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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