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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?

Laefica
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1 Answers1

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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 $$\int_0^{2\pi}\int_0^1 r-r^7\mathrm{d}r\mathrm{d}\theta=\frac 34 \pi$$

GuPe
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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? – Laefica Jan 19 '17 at 07:28
  • 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. – GuPe Jan 19 '17 at 07:30
  • @Laefica the partial derivative of a constant is $0$, so those terms fall out. – GuPe Jan 19 '17 at 07:31
  • Isn't it the partial derivative of the whole equation multiplied by the constant? – Laefica Jan 19 '17 at 08:02