Triple Integral of z sin x cos y over a Box

Separate first, then do three easy integrals

Each factor uses exactly one variable and all six limits are constants, so the triple integral is a product. Group the factors with their own limits before integrating anything.

The first factor is $\left[-\cos x\right]_0^{\pi/2} = 0 - (-1) = 1$. The second is $\left[\sin y\right]_0^{\pi/2} = 1 - 0 = 1$. The third is $\left[\frac{z^2}{2}\right]_0^2 = 2$.

Both trigonometric factors run over a quarter period where the function is non negative, which is why each comes out to exactly $1$.

The mistake: losing the sign on the sine integral

The antiderivative of $\sin x$ is $-\cos x$, and the leading minus sign gets dropped constantly. Writing $\left[\cos x\right]_0^{\pi/2} = 0 - 1 = -1$ turns the answer into $-2$, which cannot be right for an integrand that is non negative on the whole box.

Two habits prevent it.

• Differentiate your antiderivative back before substituting limits: the derivative of $-\cos x$ is $+\sin x$, so the sign is right.
• Check the sign of the integrand over the region first. Here $\sin x \geq 0$, $\cos y \geq 0$ and $z \geq 0$ throughout, so the answer must be positive.
• Keep the limits in radians. A calculator left in degree mode reads the upper limit $\frac{\pi}{2} \approx 1.5708$ as $1.5708$ degrees, a sliver of the quarter turn you wanted, and returns about $0.0215$ instead of $1$.

What changes if the box changes

Push the $y$ limit out to $\pi$ and the cosine factor becomes $\left[\sin y\right]_0^{\pi} = 0$, so the whole triple integral collapses to zero. The positive part of $\cos y$ on $\left[0, \frac{\pi}{2}\right]$ exactly cancels the negative part on $\left[\frac{\pi}{2}, \pi\right]$.

That is worth noticing before doing any work, because a factor of zero makes the other two integrals irrelevant. With a separable integrand, one vanishing factor sets the whole answer to zero.