Evaluate the double integral of (x + y)/(x^2 + y^2) over the region enclosed by the loop of the curve x^3 + y^3 = 3xy.

Step-by-step solution

1. 1

   Convert the boundary curve to polar form. With $x=r\cos\theta$, $y=r\sin\theta$ the curve $x^3+y^3=3xy$ becomes $r^3(\cos^3\theta+\sin^3\theta)=3r^2\cos\theta\sin\theta$, so for $r\neq0$

   $r_{\max}(\theta)=\frac{3\cos\theta\sin\theta}{\cos^3\theta+\sin^3\theta}$

   This vanishes at $\theta=0$ and $\theta=\pi/2$ and is positive between, so $\theta\in[0,\pi/2]$ traces exactly the loop - the bounded piece of the folium.

2. 2

   Convert the integrand and watch the Jacobian cancel.

   $\frac{x+y}{x^2+y^2}=\frac{r(\cos\theta+\sin\theta)}{r^2}=\frac{\cos\theta+\sin\theta}{r}$

   Since $dA=r\,dr\,d\theta$, the $r$ in the Jacobian cancels the $1/r$: the integrand in polar coordinates is free of $r$ entirely. That is the reason this awkward region is tractable.

3. 3

   Do the inner integral in one line. With no $r$-dependence left,

   $I=\int_0^{\pi/2}(\cos\theta+\sin\theta)\,r_{\max}(\theta)\,d\theta=\int_0^{\pi/2}\frac{3\cos\theta\sin\theta(\cos\theta+\sin\theta)}{\cos^3\theta+\sin^3\theta}\,d\theta$

4. 4

   Cancel using the sum-of-cubes identity. Because $\cos^3\theta+\sin^3\theta=(\cos\theta+\sin\theta)(1-\cos\theta\sin\theta)$ (the middle factor simplifies via $\cos^2+\sin^2=1$), the factor $\cos\theta+\sin\theta$ cancels:

   $I=\int_0^{\pi/2}\frac{3\cos\theta\sin\theta}{1-\cos\theta\sin\theta}\,d\theta$

   Note $1-\cos\theta\sin\theta\ge\tfrac12$, so the integral is proper - nothing blows up.

5. 5

   Reduce to a standard Weierstrass integral. Writing $\cos\theta\sin\theta=\tfrac12\sin2\theta$ and substituting $\varphi=2\theta$:

   $I=\frac{1}{2}\int_0^{\pi}\frac{3\sin\varphi}{2-\sin\varphi}\,d\varphi=\frac{3}{2}\left[-\pi+2\int_0^{\pi}\frac{d\varphi}{2-\sin\varphi}\right]$

   The half-angle substitution $t=\tan(\varphi/2)$ turns the remaining integral into $\int_0^{\infty}\frac{dt}{t^2-t+1}=\frac{4\pi}{3\sqrt3}$.

6. 6

   Assemble the closed form and check it numerically.

   $I=\frac{3}{2}\left(-\pi+\frac{8\pi}{3\sqrt3}\right)=\frac{4\pi}{\sqrt3}-\frac{3\pi}{2}=\frac{4\sqrt3\,\pi}{3}-\frac{3\pi}{2}$

   Numerically $7.255197-4.712389=2.5428085$, which agrees with Simpson's rule applied directly to $\int_0^{\pi/2}\frac{3\cos\theta\sin\theta}{1-\cos\theta\sin\theta}d\theta$ to seven decimals ✓.