# Double integral change of variables to polar coordinates.

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.

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 $$x=r\cdot \cos{\theta}$$.

• My problem is finding the limits for $r$ and theta.
– mat
Feb 3, 2019 at 12:09

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 $$a\cos\beta , $$0\le y\le a\sin\beta$$, and a region delimited by the hyperbola $$a, $$0\le y\le a \sin\beta$$. 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 $$c\equiv a\sin\beta$$) $$K_2=\int du \ln u \arctan\frac{c}{\sqrt{u-c^2}}$$ $$= (\ln u-3) c\sqrt{u-c^2} +u(\ln u-1) \arctan\frac{c}{\sqrt{u-c^2}} +2c^2\arctan\frac{\sqrt{u-c^2}}{c}.$$