Let $\mathbf { F } ( x , y , z ) = \mathbf { k }$ . Evaluate the surface integral $\iint _ { S } \mathbf { F } \cdot d \mathbf { S }$ , where S is that part of the plane $z = x$ that lies above the square with vertices $( 0,0 ) , ( 1,0 ) , ( 0,1 )$ , and $( 1,1 )$ and has upward orientation.
A) $1$
B) $\sqrt { 2 }$
C) $\sqrt { 3 }$
D)$2$
E)$\frac { \sqrt { 2 } } { 2 }$
F) $\frac { \sqrt { 3 } } { 2 }$
G)$\frac { 1 } { 2 }$
H)$\frac { 1 } { 4 }$

Question

Let $\mathbf { F } ( x , y , z ) = \mathbf { k }$ . Evaluate the surface integral $\iint _ { S } \mathbf { F } \cdot d \mathbf { S }$ , where S is that part of the plane $z = x$ that lies above the square with vertices $( 0,0 ) , ( 1,0 ) , ( 0,1 )$ , and $( 1,1 )$ and has upward orientation.
A) $1$ 
B) $\sqrt { 2 }$
C) $\sqrt { 3 }$ 
D)$2$
E)$\frac { \sqrt { 2 } } { 2 }$ 
F) $\frac { \sqrt { 3 } } { 2 }$ 
G)$\frac { 1 } { 2 }$
H)$\frac { 1 } { 4 }$

Quizplus · Accepted Answer

The surface integral $\iint _ { S } \mathbf { F } \cdot d \mathbf { S }$ for a constant vector field $\mathbf { F } = \mathbf { k }$ over a plane $z = x$ can be simplified to calculating the flux of $\mathbf { F }$ through the surface. Since $\mathbf { F } = \mathbf { k }$, it only has a component in the $z$-direction. The surface area element $d\mathbf{S}$ has a magnitude equal to the area of the projection of the surface onto the $xy$-plane, which is the square with vertices $(0,0), (1,0), (0,1),$ and $(1,1)$, and a direction normal to the surface. The normal vector to the plane $z = x$ is $\frac{1}{\sqrt{2}}(\mathbf{i} - \mathbf{k})$, but since we are only interested in the $z$-component of $\mathbf{F}$, and the surface has an upward orientation, the relevant component is the $z$-component of the normal vector, which is $-\frac{1}{\sqrt{2}}$. The area of the square is $1$, so the integral simplifies to the product of the magnitude of $\mathbf{F}$ in the $z$-direction (which is $1$), the $z$-component of the normal ($-\frac{1}{\sqrt{2}}$), and the area of the square ($1$), corrected for the direction to ensure an upward orientation. The correct calculation should consider the absolute value of the normal's $z$-component, leading to $\frac{1}{\sqrt{2}} \times 1 \times 1 = \frac{1}{\sqrt{2}}$, but considering the upward orientation directly, the flux is simply the area of the square, which is $1$, since the vector field is perpendicular and uniform across the surface.

Let F(x,y,z)=k\mathbf { F } ( x , y , z ) = \mathbf { k }F(x,y,z)=k

Let $\mathbf { F } ( x , y , z ) = \mathbf { k }$