Flux/Surface Integral across a Plane

[FAS1998]

Homework Statement

I attached an image of the multi-part problem on this post. I got correct answers to every question other than the last one.

Homework Equations

The Attempt at a Solution

I believe the last part is a surface integral problem.

F is given and I found n is previous parts of the problem.

And I think F (dot) n should be 11/sqrt(6).

I’m not sure what to do after that. I don’t remember my calc 3 very well and I never really had a good understanding of surface integrals and flux.

I think I might need some equation involving dS to turn the surface integral into a double integral, but I’m not quite sure.

 

[kuruman]

OK, you got $\vec F \cdot \hat n = \frac{11}{\sqrt{6}}$ correctly. Note that it is constant therefore $\int \int \vec F \cdot \hat n~dS = \frac{11}{\sqrt{6}}\int \int dS$.
So ...

 

[FAS1998]

OK, you got $\vec F \cdot \hat n = \frac{11}{\sqrt{6}}$ correctly. Note that it is constant therefore $\int \int \vec F \cdot \hat n~dS = \frac{11}{\sqrt{6}}\int \int dS$.
So ...

I’m not sure how to do that calculation. Do I replace dS with something else to get a double integral?

 

[kuruman]

You can replace it with $dx~dy$ if that makes you feel any better, but what does $\int \int dS$ actually represent? Hint: You already know the answer.

 

[FAS1998]

You can replace it with $dx~dy$ if that makes you feel any better, but what does $\int \int dS$ actually represent? Hint: You already know the answer.

I used sqrt(6) as $\int \int dS$ and got 11 as the correct answer.

So does that mean that calculating the surface integral of a given surface gives the area of that surface?

 

[kuruman]

So does that mean that calculating the surface integral of a given surface gives the area of that surface?

$\int \int dS$ is shorthand for the following in plain English: "Consider an element of area $dS$ on the triangle. Add all such elements over the entire surface of the triangle." So if you follow the instructions and add all the weensy elements together, what do you get? The total area of the triangle! This shortcut works only because $\vec F \cdot ~\hat n$ is constant and can be taken out of the integral. If it depended on $x$ and/or $y$, you would have to replace $dS$ with $dx~dy$ and do the double integral.