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I just want to check if what I've done is correct.

We know by the Divergence Theorem that the integral of the divergence over a volume yields the value of the function at the boundary, which is a surface:

$$\int_V (\nabla \cdot \vec F )dV= \int_S \vec F \cdot d \vec S$$

Then we can use it so as to compute the flux through a surface:

$$\nabla \cdot \vec F = -2z$$

$$-2\iiint zdV = -2\int_{0}^{1}\int_{1}^{x^2}\int_{0}^{1}zdV = \frac{32}{3}$$

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  • $\begingroup$ Not quite, $z$ varies between $0$ and $4$ $\endgroup$ – Rafa Budría Apr 2 at 10:18
  • $\begingroup$ @RafaBudría Typo, my bad. The integral is solved using the boundaries: $$-2\iiint zdV = -2\int_{0}^{1}\int_{1}^{x^2}\int_{0}^{4}zdV = \frac{32}{3}$$ $\endgroup$ – JD_PM Apr 2 at 11:07

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