer 1

Yes, using polar coordinates is a good idea. We find $$\iint_T\frac{dxdy}{(1+x^2+y^2)^2}=\int_{\theta=0}^{\pi/3}d\theta\int_{\rho=0}^{f(\theta)}\frac{\rho d\rho}{(1+\rho^2)^2} =-\frac{1}{2}\int_{\theta=0}^{\pi/3}\left[\frac{1}{1+\rho^2 }\right]_{\rho=0}^{f(\theta)}\,d\theta$$ where the upperbound $\rho=f(\theta)$ can be obtained from the line joining the points $(1,\sqrt{3})$ and $(2,0)$, $$\rho\sin(\theta)=y=\sqrt{3}(2-x)=\sqrt{3}(2-\rho\cos(\theta))$$ and therefore $$\rho=f(\theta)=\frac{2\sqrt{3}}{\sin(\theta)+\sqrt{3}\cos(\theta)} =\frac{\sqrt{3}}{\sin(\theta+\pi/3)}.$$ Can you take it from here?

Answer 2

As you wrote, $\theta$ can take any value from $0$ to $\frac\pi3$. For each such $\theta$, $\rho$ can take any value from $0$ to $r$, where $r$ is such that $(r\cos\theta,r\sin\theta)$ belongs to the segment joining $(2,0)$ to $\left(1\,\sqrt3\right)$. This segment is part of the line $y=2\sqrt3-\sqrt3x$. So, you solve the equation$$r\sin\theta=2\sqrt3-\sqrt3r\cos\theta$$and you will get that$$r=\frac{\sqrt3}{\sin\left(\theta+\frac\pi3\right)}.$$So, your integral is equal to$$\int_0^{\pi/3}\int_0^{\sqrt3/\sin\left(\theta+\pi/3\right)}\frac\rho{(1+\rho^2)^2}\,\mathrm d\rho\,\mathrm d\theta.$$

Answer 3

Let us have a solution based on an alternative idea. We consider on the triangle $T$ the one-form $$ \omega=\frac 12\cdot \frac {x\; dy - y\; dx}{1+x^2+y^2}\ . $$ Then $$ \begin{aligned} 2d\omega &= \frac\partial{\partial x}\left(\frac x{1+x^2+y^2}\right) dx\wedge dy + \frac\partial{\partial x}\left(\frac {-y}{1+x^2+y^2}\right) dy\wedge dx \\ &=\frac 2{(1+x^2+y^2)^2}\; dx\wedge dy\ . \end{aligned} $$ We apply Stokes now. We parametrize the boundary of $T$ using the maps

• $t\to(t,0)$ for $t$ from $0$ to $2$, and there will be no contribution because of $y=0$,
• $t\to(2-t,t\sqrt 3)$ for $t$ from $0$ to $1$,
• $t\to(t,t\sqrt 3)$ for $t$ from $1$ to $0$, and there will be no contribution, because $x\; dy-y\; dx$ becomes $t\;(t\sqrt 3)'\; dt -(t\sqrt 3)\; t'\; dt$,

and compute explicitly: $$ \begin{aligned} &\int_{\partial T} \frac {x\;dy}{1+x^2+y^2} = \int_0^2\frac {t\cdot 0'\; dt}{1+t^2+0^2} \\ &\qquad\qquad\qquad + \int_0^1\frac {(2-t)\; (t\sqrt 3)'\; dt}{1+(2-t)^2+3t^2} + \int_1^0\frac {t\; (t\sqrt 3)'\; dt}{1+t^2+3t^2} \ , \\[3mm] &\int_{\partial T} \frac {y\;dx}{1+x^2+y^2} = \int_0^2\frac {0\cdot t'\; dt}{1+t^2+0^2} \\ &\qquad\qquad\qquad + \int_0^1\frac {t\sqrt 3\; (2-t)'\; dt}{1+(2-t)^2+3t^2} + \int_1^0\frac {t\sqrt 3\; t'\; dt}{1+t^2+3t^2} \ , \\[3mm] &\iint_T\frac {dx\; dy}{(1+x^2+y^2)^2}= \iint_T d\omega \\ &\qquad= \int_{\partial T} \omega \\ &\qquad =\frac 12\int_0^1 \frac {(2-t)\cdot(t\sqrt 3)'-(t\sqrt 3)\; (2-t)'}{1+(2-t)^2+3t^2} \; dt \\ &\qquad=\frac {\sqrt 3}2\int_0^1 \frac {(2-t)+t}{(2t-1)^2+2^2} \; dt =\color{blue}{\frac {\sqrt 3}2\arctan\frac 12}\ . \end{aligned} $$

(Note: All details are included for didactical reasons, now please remove all details to have a two lines computation, given the formula for $d\omega$ and the cancellations on the first and third line path parametrizing $\partial T$.)

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A sage numerical check using Fubini...

sage: var('x,y');
sage: f = 1 / (1 + x^2 + y^2)^2
sage: assume(x>0)
sage: assume(x<2)
sage: J1 = integral( integral(f, y, 0,    x *sqrt(3)), x, 0, 1)
sage: J2 = integral( integral(f, y, 0, (2-x)*sqrt(3)), x, 1, 2)

sage: (J1+J2).n()
0.401530607798613
sage: ( sqrt(3)/2*atan(1/2) ).n()
0.401530607798613

Answer 4

Here is an alternative to Robert's nice way to find how $r(\theta)$ depends on $\theta$.

Let $A(0,0)$, $B(2,0)$ and $C(1,\sqrt{3})$ be the three vertices of the triangle. Imagin, or see the picture below, a ray starting from $A$ and intersects with the side $CB$ at $D$. Suppose the angle $\angle DAB=\theta$. You want to find the length of $AD$ in terms of $\theta$. You can apply the law of sines here to the triangle $ABD$: $$ \frac{\sin(\pi-\theta-\pi/3)}{2}=\frac{\sin (\pi/3)}{f(\theta)}, $$ Since $\sin(\pi-a)=\sin(a)$, and $\sin(\pi/2) = \sqrt{3}/2$, it follows that $$ f(\theta) = \frac{\sqrt{3}}{\sin(\theta+\pi/3)}\;. $$

Let us continue the calculations that are done in Robert's answer:

$$\iint_T\frac{dxdy}{(1+x^2+y^2)^2} =\int_{0}^{\pi/3}\left(\int_{0}^{f(\theta)}\frac{\rho d\rho}{(1+\rho^2)^2}\right)\;d\theta =-\frac{1}{2}\int_{0}^{\pi/3} \left[ \frac{1}{1+\rho^2 } \right]_{\rho=0}^{\rho=f(\theta)}\,d\theta=:\frac12 I\;. $$ where $$ I=-\int_{0}^{\pi/3} \frac{1}{1+f^2(\theta) }-1\,d\theta =\int_{0}^{\pi/3} \frac{f^2(\theta)}{1+f^2(\theta) }\,d\theta =\int_{0}^{\pi/3}\frac{3}{3+\sin^2(\theta+\pi/3)}\;d\theta=:3J\;. $$ Up to this point, you can go directly to the general method of Weierstrass substitution. But in this specific case, some trig substitutions makes the integral easier.

Observe that $\cos(\pi/2-a)=\sin(a)$. So $$ \begin{align} J &= \int_{0}^{\pi / 3} \frac{1}{\cos ^{2}\left(\frac{\pi}{6}-x\right)+3} dx = \int_{-\pi / 6}^{\pi / 6} \frac{1}{\cos ^{2}(u)+3} du = \int_{-\pi / 6}^{\pi / 6} \frac{\sec ^{2}(u)}{3 \sec ^{2}(u)+1} du\\ &= \int_{-\pi / 6}^{\pi / 6} \frac{\sec ^{2}(u)}{3 \tan ^{2}(u)+4} du \quad (\sec^2u = \tan^2u+1)\\ &= \int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{1}{3 s^{2}+4} ds \quad (d(\tan u)=\sec^2u\;du)\\ &= \frac{1}{4} \int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{1}{\frac{3 s^{2}}{4}+1} ds =\frac{\sqrt{3}}{6} \int_{-1 / 2}^{1 / 2} \frac{1}{p^{2}+1} d p\\ &=\frac{\sqrt{3}}{3} \tan ^{-1}\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{3} \cot ^{-1}(2)\;. \end{align} $$ So the result is $$ \frac32J = \frac{\sqrt{3}}{2} \cot ^{-1}(2)\;. $$