Flux through a square and rectangle

[bowlbase]

Homework Statement

Flux:
a. Calculate the flux of the vector v₁ = (1, 3 5) through a 2×2 square in the x-z plane (i.e., y = 0).
b. Calculate the flux of the vector v₂=(z, y, -x) through this rectangle:0≤ x ≤3, 0≤ y ≤ 2, z = 0..

The Attempt at a Solution

I guess flux is suppose to be some kind of surface integral though I don't recall doing these in calculus before. That's not to say I didn't.. but I simply don't remember them. I've looked up the integral methods and this is what I got:
A. $\int \vec{v_1}da$
z=0, da=dxdy$\hat{z}$
$\int^2_0 5 dxdy$ = 5(2)(2)=20

B. $\int \vec{v_3}da$
again z=0, da=dxdy$\hat{z}$
$\int -x dxdy$
$\int^3_0 -x dx$= -9/2
$\int^2_0 dy$=2
=-5/2

First thing that strikes me is the negative in B. I don't think that I should have a negative flux if I'm understanding this right. Can anyone help me understand this?

 

[tiny-tim]

hi bowlbase!

First thing that strikes me is the negative in B. I don't think that I should have a negative flux if I'm understanding this right. Can anyone help me understand this?

it depends whether you regard the normal as being in the positive or negative z direction

since the question doesn't say, it doesn't matter

a. Calculate the flux of the vector v₁ = (1, 3 5) through a 2×2 square in the x-z plane (i.e., y = 0).

The Attempt at a Solution

I guess flux is suppose to be some kind of surface integral though I don't recall doing these in calculus before. That's not to say I didn't.. but I simply don't remember them. I've looked up the integral methods and this is what I got:
A. $\int \vec{v_1}da$
z=0, da=dxdy$\hat{z}$
$\int^2_0 5 dxdy$ = 5(2)(2)=20

correct method, but …
i] you've misread the question … the normal is in the y direction!
ii] please always write ∫∫ not ∫ for a double integral, and you're less likely to make mistake

b. Calculate the flux of the vector v₂=(z, y, -x) through this rectangle:0≤ x ≤3, 0≤ y ≤ 2, z = 0..
…
B. $\int \vec{v_3}da$
again z=0, da=dxdy$\hat{z}$
$\int -x dxdy$
$\int^3_0 -x dx$= -9/2
$\int^2_0 dy$=2
=-5/2

no, you seem to have evaluated ∫∫ xdxdy as a sum of two integrals

you need to integrate wrt x first, then integrate that result wrt y

 

[bowlbase]

A) its funny that I have done it correctly on paper but incorrectly here. I'm not paying enough attention I guess.

B) no idea why I added those. I'm glad that I at least had the correct idea.

thanks for the help!