Evaluate the surface integral $\iint _ { S } x d S$ , 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)2
E)$\frac { \sqrt { 2 } } { 2 }$
F)$\frac { \sqrt { 3 } } { 2 }$
G)$\frac { \sqrt { 2 } } { 3 }$
H) $\frac { \sqrt { 3 } } { 3 }$

Question

Evaluate the surface integral $\iint _ { S } x d S$ , 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)2
E)$\frac { \sqrt { 2 } } { 2 }$ 
F)$\frac { \sqrt { 3 } } { 2 }$ 
G)$\frac { \sqrt { 2 } } { 3 }$
H) $\frac { \sqrt { 3 } } { 3 }$

Quizplus · Accepted Answer

The surface integral is evaluated over the region where $z = x$ above the square in the xy-plane. The surface element $dS$ can be expressed in terms of $dxdy$ using the normal vector. The integral simplifies to $\iint_{[0,1]\times[0,1]} x \sqrt{2} \, dx \, dy$, which evaluates to $\frac{\sqrt{2}}{2}$.

Evaluate the Surface Integral ∬SxdS\iint _ { S } x d S∬S​xdS

Evaluate the Surface Integral $\iint _ { S } x d S$