Evaluate the surface integral $\iint _ { S } ( x + y + z ) d S$ , where S is that part of the plane $z = x$ that lies above the square with vertices $( 0,0 ) , ( 1,0 ) , ( 0,1 )$ , and $( 1,1 )$ .
A) $1$
B)$\sqrt { 2 }$
C) $\sqrt { 3 }$
D)$\frac { 3 } { 2 }$
E)$\frac { \sqrt { 2 } } { 2 }$
F)$\frac { \sqrt { 3 } } { 2 }$
G)$\frac { \sqrt { 2 } } { 3 }$
H)$\frac { 3 \sqrt { 2 } } { 2 }$

Question

Evaluate the surface integral $\iint _ { S } ( x + y + z ) d S$ , where S is that part of the plane $z = x$ that lies above the square with vertices $( 0,0 ) , ( 1,0 ) , ( 0,1 )$ , and $( 1,1 )$ .
A) $1$ 
B)$\sqrt { 2 }$ 
C) $\sqrt { 3 }$ 
D)$\frac { 3 } { 2 }$ 
E)$\frac { \sqrt { 2 } } { 2 }$
F)$\frac { \sqrt { 3 } } { 2 }$
G)$\frac { \sqrt { 2 } } { 3 }$ 
H)$\frac { 3 \sqrt { 2 } } { 2 }$

Quizplus · Accepted Answer

The surface integral can be evaluated by parameterizing the surface and using the formula for surface integrals. The plane $z = x$ can be parameterized as $(x, y, x)$. The integral becomes $\int_0^1 \int_0^1 (x + y + x) \sqrt{2} \, dy \, dx$, which evaluates to $\frac{3\sqrt{2}}{2}$.

Evaluate the Surface Integral ∬S(x+y+z)dS\iint _ { S } ( x + y + z ) d S∬S​(x+y+z)dS

Evaluate the Surface Integral $\iint _ { S } ( x + y + z ) d S$