The surface integral ∫ ∫ xdydz + y ^ 2dzdy + zdxdy, ∑ is the upper side of the triangle on the plane where x + y + Z = 1 is cut by the coordinate plane

The surface integral ∫ ∫ xdydz + y ^ 2dzdy + zdxdy, ∑ is the upper side of the triangle on the plane where x + y + Z = 1 is cut by the coordinate plane

Find the surface integral ∫ ∫ xdydz + y ^ 2dzdx + zdxdy, where ∑ is the upper side of the triangle on the plane where x + y + Z = 1 is cut by the coordinate plane
Patching:
Σ 1: x = 0, back side
Σ 2: y = 0, left
Σ 3: z = 0, lower side
∫∫(Σ+Σ1+Σ2+Σ3) xdydz + y^2dzdy + zdxdy
= ∫∫∫Ω (1 + 2y + 1) dV
= 2∫∫∫Ω (1 + y) dV
= 2∫(0→1) dx ∫(0→1 - x) dy ∫(0→1 - x - y) (1 + y) dz
= 5/12
∫∫Σ1 xdydz + y^2dzdy + zdxdy = 0
∫∫Σ2 xdydz + y^2dzdy + zdxdy = 0
∫∫Σ3 xdydz + y^2dzdy + zdxdy = 0
So ∫ ∫ ∑ xdydz + y ^ 2dzdy + zdxdy = 5 / 12
Use the original method to solve: (skillfully, so that you can see how much you know about surface integral)
Find the surface integral ∫ ∫ xdydz + y ^ 2dzdx + zdxdy, where ∑ is the upper side of the triangle on the plane where x + y + Z = 1 is cut by the coordinate plane
∫∫Σ xdydz + y^2dzdx + zdxdy = ∫∫Σ x dydz + ∫∫Σ y^2 dzdx + ∫∫Σ z dxdy
In YZ plane, ∫ ∫ ∑ x dydz, x = 1 - Y - Z, take the front side
=The area bounded by ∫ ∫ D (1 - Y - z) dydz, y + Z = 1 and YZ coordinate plane
= ∫(0→1) dy ∫(0→1 - y) (1 - y - z) dz
= 1/6
In ZX plane, ∫∫Σy ^ 2 dzdx, y = 1 - Z - x, take the right side
= ∫∫D (1 - z - x)^2 dzdx
= ∫∫D (z^2 + x^2 + 2zx - 2z - 2x + 1) dzdx
= ∫(0→1) dx ∫(0→1 - x) (z^2 + x^2 + 2zx - 2z - 2x + 1) dz
= 1/12
In the XY plane, ∫ ∫ Z DXDY, z = 1 - X - y, take the upper side
= ∫∫D (1 - x - y) dxdy
= ∫(0→1) dx ∫(0→1 - x) (1 - x - y) dy
= 1/6
So ∫ ∫ ∑ xdydz + y ^ 2dzdx + zdxdy = 1 / 6 + 1 / 12 + 1 / 6 = 5 / 12