Possible shortcut to integrate the double integral $\iint \frac{1}{x^2+y^2-2x+6y+20} dA$ over a region

Question

This problem was on math homework. Integrate the following double integral over the region D = {($x, y$) | $x^2+y^2 \le 2x-6y+15$} $$\iint \frac{1}{x^2+y^2-2x+6y+20}dA$$ This problem was supposed to be solved with a change of variables to polar coordinates, but I realized that the domain and the integral looked very similar. So, I used the substitution $u = x^2+y^2-2x+6y+20$, which has a minimum value of $10$. The domain becomes $u \le 35$. Thus, the bounds for $u$ are $10$ to $35$. My integral becomes $$\int_{10}^{35} \frac{1}{u} du.$$ Which evaluates to $ln(7/2)$. The actual answer is $\pi ln(7/2)$. The discrepancy comes from the fact that I did not convert $dA$ to $du$ properly. I tried using a Jacobian, but that only works if I used a two variable substitution. So, I'm not sure what I have to do to get my shortcut to work, assuming it's even possible to get it to.

Answer 1

Your $u=x^2+y^2-2x+6y+20$ can be completed for instance by $v=y,$ so $$I:=\iint_{x^2+y^2 \le 2x-6y+15}\frac1{x^2+y^2-2x+6y+20}dxdy$$$$=2\iint_{(v+3)^2+10\le u\le35}\frac1u\frac{dudv}{2\sqrt{u-10-(v+3)^2}}$$ (the $2$ in front of the RHS stands for the fact that $x-1$ is determined only up to sign by $(u,v)$).

Letting $s=\frac{v+3}{\sqrt{u-10}},$ $$\int_{(v+3)^2\le u-10}\frac{dv}{2\sqrt{u-10-(v+3)^2}}=\int_{-1}^1\frac{ds}{2\sqrt{1-s^2}}=\frac\pi2$$ hence $$I=2\int_{10}^{35}\frac\pi2\frac{du}u=\pi\ln\frac72. $$

Answer 2

Here is an artificial way to proceed, so that the missing $\pi$-factor from the posted question falls transparently. To find the right scene for the tricky substitution, i need a more symmetric setting, so let me use first the linear, cheap substitution: $$ \begin{aligned} x-1 &=5X\ ,\\ y+3 &=5Y\ .\\[2mm] \end{aligned} $$ We will then use: $$ \bbox[yellow]{\qquad u =x^2 +y^2 -2x+6y+20=25(X^2+Y^2)+10\ .\qquad } $$ The domain of integration is $(x-1)^2+(y+3)^6\le 5^2$, i.e. $X^2+Y^2\le 1$. I will not pass to polar coordinates. Instead, consider a new variable, $s$, to complete $u$ to a adequate pair for the change of variables, thus to pass from $(x,y)$ to $(X,Y)$, and finally to $(s,u)$: $$ \bbox[yellow]{\qquad s=\arctan\frac XY\qquad\ .\ } $$ It takes values between $\arctan(-\infty)=-\frac \pi2$ and $\arctan(\infty)=\frac \pi2$. To have a bijection at the last step, we may restrict in between to $Y>0$. This is ok, since we integrate an even function w.r.t. $Y$. Then we have formally: $$ \begin{aligned} dx&=5\; dX\ , &du &=2\cdot 25\cdot(X\;dX+Y\;dY)& \\ dy&=5\; dY\ , &ds &=\frac 1{X^2+Y^2}\cdot(Y\; dX-X\;dY) \\[3mm] dx\wedge dy&=25\; dX\wedge dY\ , & ds\wedge du &= \frac {2\cdot 25}{X^2+Y^2}\cdot(Y^2\; dX\wedge dY-X^2\;dY\wedge dX)\\ &&&= 2\cdot 25\cdot dX\wedge dY\ , \end{aligned} $$ and we can now calculate the wanted integral, denote it by $J$: $$ \begin{aligned} J &=\iint_D\frac {dx\; dy}{x^2 +y^2-2x+6y+20} \\ &=\iint_{X^2+Y^2\le 1}\frac {25\; dX\; dY}{25(X^2 +Y^2)+10} \\ &=2\iint_{\substack{X^2+Y^2\le 1\\ Y>0}}\frac {25\; dX\; dY}{25(X^2 +Y^2)+10} \\ &=2\iint_{\substack{u\in[25\cdot 0+10,\ 25\cdot 1+10]\\s\in[-\pi/2,\ +\pi/2]}} \frac {\frac 12\; ds\; du}u =\int_{0+10}^{25+10}\frac {du}u=\log 35-\log 10 \\ &=\bbox[yellow]{\color{maroon}{\boxed{\quad \log\frac{35}{10}\quad}}}\ . \end{aligned} $$ $\square$

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Yes, using this trick, we integrate in between a clean $du/u$. (And well, i'm afraid somebody will still say i was using polar coordinates... But else, there is no way to involve a clean $du/u$, respecting the wish from the question, making the "almost coincidence" work.)