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

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.

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What is $s$, a constant? –  Ron Gordon Mar 19 at 9:05
I assumed so. It is not a typo on my part. –  XxGaMbiT Mar 19 at 9:14

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

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

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