# How would one plot a multivariable function over several disjoint domains?

I would like to plot Scherk's first surface which is defined on a checkerboard of the plane, e.g. $-\frac{\pi}{2} < x < \frac{\pi}{2}, -\frac{\pi}{2} < y < \frac{\pi}{2}$ and $\frac{\pi}{2} < x < \frac{3\pi}{2}, \frac{\pi}{2} < y < \frac{3\pi}{2}$.

The patch is $\mathbf{X}(u,v) = (u,v,f(u,v))$ where $f = \ln\left( \frac{\cos(x)}{\cos(y)}\right)$.

• Consider plotting the implicit equation for Scherk's minimal surface instead: ContourPlot3D[Exp[z] Cos[y] == Cos[x], {x, -π/2, 3 π/2}, {y, -π/2, 3 π/2}, {z, -π/2, 3 π/2}]. – J. M.'s technical difficulties May 31 '16 at 16:22
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## 1 Answer

You can use RegionFunction to set the region:

Plot3D[Log[Cos[x]/Cos[y]], {x, -(π/2), 3 π/2}, {y, -(π/2),3 π/2},
RegionFunction ->
Function[{x, y, z}, -(π/2) < x < π/2 && -(π/2) < y < π/2 || π/
2 < x < (3 π)/2 && π/2 < y < (3 π)/2]]

Or you can construct the region and plot inside it:

reg = ImplicitRegion[-(π/2) < x < π/2 && -(π/2) < y < π/2
|| π/2 < x < (3 π)/2 && π/2 < y < (3 π)/2, {x, y}];
Plot3D[Log[Cos[x]/Cos[y]], {x, y} ∈ reg]
• Or, reg = RegionUnion[Cuboid[-Pi/2 {1, 1}, Pi/2 {1, 1}], Cuboid[Pi/2 {1, 1}, 3 Pi/2 {1, 1}]]. – rcollyer May 31 '16 at 16:43