Evaluate the double integral of x/(x^2 + y^2) over the region D bounded by the parabola 2y = x^2 and the line y = x.

Step-by-step solution

1. 1

   Find where the two curves meet. From $2y=x^2$ we get $y=\tfrac{x^2}{2}$. Setting $\tfrac{x^2}{2}=x$ gives $x^2-2x=0$, so $x=0$ or $x=2$; the intersection points are $(0,0)$ and $(2,2)$.

2. 2

   Decide which curve is on top. At $x=1$ the line gives $y=1$ and the parabola $y=\tfrac12$, so the line is the upper boundary. The region is $0\le x\le 2,\qquad\frac{x^2}{2}\le y\le x.$

3. 3

   Integrate in $y$ first. Treating $x$ as constant, $\int\frac{x\,dy}{x^2+y^2}=\arctan\frac{y}{x}.$ Choosing $y$ as the inner variable is what makes the $x$ in the numerator do useful work.

4. 4

   Substitute the bounds. At $y=x$ the argument is $1$; at $y=\tfrac{x^2}{2}$ it is $\tfrac{x}{2}$. So $I=\int_0^2\left(\arctan 1-\arctan\frac{x}{2}\right)dx=\frac{\pi}{2}-\int_0^2\arctan\frac{x}{2}\,dx.$

5. 5

   Evaluate the remaining integral by parts. With $u=\tfrac{x}{2}$ it becomes $2\int_0^1\arctan u\,du$, and $\int\arctan u\,du=u\arctan u-\tfrac12\ln(1+u^2)$, giving $2\left(\frac{\pi}{4}-\frac{\ln 2}{2}\right)=\frac{\pi}{2}-\ln 2.$

6. 6

   Combine. $I=\frac{\pi}{2}-\left(\frac{\pi}{2}-\ln 2\right)=\ln 2\approx 0.693147,$ which matches a direct numerical evaluation of $\int_0^2\left(\tfrac{\pi}{4}-\arctan\tfrac{x}{2}\right)dx$.