Find the volume of the region bounded by the surface sqrt(x) +sqrt(y) +sqrt(z)=1 and the coordinate pla

find the volume of the region bounded by the surface sqrt(x) +sqrt(y) +sqrt(z)=1 and the coordinate planes

Use the transformation x = u^2, y = v^2, z = w^2 (with u,v,w non-negative).

So, the region transforms to that bounded by the plane u + v + w = 1 and the coordinate planes in the first octant.

Moreover, the Jacobian ∂(x,y,z)/∂(u,v,w) equals
|2u 0 0|
|0 2v 0| = 8uvw.
|0 0 2w|

Hence, the volume ∫∫∫ 1 dV equals by change of coordinates
∫(u = 0 to 1) ∫(v = 0 to 1 - u) ∫(w = 0 to 1 - u - v) 1 * (8uvw dw dv du).

Simplifying (after much algebra) yields
∫(u = 0 to 1) ∫(v = 0 to 1 - u) ∫(w = 0 to 1 - u - v) 8uvw dw dv du
= ∫(u = 0 to 1) ∫(v = 0 to 1 - u) 4uvw^2 {for w = 0 to 1 - u - v} dv du
= ∫(u = 0 to 1) ∫(v = 0 to 1 - u) 4uv((1 - u) - v)^2 dv du
= ∫(u = 0 to 1) ∫(v = 0 to 1 - u) 4uv((1 - u)^2 - 2(1 - u)v + v^2) dv du
= ∫(u = 0 to 1) ∫(v = 0 to 1 - u) [4u(1 - u)^2 v - 8u(1 - u)v^2 + 4uv^3) dv du
= ∫(u = 0 to 1) [2u(1 - u)^2 v^2 - (8/3)u(1 - u)v^3 + uv^4] {for v = 0 to 1 - u} du
= ∫(u = 0 to 1) (1/3)u(1 - u)^4 du
= ∫(t = 1 to 0) (1/3)(1 - t)t^4 * -dt, letting t = 1 - u
= (1/3) ∫(t = 0 to 1) (t^4 - t^5) dt
= (1/3)(1/5 - 1/6)
= 1/90.

I hope this helps!