Double Integral of sin(x + y) Over a Square

Treat x as a constant shift inside the sine

During the inner integration $x$ is frozen, so $\sin(x+y)$ is a sine wave in $y$ shifted by the constant $x$. Its antiderivative in $y$ is $-\cos(x+y)$, with no chain-rule factor because the coefficient on $y$ is $1$.

Now use $\cos\!\left(x + \frac{\pi}{2}\right) = -\sin x$ to clean it up.

The mistake: distributing the sine over the sum

The wrong move is to write $\sin(x+y) = \sin x + \sin y$. Sine is not additive, and the correct identity is the angle addition formula.

Integrating the bogus $\sin x + \sin y$ over the same square gives $\frac{\pi}{2}\cdot 1 + 1\cdot\frac{\pi}{2} = \pi \approx 3.1416$, against the true $2$. That is an error of about $1.14$, produced entirely by the false identity.

The angle addition formula is also a legitimate second route to the answer. Expanding and integrating term by term over the square gives $1\cdot 1 + 1\cdot 1 = 2$, which is a useful independent check on the first method.

A sanity check on the size of the answer

On this square $x+y$ ranges from $0$ to $\pi$, so $\sin(x+y)$ stays between $0$ and $1$ and never goes negative. The volume must therefore be positive and no larger than the base area.

• Base area: $\left(\frac{\pi}{2}\right)^{2} \approx 2.467$.
• Maximum height: $1$, attained along the anti-diagonal $x + y = \pi/2$.
• So the answer must lie in $(0, 2.467)$, and $2$ does.

If the square were extended to $[0,\pi]^2$ the integrand would go negative on part of the region and the integral would be signed volume, not volume. Bounds decide which of the two you are computing.