Evaluate double integral_S F vector . d vector S, where S is the surface z = 5 - x^2 - y^2 above...


Evaluate {eq}\displaystyle \iint_S \vec F \cdot d \vec S {/eq}, where S is the surface {eq}z = 5 - x^2 - y^2 {/eq} above {eq}z = 1 {/eq} oriented upward and {eq}\displaystyle \vec F = \langle -2 y z,\ z^2,\ 3 x \rangle {/eq}.

Divergence theorem

Suppose {eq}V {/eq} is a subset of {eq}\mathbb{R}^{3} {/eq} which is compact and has a piecewise smooth boundary {eq}S {/eq} such that {eq}\partial V {/eq} identical to {eq}S {/eq}. If {eq}F {/eq} is a continuously differentiable vector field defined on a neighborhood of {eq}V {/eq}, then we have

$$\iint_S \vec F \cdot d \vec S = \iiint_V \nabla\vec F \cdot dV\\ $$

Answer and Explanation:

Straightforward by divergence theorem, we find

$$\begin{align} \iint_S \vec F \cdot d \vec S &= \iiint_V \nabla\vec F \cdot dV\\ &= \iiint_V \Big(\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}+\frac{\partial F}{\partial z}\Big) dV\\ &= \iiint_V \Big(0+0+0\Big) dV\\ &=0 \end{align} $$

Learn more about this topic:

Calculating Integrals of Simple Shapes

from Math 104: Calculus

Chapter 13 / Lesson 1

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