Evaluate the double integral of x cos(x + y) over the triangular region with vertices (0, 0), (π, 0) and (π, π).

Step-by-step solution

1. 1

   Turn the three vertices into boundary equations. The side from $(0,0)$ to $(\pi,0)$ lies on $y=0$; the side from $(\pi,0)$ to $(\pi,\pi)$ is the vertical line $x=\pi$; the side from $(0,0)$ to $(\pi,\pi)$ is the line $y=x$. So the triangle is the set of points with $0\le x\le\pi$ and $0\le y\le x$.

2. 2

   Choose the order $dy\,dx$. Integrating in $y$ first keeps the $x$ limits constant ($0$ to $\pi$) and makes the $y$ limits the two simple functions $0$ and $x$. It also helps that $x$ is a constant during the inner integration, so the factor $x$ slides outside:

   $\int_0^\pi\!\!\int_0^x x\cos(x+y)\,dy\,dx.$

3. 3

   Do the inner integral. With $x$ fixed, $\int\cos(x+y)\,dy=\sin(x+y)$:

   $x\Big[\sin(x+y)\Big]_{y=0}^{y=x}=x\big(\sin 2x-\sin x\big).$

4. 4

   Handle $\int_0^\pi x\sin 2x\,dx$ by parts. Take $u=x$, $dv=\sin 2x\,dx$, so $du=dx$ and $v=-\tfrac{\cos 2x}{2}$:

   $\left[-\frac{x\cos 2x}{2}\right]_0^\pi+\frac12\int_0^\pi\cos 2x\,dx=-\frac{\pi}{2}+\left[\frac{\sin 2x}{4}\right]_0^\pi=-\frac{\pi}{2}.$

5. 5

   Handle $\int_0^\pi x\sin x\,dx$ the same way. With $u=x$, $v=-\cos x$:

   $\Big[-x\cos x\Big]_0^\pi+\int_0^\pi\cos x\,dx=\pi+\Big[\sin x\Big]_0^\pi=\pi.$

6. 6

   Combine the two pieces. The outer integral is the difference of the results:

   $\iint_R x\cos(x+y)\,dA=-\frac{\pi}{2}-\pi=-\frac{3\pi}{2}\approx -4.712.$

   The answer is negative, which is expected: on most of this triangle $x+y$ lies between $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}$, where $\cos(x+y)$ is negative, and the weight $x$ is largest exactly there. Numerical integration returns $-4.71239$, matching $-\tfrac{3\pi}{2}$.