Calculation of surface integral

Question

Question

Calculate $ \int x+z ~dS $ on the domain bounded by the first octant and the plane $x+y+z=1$

Attempt

$$ \int_S x+z~dS = \sqrt3\int_0^1 \int_0^{y-1} x +z ~dxdy $$ Since on the $xy$ plane $z=0$ then $$ \sqrt3\int_0^1 \int_0^{y-1} x +z ~dxdy = \sqrt3\int_0^1 \int_0^{y-1} x ~dxdy = A $$

Attempt 2

$$ \int_S x+z~dS = \sqrt3\int_0^1 \int_0^{y-1} x +z ~dxdy = \sqrt3\int_0^1 \int_0^{y-1} x+1-y-x ~dxdy =\sqrt3\int_0^1 \int_0^{y-1} 1-y ~dxdy = B $$

However $A\neq B $. So can somebody please tell me what mistake I am making in any of the methods?

Answer 1

To calculate the integral we need to divide it in the 4 boundaries

• $S_1 \quad x=0$
• $S_2 \quad y=0$
• $S_3 \quad z=0$
• $S_4 \quad x+y+z=1$

then

$$ \int_S x+z~dS = \int_{S_1} z~dS+\int_{S_2} x+z~dS+\int_{S_3} x~dS+\int_{S_4} (1-y)~dS $$

You attempt 2 seems indeed to be correct.