Change of variable with circle and ln

Question

I don't know how to solve: $$\int \!\!\! \int_D x^3y^2 \ln (x^2+y^2) \, dxdy$$ over the area: $$D=\{ (x,y): 4 \le x^2+y^2 \le 25, \,\, x,y \ge 0 \}$$

Obviously we are doing a double integration over the first quadrant of two circles, one with radius $2$ and one with radius 4.

I changed $$\left[ \begin{align} dx\,dy=r\,dr\,d\theta\\ x=r \cos\theta\\ y=r \sin\theta\\ \end{align} \right] $$

Now we have: $$ 2\leq r \leq 5$$ $$ 0\leq \theta \leq \frac{\pi}{4} $$

$$\int\int_D (r\cos\theta)^3 (r\sin\theta)^2\ln (r^2) \ r \, dr\ d\theta$$

I replaced $\sin^2$ by $1-\cos^2$

$$\int\int_D r^6 \cos^3\theta (1-\cos^2\theta) \ln (r^2) \, dr\, d\theta$$

$$\int\int_D r^6 \ \left(\cos^3\theta-\cos^5\theta)\right)\ln (r^2)\,dr\,d\theta$$

And after that I tried a double integration with the hope that I'll be back to $\cos^3-\cos^5$ after two cycles but $\ln(r^2)$ is crumping, so I am NOT getting anywhere with that technique. What shall I do?

Answer 1

What you should do is:$$\cos^3\theta\sin^2\theta=\cos\theta(1-\sin^2\theta)\sin^2\theta.$$Now, you can do the substitution $\sin\theta=u$ and $\cos\theta\,\mathrm d\theta=\mathrm du$.

Now, you have\begin{multline}\int_2^5\int_0^{\frac\pi2}r^6\cos\theta(\sin^2\theta-\sin^4\theta)\ln(r^2)\,\mathrm dr\,\mathrm d\theta=\\=\left(\int_2^52r^6\ln r\,\mathrm dr\right)\left(\int_0^{\frac\pi2}(\sin^2\theta-\sin^4\theta)\cos\theta\,\mathrm d\theta\right).\end{multline}You can compute the first integral using the fact that$$\int r^n\ln(r)\,\mathrm dr=-\frac1{(n+1)^2}r^{n+1}+\frac1{n+1}r^{n+1}\ln(r).$$And, using the substitution that I mentioned above, the second one becomes$$\int_0^1u^2-u^4\,\mathrm du.$$

Answer 2

Completing Jsé Carlos Santos' answer:

For the $\;\displaystyle\int r^6\ln r\,\mathrm dr$ part, proceed to an integration by parts.

More generally, it is good to know integration by parts enables you to establish a recurrence relation between $$I_{p,q}=\int r^p \ln^q r\,\mathrm dr $$ and $I_{p,q-1}$.