Thread: help with finding dA in computing the flux of the vector field F

1. help with finding dA in computing the flux of the vector field F

Compute the flux of the vector field F(x, y, z) = 2xj + yk through the surface S, where S is the part of the surface z = -y + 4 above the square 0 ≤ x ≤ 3, 0 ≤ y ≤ 1, oriented upward.

Again, having trouble finding dA. is it not dA = (0i + j + k)dxdy?

2. Re: help with finding dA in computing the flux of the vector field F

Do you mean "dA", which is a scalar differential or " $\vec{n} dA= d\vec{S}$", a vector differential? Since you are talking about the flux of a vector function through the surface, you must mean the latter and you want to find $\int\int \vec{F}\cdot d\vec{S}$.

You can write the surface, z= 4- y as x= x, y= y, z= 4- y or as the vector equation $\vec{r}(x, y)= x\vec{i}+ y\vec{j}+ (4- y)\vec{k}$. The derivatives, $\vec{r}_x= \vec{i}$ and $\vec{r}_y= \vec{j}- \vec{k}$ lie in the tangent plane at each point and their cross product,
$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 0 \\ 0 & 1 & -1\end{array}\right|= -\vec{j}+ \vec{k}$
with "upward z component" or $\vec{j}- \vec{k}$ with "downward z component". Which of those you use depends upon how you want the surface oriented- whether you want to think of the flux as positive or negative.

Taking the "upward z compontent" $d\vec{S}= (\vec{j}- \vec{k})dxdy$ so you have a sign wrong.

3. Re: help with finding dA in computing the flux of the vector field F

Ok. thanks. I used the scalar differential <0, 1, -1>dxdy and then tried to calculate the flux by
∫_s F · dA = ∫_R <0, 2x, y> · <0, 1, -1> dxdy
= ∫(y = 0 to 1)∫(x = 0 to 3) (2x - y) dxdy

and got 15/2 but that was wrong.
I also tried <0, -1, 1>dxdy and got -15/2 and that was also incorrect. what did I do wrong?