Double integral change of variables to polar coordinates.

Question

Sketch the region represented in the following double integral: $I =∬ln(x^2+y^2)dxdy$

In the region:

$x:ycotβ$ to $\sqrt(a^2+y^2)$

$y:0$ to $asinβ$

where a > 0 and 0 < β < π/2.

By changing to polar coordinates or otherwise, show that I = $a^2β(ln a − 1/2)$.

So far I have found that the sketch yields a triangle inscribed in the circle of radius $a$ in the first quadrant. I am having trouble converting to polar coordinates.

Answer 1

Hint: In the polar coordinate system $x^2+y^2 \Rightarrow r^2$ and $dxdy \Rightarrow r\cdot drd\theta$. Also, we have that $y=r\cdot \sin{\theta}$ and $x=r\cdot \cos{\theta}$.

Answer 2

The condition $0\le y\le a\sin\beta$, $0\le \beta\le\pi/2$ selects a horizontal stripe of width $a\sin\beta$ above the horizontal. The condition $x\ge y\cot\beta$ means to select a triangular area below the slanted diagonal straight line $x\sin\beta=y\cos\beta$. The intersection with the horizontal stripe leaves a right-infinite stripe ($x\ge 0$) with a triangular section at the left. The line $x\sin\beta=y\cos\beta$ intersects the horizontal upper rim of the stripe $y=a\sin\beta$ at $x=a\cos\beta$.

The condition $x^2\le a^2+y^2$ selects an area left from a hyperbolic curve opening to the right with apex at $(a,0)$ and tangent approaching the diagonal $x=y$. The hyperbola intersects the upper rim of the stripe $y=a\sin\beta$ at $x=a\sqrt{1+\sin^2\beta}$.

Because $\sqrt{1+\sin^2\beta}\ge \cos \beta$, the hyperbola intersects the upper rim at larger $x$, so this splits the area into a triangular region $0\le x\le a\cos\beta$, $0\le y\le x\tan \beta$, a rectangular region $a\cos\beta <x < a$, $0\le y\le a\sin\beta$, and a region delimited by the hyperbola $a<x<\sqrt{1+\sin^2\beta}$, $0\le y\le a \sin\beta$. As usual we use polar coordinates $x=r\cos\phi$, $y=r\sin\phi$ and Jacobian $r$. It is easier to split the area into a (i) circular section with a circle of radius $a$ centered at the origin, which intersects $(a,0)$ at $\phi=0$ as well as $(a\cos\beta, a\sin\beta)$ at $\phi=\beta$, and (ii) a region outside that circle and left to the hyperbola: $$ I=\int d\phi\int rdr \ln r^2 = 2\int d\phi r dr \ln r $$ $$ =2\int_0^\beta d\phi \int_0 r dr \ln r =2(I_1+I_2). $$ Circle section: $$ I_1=\int_0^\beta d\phi \int_0^a r dr \ln r $$ $$ =\int_0^\beta d\phi [\frac{r^2}{2}\ln r-\frac14 r^2]\mid_{r=0}^a $$ $$ $$ $$ = \int_0^\beta d\phi[\frac{a^2}{2}\ln a-\frac14 a^2] = \frac{\beta}{2}a^2(\ln a-\frac12). $$ The OP's solution is apparently missing the contribution from region (ii) or having a sign error in the formula of the posted problem (ie., using a circle instead of a hyperbola).

In the second region the upper limit of $r$ is given by the intersection of the hyperola $x^2=a^2+y^2$ and the stripe $y=a\sin\beta$ at $x=a\sqrt{1+\sin^2\beta}$, which is $r^2=a^2(1+\sin^2\beta)+a^2\sin^2\beta$. $\phi$'s lower limit is set by the parabola, $x^2=r^2-y^2=a^2+y^2$, so $y=\sqrt{(r^2-a^2)/2}$, $x=\sqrt{(r^2+a^2)/2}$. $\phi$'s upper limit is given by the stripe $y=a\sin\beta$, $x=\sqrt{r^2-y^2}=\sqrt{r^2-a^2\sin^2\beta}$, $\tan\phi=y/x=a\sin\beta/\sqrt{r^2-a^2\sin^2\beta}$, $$ I_2= \int_a^{a\sqrt{1+2\sin^2\beta}} r dr \ln r \int_{\arctan[a\sin\beta/\sqrt{r^2-a^2\sin^2\beta}]}^{\arctan[\sqrt{(r^2-a^2)/2}/\sqrt{(r^2+a^2)/2}]} d\phi $$ $$ = \int_a^{a\sqrt{1+2\sin^2\beta}} r dr \ln r [ \arctan[\sqrt{(r^2-a^2)/2}/\sqrt{(r^2+a^2)/2}] - \arctan[a\sin\beta/\sqrt{r^2-a^2\sin^2\beta}] ] $$ $$ = \frac12 \int_{a^2}^{a^2(1+2\sin^2\beta)} du \ln \sqrt u [ \arctan\frac{\sqrt{u-a^2}}{\sqrt{u+a^2}} - \arctan\frac{a\sin\beta}{\sqrt{u-a^2\sin^2\beta}} ] $$ $$ = \frac14 \int_{a^2}^{a^2(1+2\sin^2\beta)} du \ln u [ \arctan\frac{\sqrt{u-a^2}}{\sqrt{u+a^2}} - \arctan\frac{a\sin\beta}{\sqrt{u-a^2\sin^2\beta}} ] . $$ By partial integration $$ K_1=\int du \ln u \arctan\frac{\sqrt{u-a^2}}{\sqrt{u+a^2}} $$ $$ = a^2\ln u \frac{1}{2u\sqrt{u^2-a^4}} -\int du u(\ln u-1) a^2\ln u \frac{1}{2u\sqrt{u^2-a^4}} $$ $$ = a^2\ln u \frac{1}{2u\sqrt{u^2-a^4}} -\frac{a^2}{2}\int du (\ln u-1) \ln u \frac{1}{\sqrt{u^2-a^4}}, $$ which I cannot solve, so the answer is incomplete at that point. By partial integration (for $c\equiv a\sin\beta$) $$ K_2=\int du \ln u \arctan\frac{c}{\sqrt{u-c^2}} $$ $$ = (\ln u-3) c\sqrt{u-c^2} +u(\ln u-1) \arctan\frac{c}{\sqrt{u-c^2}} +2c^2\arctan\frac{\sqrt{u-c^2}}{c}. $$