Show that the integral $$\int_{0}^{2a}\int_{\sqrt{2ax-x^2}}^{\sqrt{4ax-x^2}}\left(1+\frac{y^2}{x^2}\right)\ dy\ dx=\left(\pi+\frac{8}{3}\right)a^2$$ by changing the coordinates $x,y$ to $r$,$\theta$ where, $x=r\cos^2\theta , y=r\sin\theta \cos\theta$.
Solution:
$\int_{0}^{2a}\int_{\sqrt{2ax-x^2}}^{\sqrt{4ax-x^2}}(1+\frac{y^2}{x^2})\ dy\ dx
{=
{\int_{0}^{2a}\int_{0}^{\sqrt{4ax-x^2}}(1+\frac{y^2}{x^2})\ dy\ dx}
-{\int_{0}^{2a}\int_0^{\sqrt{2ax-x^2}}(1+\frac{y^2}{x^2})\ dy\ dx}
}=I_1+I_2.$
The Jacobian is $J=r cos^2(\theta)$ for the transformation $x=r\cos^2\theta , y=r\sin\theta \cos\theta$.
Hints is given in the link Evaluate the integral $\int_{0}^{2a}\int_{\sqrt{2ax-x^2}}^{\sqrt{4ax-x^2}}(1+\frac{y^2}{x^2})\ dy\ dx$ by changing the coordinates to r,$\theta$
Q1: Please help me to solve the problem. How to transform $I_1,I_2$ using the given transformation $x=r\cos^2\theta, y=r\sin\theta \cos\theta$. I evaluate concern Jacobian, but unable to understand the limits in $r-\theta$ plane.
Q2: I understand the Fig in x-y plane which is the common portion of two circles in the 1st quadrant which are $(x-a)^2+y^2=a^2$ (inner) and $(x-2a)^2+y^2=(2a)^2$. But I cannot understand the Fig in $r-\theta$ plane. Please explain the transformation $x=rcos^2θ,\,y=rsinθ\,cosθ$.
Best Answer
The following pictures explains the idea of the "non-standard" substitution.
We have two disks in the picture. The one in dark red with diameter between the origin and $(2a,0)$, and the one in light red with diameter between the origin and $(4a,0)$. Consider now a radar line moving around the origin. Fix its position to build the angle $t$ with the $Ox$-axis. It hits the two circles, the boundaries of the disks in (the origin and) the two points $A,B$. By homotethy, $2OA=OB$. Which is the distance between $O$ and $A$? It is the one leg of the right triangle with right angle in $A$ and opposite side the diameter of the first disk, of length $2a$. So $$ \begin{aligned} OA &= 2a\cos t\ ,\\ OB &= 4a\cos t\ . \end{aligned} $$ Which are the affixes / complex images of the points $A,B$. We denote these complex numbers also by $A,B$: $$ \begin{aligned} A &= 2a\cos t\cdot e^{it}=2a\cos t\cdot (\cos t+i\sin t)\ ,\\ B &= 4a\cos t\cdot e^{it}=4a\cos t\cdot (\cos t+i\sin t)\ . \end{aligned} $$ So it is natural to consider the substitution for a point $(x,y)$ running in the disk difference (light red, not dark red, only) $\Delta$: $$ (x,y)=r\cos t(\cos t, \sin t)\ . $$ Which is exactly the suggested substitution. From the picture, the domain $\Delta$ for $(x,y)$ corresponds to $(r,t)\in[2a,4a]\times [-\pi/2,\; \pi/2]$. We obtain for instance as an warming up exercise: $$ \begin{aligned} \int_\Delta\left(1 +\frac{y^2}{x^2}\right)\;dx\; dy &= \int_{2a}^{4a}dr \int_{-\pi/2}^{+\pi/2} \left(1 +\frac{\sin^2 t}{\cos^2 t}\right)\;r\cos^2 t\; dt \\ &= \left( \int_{2a}^{4a}r\;dr \right) \left( \int_{-\pi/2}^{+\pi/2} (\cos^2 t+\sin ^2 t) \;dt \right)=6\pi a^2\ . \end{aligned} $$ However, the given integral $J$ is not an integral on the full $\Delta$. Fubini gives for the posted integral the constraints $x\in[0,2a]$, and $y\ge 0$, chosen so that $y^2$ is between $2ax-x^2$ and $4ax-x^2$. This means $x\in[0,2a]$, and $x^2 -2ax+a^2+y^2\ge a^2$, and $x^2 -4ax+4a^2+y^2\le (2a)^2$. So the given integral is an integral on the intersection $x\le 2a$, $y\ge 0$ with $\Delta$. We have two cases.
As we saw in the warming up, $(1\ +\ y^2/x^2)\; dx\; dy$ gets translated into $r\; dr\; dt$. So we can compute: $$ \begin{aligned} J &= \int_{\substack{(x,y)\in\Delta\\x\le 2a\\ y\ge 0}} \left(1 +\frac{y^2}{x^2}\right)\;dx\; dy \\ &= \int_0^{\pi/4} \;dt \int_{2a}^{2a/\cos^2 t} r\; dr + \int_{\pi/4}^{\pi/2} \;dt \int_{2a}^{4a} r\; dr \\ &= \int_0^{\pi/4} 2a^2\left(\frac 1{\cos^4 t}-1\right)\;dt + \int_{\pi/4}^{\pi/2} 6a^2\;dt \\ &=\frac 16(16-3\pi)a^2 + \frac 14 6a^2 =a^2\left(\frac 83-\frac \pi 2+\frac{3\pi}2\right) = \bbox[lightblue]{\ a^2\left(\frac 83 + \pi\right)\ } \ . \end{aligned} $$ $\square$
Note: The expression $dt/\cos^4t$ was easy to integrate. Use the substitution $u$ for the tangent of $t$. Then $1+u^2=1/\cos^2 t$, and $du$ is $dt/\cos^2t$, so we have to integrate $(1+u^2)\; du$. The denominator $3$ appears...