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### Section 6-6 : Divergence Theorem

1. Use the Divergence Theorem to evaluate $$\displaystyle \iint\limits_{S}{{\vec F\centerdot d\vec S}}$$ where $$\vec F = y{x^2}\,\vec i + \left( {x{y^2} - 3{z^4}} \right)\,\vec j + \left( {{x^3} + {y^2}} \right)\,\vec k$$ and $$S$$ is the surface of the sphere of radius 4 with $$z \le 0$$ and $$y \le 0$$. Note that all three surfaces of this solid are included in $$S$$. Solution
2. Use the Divergence Theorem to evaluate $$\displaystyle \iint\limits_{S}{{\vec F\centerdot d\vec S}}$$ where $$\vec F = \sin \left( {\pi x} \right)\,\vec i + z{y^3}\,\vec j + \left( {{z^2} + 4x} \right)\,\vec k$$ and $$S$$ is the surface of the box with $$- 1 \le x \le 2$$, $$0 \le y \le 1$$ and $$1 \le z \le 4$$. Note that all six sides of the box are included in $$S$$. Solution
3. Use the Divergence Theorem to evaluate $$\displaystyle \iint\limits_{S}{{\vec F\centerdot d\vec S}}$$ where $$\vec F = 2xz\vec i + \left( {1 - 4x{y^2}} \right)\,\vec j + \left( {2z - {z^2}} \right)\,\vec k$$ and $$S$$ is the surface of the solid bounded by $$z = 6 - 2{x^2} - 2{y^2}$$ and the plane $$z = 0$$ . Note that both of the surfaces of this solid included in $$S$$. Solution