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I've been given a region bounded by the values $z = x^2 + y^2$ and $z = 1$, with a force of $F = 2i + xj + z^3k$.

So with Gauss' theorem I have been able to deduce:

$$\int\int_S -F1f_x - F2f_y + F3dS = \int\int\int_V \frac{\delta}{\delta{x}}F1 + \frac{\delta}{\delta{y}}F2 + \frac{\delta}{\delta{z}}F3dV$$

$$-\int\int_S 2 + z^3dS = \int\int\int_V 2z^2dV$$ S negative as normal is downwards. Bounds for the Surface $S$ would then be: $0 \le r \le 1$ and $0 \le \theta \le 2\pi$, and volume V will be $0 \le r \le 1$, $r^2 \le z \le 1$ and $0 \le \theta \le 2\pi$. As $z = x^2 + y^2 = r^2, z^3 = r^6$.

So now I have: $$-\int_0^{2\pi}\int_0^1 (-2 + r^6)rdrd\theta = \int_0^{2\pi}\int_0^1\int_{r^2}^1 2z^2rdzdrd\theta$$

Which ends up as

$$\frac{7\pi}{4} = \frac{\pi}{2}$$

Not even close. What have I done wrong?

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    The boundary of the region $V$ consists of $S$ together with the disk of radius $1$ at $z=1$. – 2017-01-19

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We know that $\nabla \cdot F =3z^2$, therefore we need $$\iiint _{x^2+y^2}^1 3z^2 \mathrm{d}z\mathrm{d}y\mathrm{d}x=\iint 1-(x^2+y^2)^3\mathrm{d}y\mathrm{d}x$$ Because $x^2+y^2$ needs to be less than one, we are integrating over the unit circle and can therefore use polar coordinates to get $$\int_0^{2\pi}\int_0^1 r-r^7\mathrm{d}r\mathrm{d}\theta=\frac 34 \pi$$

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    Hey, how come it becomes 1 - r^7, and not 2 - r^7? F1fx should result in a 2, shouldn't it? – 2017-01-19
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    By the divergence theorem, both sides are equal to each other. One uses this statement to make the computations easier. So there is nothing to be balanced, usually, computing surface integrals is harder than volume integrals, so one can use this theorem as a shortcut. – 2017-01-19
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    @Laefica the partial derivative of a constant is $0$, so those terms fall out. – 2017-01-19
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    Isn't it the partial derivative of the whole equation multiplied by the constant? – 2017-01-19