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Let $\omega \subset \mathbb{R}^3 \ be \ the\ solid \ given \ by \ \{(x,y,z):x^2+y^2+z^2\leq1,x+2y-z\geqslant0\}. Consider \ the \ gradient \ field \ F(x,y,z)=(2x-x^2y,xy^2,-z). \ Knowing\ that \ in \ \mathbb{R}^3 \ the \ volume \ of \ a \ ball \ is \ \frac{4}{3}\pi r^3. Use\ the\ divergence\ theorem\ to\ calculate\ \iint_{\partial\omega}dS $

I can simply calculate $div F=1$. My question is how I get the values that x,y and z will vary(be limited) in the triple integral? Without knowing this I cant solve the triple integral.

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As you write correctly, we have $$ \int_{\partial\omega} \langle F,n\rangle\,dS = \int_{\omega}\def\div{\mathop{\rm div}} \div F\, d(x,y,z) $$ Your domain of integration is $\omega$, that is the part of the unit ball, where $x-2y+z \ge 0$. Note that $x-2y+z = 0$ describes a plane through the origin, that is, it splits the unit ball in two halves. Therefore $\omega$ is half a unit ball, hence its volume is $\lambda(\omega) = \frac 12 \cdot \frac 43\pi =\frac 23\pi$. Hence $$ \int_{\partial\omega} \langle F,n\rangle\,dS = \int_{\omega}\def\div{\mathop{\rm div}} \div F\, d(x,y,z) = \frac 23 \pi$$