Surface integral of vector over square plane

Problem 2.25 D.K.Cheng

The unit normal vector for the plane is found as the cross product of the two vectors spanning the plane. $$\mathbf{a_n}=\frac{1}{\sqrt{2}}\bigg(\mathbf{a_y}+\mathbf{a_z}\bigg) $$ The dot product of the vector field and the plane's normal vector is $$\mathbf{F} \cdot \mathbf{a_n}=\frac{5}{\sqrt{2}} $$ The last thing to do is evaluate the integral $\displaystyle\int^d_c\int^b_a \frac{5}{\sqrt{2}} \text{d}s\:$ however, I'm uncertain what the limits of the integrals should be.

The plane doesn't lie in either the xy/yz/xz-plane, so I can't just set x/y/z = $0$, set the limits to $a=c=0$, $b=d=2$ and then integrate.

How do I determine what the limits should be? The answer to the problem is supposedly $\displaystyle\int_S \mathbf{F} \cdot \text{d}\mathbf{s} = 20$

Solution 1:

The surface is in the plane $y + z = 2$ with normal vector $(0, 1, 1)$.

Also $S$ is not a square surface but a rectangular surface though its projection $E$ in xy-plane is a square with bounds $~0 \leq x \leq 2, 0 \leq y \leq 2$

As you have normalized the normal vector, note that $dS = \sqrt 2 ~ dA$ where $dA$ is the area element of the projection in xy-plane.

So the integral is,

$$\displaystyle \iint_S \frac{5}{\sqrt2} ~ dS = \iint_E 5 ~ dA = 20.$$