0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$$\\$$

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.

$\endgroup$
0
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.