Evaluate the double integral of sin x − cos y over the square 0 ≤ x ≤ 2π, 0 ≤ y ≤ 2π.

Step-by-step solution

1. 1

   Split the integrand into two independent pieces. Integration is linear and the region is a rectangle with constant limits, so

   $\iint(\sin x-\cos y)\,dA=\iint \sin x\,dA-\iint\cos y\,dA.$

   Each piece depends on only one variable, which makes both of them products of one-dimensional integrals.

2. 2

   Evaluate the first piece. Since $\sin x$ does not involve $y$, the $y$-integration just multiplies by the length $2\pi$:

   $\int_{0}^{2\pi}\!\int_{0}^{2\pi}\sin x\,dx\,dy=2\pi\int_{0}^{2\pi}\sin x\,dx=2\pi\bigl[-\cos x\bigr]_{0}^{2\pi}=2\pi(-1+1)=0.$

3. 3

   Evaluate the second piece the same way.

   $\int_{0}^{2\pi}\!\int_{0}^{2\pi}\cos y\,dx\,dy=2\pi\int_{0}^{2\pi}\cos y\,dy=2\pi\bigl[\sin y\bigr]_{0}^{2\pi}=2\pi(0-0)=0.$

4. 4

   Combine.

   $\iint_{[0,2\pi]^{2}}(\sin x-\cos y)\,dA=0-0=0.$

5. 5

   Say why this was inevitable. The side length $2\pi$ is exactly one full period of both $\sin$ and $\cos$, and every sinusoid integrates to zero over a whole period — the positive and negative humps have equal area. Any interval of length $2\pi$ would have given the same answer; had the limits been $0$ to $\pi$, the sine term would have contributed $2\pi\cdot 2=4\pi$ instead.