Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Use the Fundamental Theorem to evaluate the integral of $ ze^{x^2} dydz + 3ys dydz + (2-yz^7)dxdy $ over the surface of the unit cube, except the bottom face.

share|improve this question
    
What is $s$, a constant? –  Ron Gordon Mar 19 '13 at 9:05
    
I assumed so. It is not a typo on my part. –  GaMbiT Mar 19 '13 at 9:14

1 Answer 1

up vote 1 down vote accepted

By "unit cube," I assume you mean $[0,1]^3$. In any case, use the divergence theorem to get the net surface integral, i.e., the net outward flow across all faces of the cube. Then subtract the specific surface contribution from the bottom face to get the quantity you seek.

The vector field described above is

$$\vec{F} = (z\, e^{x^2}, 3 s y, 2-y z^7)$$

Then its divergence is

$$\vec{\nabla}\cdot \vec{F} = 2 x z e^{x^2} + 3 s - 7 y z^6 $$

The net surface integral is then the integral of the divergence over the unit cube. I assume you can do this relatively simple integral; I get

$$\iiint_{[0,1]^3} dx\,dy\,dz\: \vec{\nabla}\cdot \vec{F} = \frac{1}{2} (e-3) + 3 s$$

Then you must subtract out the contribution from the bottom face, i.e. $z=0$. Since the flow is outward, i.e., down in the negative $z$ direction, you add back in the integral

$$2 \iint_{[0,1]^2} dx \,dy 2 = 2$$

so the quantity you seek is

$$\frac{1}{2} (e+1) + 3 s$$

share|improve this answer
    
*(e - 1), instead of (e + 1)? Thank you very much for the help, by the way. –  GaMbiT Apr 19 '13 at 14:18
1  
@XxGaMbiT: No, because of the factor of $1/2$. –  Ron Gordon Apr 19 '13 at 14:19
    
Oh, right. Careless mistake. Sorry. –  GaMbiT Apr 19 '13 at 14:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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