Find the flux of a given vector field upwards across the portion of a plane that lies above a square

Question

This is the question I'm doing for revision for my finals retake

Find the flux of the vector field f = 2xyi + 2yzj + 2xzk

upwards across the portion of the plane

x + y + z = 2a that lies above the square

0 ≤ x ≤ a and 0 ≤ y ≤ a in the xy-plane

I have no problem with the general way to calculate flux, I'm just really bad with finding limits for the equations. There being a z in one equation and not in another is throwing me off a bit. I'm a bit out of practice with Maths as you can probably tell. Thanks in advance.

Answer 1

$$\textbf{Flux} = \iint_S \textbf{F} \cdot d \vec{S} = \iint_D \textbf{F}(G(u,v)) \cdot \vec{n}(u,v) \ du \ dv = \int_{0}^a \int_{0}^a \textbf{F}(G(x,y)) \cdot \vec{n}(x,y) \ dx \ dy$$

$$\\$$

The limits for the domain were already given. Here we can write $G(x,y) = (x,y,2a-x-y)$ and $\vec{n} = G_X \times G_y$ in which you should just get the normal for the plane i.e $\vec{n} = \langle 1,1,1 \rangle$ and now just plug in everything.

Answer 2

The plane is providing a (linear) link among the three variables $x,y,z$, thereby reducing the degree-of-freedom from $3$ to $2$, meaning you can express any of the coordinates as a function of the other two remaining. Since you are given the limits in $x,y$ you may want to express $z$ as $z(x,y)$, i.e. $z=2a-x-y$ and plug that into $\mathbf{f}$ to get $\mathbf{f}(x,y)$. Can you proceed from here?