Find the volume of the solid bounded by: z=cos x cos y

Find the volume of the solid bounded by: $$ z=\cos x\cos y,\qquad z=0,\qquad |x+y|<\frac\pi2,\qquad |x-y|<\frac\pi2. $$

Step 1 · Substitute and compute the Jacobian

Let $u=x+y,\,v=x-y$, so $x=\dfrac{u+v}2,\,y=\dfrac{u-v}2$. The region becomes the square $-\frac\pi2<u<\frac\pi2,\,-\frac\pi2<v<\frac\pi2$. The Jacobian is:

$$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}\frac12&\frac12\\\frac12&-\frac12\end{vmatrix} = -\frac14-\frac14=-\frac12, \qquad\left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\frac12. $$

Step 2 · Transform the integrand

With $A=\frac{u+v}2,\,B=\frac{u-v}2$: $A-B=v$, $A+B=u$, so $\cos A\cos B=\frac12[\cos v+\cos u]$, i.e.:

$$ \cos x\cos y = \frac12\big[\cos u+\cos v\big]. $$

Step 3 · Set up and evaluate the volume integral

$$ V = \iint_D\cos x\cos y\,dx\,dy = \frac14\int_{-\pi/2}^{\pi/2}\!\int_{-\pi/2}^{\pi/2}\big(\cos u+\cos v\big)\,du\,dv. $$

Using $\displaystyle\int_{-\pi/2}^{\pi/2}\cos u\,du=2$:

$$ V = \frac14\int_{-\pi/2}^{\pi/2}\Big[2+\pi\cos v\Big]\,dv = \frac14\Big[2v+\pi\sin v\Big]_{-\pi/2}^{\pi/2} = \frac14\big[2\pi+2\pi\big] = \pi. $$

Additional Notes

• This derivation matches the source exactly. Since $\cos x\cos y\geqslant0$ throughout this rotated square (verified numerically: no sign changes occur inside the region), $z=0$ is indeed the correct lower bound everywhere, with no need to split the integral.
• Verified numerically: a fine-grid quadrature over the rotated square gives $V\approx3.14163$, matching $\pi\approx3.14159$.