Find the flux of the vector field F across the surface \sigma by expressing \sigma...

Question:

Find the flux of the vector field F across the surface {eq}\sigma {/eq} by expressing {eq}\sigma {/eq} parametrically. F(x, y, z) = 8i - 4j + zk where {eq}\sigma {/eq} is the portion of the cylinder {eq}x^2 + y^2 = 4 {/eq} between the planes z = -2 and z = 2, oriented by outward unit normals.

Integration:

A mathematical quantity that represents the summation of differential of the function to produce the equation of the system is known as integration. It used in engineering application.

Answer and Explanation:

Given Data:

  • The vector field is: {eq}F\left( {x,y,z} \right) = 8i + 4j + zk {/eq}
  • {eq}x^2 + {y^2} = 4 {/eq}
  • {eq}z = - 2 {/eq}
  • {eq}z = 2 {/eq}


The expression for the radius is

{eq}{r^2} = {x^2} + {y^2} {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} {r^2} &= 4\\ r &= 2 \end{align*} {/eq}


The interval of the variable is

{eq}\begin{align*} 0 &\le r \le 2\\ - 2 &\le z \le 2\\ 0 &\le \theta \le 2\pi \end{align*} {/eq}


The expression for diversion of the vector field is

{eq}\nabla \cdot F = \dfrac{{dF}}{{dx}} + \dfrac{{dF}}{{dy}} + \dfrac{{dF}}{{dz}} {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} \nabla \cdot F &= \dfrac{{d\left( {8i + 4j + zk} \right)}}{{dx}} + \dfrac{{d\left( {8i + 4j + zk} \right)}}{{dy}} + \dfrac{{d\left( {8i + 4j + zk} \right)}}{{dz}}\\ &= 0 - 0 + 1\\ &= 1 \end{align*} {/eq}


The expression for flux of the vector field by Gauss Theorem is

{eq}\int {\int {\int_V {\nabla \cdot FdV = } } } \int {\int {\int_V {\nabla \cdot Frdzdrd\theta } } } {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} \int {\int {\int_V {\nabla \cdot FdV } } }&= \int_0^{2\pi } {\int_0^2 {\int_{ - 2}^2 {1 \cdot rdzdrd\theta } } } \\ &= \int_0^{2\pi } {\int_0^2 {\left[ z \right]_{ - 2}^2rdrd\theta } } \\ &= \int_0^{2\pi } {\int_0^2 {2 - \left( { - 2} \right)rdrd\theta } } \\ &= \int_0^{2\pi } {\int_0^2 {4rdrd\theta } } \\ &= 4\int_0^{2\pi } {\left[ {\dfrac{{{r^2}}}{2}} \right]_0^2d\theta } \\ &= 4\int_0^{2\pi } {\left[ {\dfrac{{{2^2}}}{2} - 0} \right]d\theta } \\ &= 4\int_0^{2\pi } 2 d\theta \\ &= 8\int_0^{2\pi } {d\theta } \end{align*} {/eq}


Integrate the above expression with respect to \theta

{eq}\begin{align*} \int {\int {\int_V {\nabla \cdot FdV } } }&= 8\int_0^{2\pi } {d\theta } \\ &= 8\left[ \theta \right]_0^{2\pi }\\ &= 8\left[ {2\pi - 0} \right]\\ &= 16\pi \end{align*} {/eq}


Thus the flux of the vector field is {eq}16\pi {/eq}


Learn more about this topic:

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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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