Section 17.4 : Surface Integrals of Vector Fields

1. Evaluate $\displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}$ where $\vec F = 3x\,\vec i + 2z\,\vec j + \left( {1 - {y^2}} \right)\vec k$ and $S$ is the portion of $z = 2 - 3y + {x^2}$ that lies over the triangle in the xy-plane with vertices $\left( {0,0} \right)$, $\left( {2,0} \right)$ and $\left( {2, - 4} \right)$ oriented in the negative $z$-axis direction. Solution
2. Evaluate $\displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}$ where $\vec F = - x\,\vec i + 2y\,\vec j - z\,\vec k$ and $S$ is the portion of $y = 3{x^2} + 3{z^2}$ that lies behind $y = 6$ oriented in the positive $y$-axis direction. Solution
3. Evaluate $\displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}$ where $\vec F = {x^2}\,\vec i + 2z\,\vec j - 3y\,\vec k$ and $S$ is the portion of ${y^2} + {z^2} = 4$ between $x = 0$ and $x = 3 - z$ oriented outwards (i.e. away from the $x$-axis). Solution
4. Evaluate $\displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}$ where $\vec F = \,\vec i + z\,\vec j + 6x\,\vec k$ and $S$ is the portion of the sphere of radius 3 with $x \le 0$, $y \ge 0$ and $z \ge 0$ oriented inward (i.e. towards the origin). Solution
5. Evaluate $\displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}$ where $\vec F = y\,\vec i + 2x\,\vec j + \left( {z - 8} \right)\,\vec k$ and $S$ is the surface of the solid bounded by $4x + 2y + z = 8$, $z = 0$, $y = 0$ and $x = 0$ with the positive orientation. Note that all four surfaces of this solid are included in $S$. Solution
6. Evaluate $\displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}$ where $\vec F = yz\,\vec i + x\,\vec j + 3{y^2}\,\vec k$ and $S$ is the surface of the solid bounded by ${x^2} + {y^2} = 4$, $z = x - 3$, and $z = x + 2$ with the negative orientation. Note that all three surfaces of this solid are included in $S$. Solution