General Notice
I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
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Section 17.6 : Divergence Theorem
- 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
- 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
- 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