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I was trying to solve for the region enclosed by y=Sqrt[x], z=1+x+y, y=0, and x=1, to evaluate a triple integral of 6xy.

So far I have tried to this code to only an error.

Integrate[(6*x*y)Boole[z=1+x+y && y==Sqrt[x] &&y==0 && x==1], {x, -∞, ∞}, {y, -∞, ∞}, {z, -∞, ∞}]

Could someone please help me understand how to calculate a region in Mathematica?

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It is right?

reg = ImplicitRegion[ y <=  Sqrt[x] && z == 1 + x + y &&  y >= 0 && x <= 1, {x, y, z}]
% // Region
Integrate[6*x*y, {x, y, z} ∈ reg]

Sqrt[3]

Another way

According to the definition of the integral along a surface.

The normal of the surface z-(1+x+y)==0 is Grad[z - (1 + x + y), {x, y, z}

Integrate[
 6 x*y*Norm[Grad[z - (1 + x + y), {x, y, z}]], {x, y} ∈ 
  ImplicitRegion[y <= Sqrt[x] && y >= 0 && x <= 1, {x, y}]]

Sqrt[3]

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  • $\begingroup$ Here's the Boole[] version of the last snippet: Integrate[6 x y Norm[Grad[z - (1 + x + y), {x, y, z}]] Boole[0 <= y <= Sqrt[x] && x <= 1], {x, -∞, ∞}, {y, -∞, ∞}] $\endgroup$ – J. M.'s ennui Jan 31 at 7:11

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