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Find the flux of the vector field \vec F(x, y, z) = \langle z, y, -x\rangle through the surface...

Question:

Find the flux of the vector field {eq}\vec F(x, y, z) = \langle z, y, -x\rangle {/eq} through the surface {eq}S {/eq}, which is the top of the sphere {eq}x^2 + y^2 + z^2 = 2 {/eq} above the plane {eq}z = 1 {/eq}.

Integration:

A mathematical quantity that represents the summation of the derivative of the function to form an equation of the motion is known as integration. It used in engineering application to solve the complex problem.

Answer and Explanation:


Given Data:

  • The vector field is: {eq}\mathop F\limits^ \to \left( {x,y,z} \right) = \left( {z,y, - x} \right) {/eq}
  • The equation of sphere is: {eq}{x^2} + {y^2} + {z^2} = 2 {/eq}
  • {eq}z = 1 {/eq}


The expression for diversion of vector field is

{eq}\begin{align*} \nabla \cdot \mathop F\limits^ \to &= \left( {\dfrac{d}{{dx}},\dfrac{d}{{dy}},\dfrac{d}{{dz}}} \right) \cdot \left( {z,y, - x} \right)\\ &= \dfrac{{dz}}{{dx}} + \dfrac{{dy}}{{dy}} + \dfrac{{d\left( { - x} \right)}}{{dz}}\\ &= 0 + 1 + 0\\ &= 1 \end{align*} {/eq}


Substitute the value in equation of sphere is

{eq}\begin{align*} {x^2} + {y^2} + {\left( 1 \right)^2} &= 2\\ {x^2} + {y^2} &= 1 \end{align*} {/eq}


The expression for radius of sphere is

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


Substitute the value and solve the above expression

{eq}\begin{align*} 1 &= {R^2}\\ R &= 1 \end{align*} {/eq}


The expression for flux of vector field is

{eq}\int {\int_S {\mathop F\limits^ \to \cdot ndS = \int {\int {\int_V {\left( {\nabla \cdot \mathop F\limits^ \to } \right)dV} } } } } {/eq}


The expression for volume in spherical coordinates

{eq}dV = RdzdRd\theta {/eq}


The domain of coordinates is

{eq}\begin{align*} 0 &\le R \le 1\\ 0 &\le z \le 1\\ 0 &\le \theta \le 2\pi \end{align*} {/eq}


The expression for flux of vector field in spherical coordinates is

{eq}\int {\int_S {\mathop F\limits^ \to \cdot ndS = \int_0^{2\pi } {\int_0^1 {\int_0^1 {1 \cdot RdRdzd\theta } } } } } {/eq}

Solve the above expression

{eq}\begin{align*} \phi &= \int_0^{2\pi } {\int_0^1 {\left[ {\dfrac{{{{\left( 1 \right)}^2}}}{2} - 0} \right]} } dzd\theta \\ &= \int_0^{2\pi } {\int_0^1 {\dfrac{1}{2}} } dzd\theta \\ & = \dfrac{1}{2}\int_0^{2\pi } {d\theta } \int_0^1 {dz} \\ &= \dfrac{1}{2}\left[ \theta \right]_0^{2\pi }\left[ z \right]_0^1\\ &= \dfrac{1}{2}\left[ {2\pi - 0} \right]\left[ {1 - 0} \right]\\ &= \pi \end{align*} {/eq}


Thus the flux of vector field is {eq}\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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