Troubleshooting Flux Out of a Cube: Evaluating a Double Integral

[squiggles0]

I am trying to work through some examples we have been given on flux out of a cube but am having difficulty in seeing how one one line of the answer becomes the next.

The question is analysing the flux out of a cube by looking at each side individually and working out the surface integrals.

Looking at side a a double integral is formed, which I can get to and understand, but I don't not understand how this has been evaluated.

The formula is in the attached .bmp

I get the answer of 1/3 when I try to evaluate this. I do not see why in the second line the Int dz has been separated out, or where the 1/2 has appeared from.

Many thanks for any help.

PS. apologies for not writing it in LaTeX but I don't know how yet and was hoping for a quick answer.

 

[Kurdt]

It will probably take even longer if you don't write in latex as attachments have to be approved. You can click on latex images to see the coding used and there is a small pdf that can be accessed from a link in the code window aswell. If you want to have a look at some latex examples then do so in the following thread:

https://www.physicsforums.com/showthread.php?t=8997

 

[squiggles0]

Thanks for the info and the link Kurdt. I have now written the formula up in LaTeX.

$$flux = \int_{0}^{1}\int_{0}^{1} y^2\; dy dz$$

$$= \frac{1}{2} \left[ y^3 \right]_{0}^{1} \int_{0}^{1} dz$$

$$= \frac{1}{2}$$

 

[malawi_glenn]

No primitive function of $$y^2$$ is $$y^3 /3$$

 

[squiggles0]

Sorry, but what do you mean by primitive function?
I was performing one of the intergrations then the other. So first:
$$\int_{0}^{1} y^2\; dy$$
Which gives $$\left[ \frac{y^3}{3} \right]_{0}^{1}$$, but this is wrong?

 

[Kurdt]

The primitive function is a function that gives the integrand when differentiated. Or the primitive function is the indefinite integral of the integrand.

It appears that the example is wrong and that is why you're getting confused. It should be as malawi glenn said.

 

[HallsofIvy]

Sorry, but what do you mean by primitive function?
I was performing one of the intergrations then the other. So first:
$$\int_{0}^{1} y^2\; dy$$
Which gives $$\left[ \frac{y^3}{3} \right]_{0}^{1}$$, but this is wrong?

A "primitive function" or just "primitive" is an anti-derivative. What you have here is correct but what you had earlier was

Thanks for the info and the link Kurdt. I have now written the formula up in LaTeX.

$$flux = \int_{0}^{1}\int_{0}^{1} y^2\; dy dz$$

$$= \frac{1}{2} \left[ y^3 \right]_{0}^{1} \int_{0}^{1} dz$$

$$= \frac{1}{2}$$

and that's clearly incorrect- it should be 1/3.

 

[squiggles0]

I couldn't see how this result had been arrived at, but it is repeated later on as well. I'm guessing it was just copied from earlier then, just wasn't sure if i was missing out something or if it really was a mistake.
Thanks for the confirmation guys.

 

[Kurdt]

Of course books or notes are never infallible. Well done for spotting it.