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Compute the double integral over S of F*dS for the given oriented surface. F = (x, y, z), part of...

Question:

Compute {eq}\; \iint_S F \cdot \, \mathrm{d}S \; {/eq} for the given oriented surface.

{eq}F = \left \langle x, y, z \right \rangle {/eq}, part of sphere {eq}x^2 + y^2 + z^2 = 1 {/eq}, where {eq}\frac{1}{2} \leq z \leq \frac{\sqrt{3}}{2} {/eq}, inward-pointing normal.

Calculus:

A mechanical quantity that represents the continuous variation in the function or equation with other variables is known as calculus. The integration and differentiation is part of the calculus analysis.

Answer and Explanation:


Given Data:

  • The function is: {eq}F = \left( {x,y,z} \right) {/eq}
  • The equation of sphere is: {eq}{x^2} + {y^2} + {z^2} = 1 {/eq}
  • {eq}\dfrac{1}{2} \le z \le \dfrac{{\sqrt 3 }}{2} {/eq}


Rearrange the equation of sphere

{eq}\begin{align*} {x^2} + {y^2} + {z^2} &= 1\\ {z^2} &= 1 - {x^2} - {y^2} \cdots\cdots\rm{(I)} \end{align*} {/eq}


Differentiate the above expression with respect to x

{eq}\begin{align*} \dfrac{{d{z^2}}}{{dx}} &= \dfrac{{d\left( {1 - {x^2} - {y^2}} \right)}}{{dx}}\\ 2z\dfrac{{dz}}{{dx}} &= - 2x\\ \dfrac{{dz}}{{dx}} &= - \dfrac{x}{z} \end{align*} {/eq}


Differentiate the expression (I) with respect to y

{eq}\begin{align*} \dfrac{{d{z^2}}}{{dy}} &= \dfrac{{d\left( {1 - {x^2} - {y^2}} \right)}}{{dy}}\\ 2z\dfrac{{dz}}{{dy}} &= - 2y\\ \dfrac{{dz}}{{dy}} &= - \dfrac{y}{z} \end{align*} {/eq}


The expression for surface area is

{eq}dS = \sqrt {1 + {{\left( {\dfrac{{dz}}{{dx}}} \right)}^2} + {{\left( {\dfrac{{dz}}{{dy}}} \right)}^2}} dxdy {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} dS &= \sqrt {1 + {{\left( { - \dfrac{x}{z}} \right)}^2} + {{\left( { - \dfrac{y}{z}} \right)}^2}} dxdy\\ &= \sqrt {1 + \dfrac{{{x^2}}}{{{z^2}}} + \dfrac{{{y^2}}}{{{z^2}}}} dxdy\\ &= \sqrt {\dfrac{{{z^2} + {x^2} + {y^2}}}{{{z^2}}}} dxdy\\ &= \sqrt {\dfrac{1}{{{z^2}}}} dxdy\\ &= \dfrac{1}{z}dxdy \end{align*} {/eq}


The expression for radius is

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


Substitute the value and solve the above expression

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


The limit of variable in polar coordinates is

{eq}\begin{align*} 0 &\le \theta \le 2\pi \\ \dfrac{1}{2} &\le \sqrt {1 - {R^2}} \le \dfrac{{\sqrt 3 }}{2}\\ \dfrac{1}{4} &\le 1 - {R^2} \le \dfrac{3}{4}\\ \dfrac{3}{4}& \le {R^2} \le \dfrac{1}{4}\\ \dfrac{{\sqrt 3 }}{2} &\le R \le \dfrac{1}{2} \end{align*} {/eq}


The expression for surface integral is

{eq}\begin{align*} \int {\int_S {F \cdot dS} } & = \int_{\frac{{\sqrt 3 }}{2}}^{\frac{1}{2}} {\int_0^{2\pi } {\dfrac{1}{{\sqrt {1 - {R^2}} }}} } d\theta RdR\\ &= \int_{\frac{{\sqrt 3 }}{2}}^{\frac{1}{2}} {\left[ \theta \right]_0^{2\pi }\dfrac{1}{{\sqrt {1 - {R^2}} }}RdR} \\ &= \int_{\frac{{\sqrt 3 }}{2}}^{\frac{1}{2}} {\left[ {2\pi - 0} \right]\dfrac{1}{{\sqrt {1 - {R^2}} }}RdR} \\ &= 2\pi \int_{\frac{{\sqrt 3 }}{2}}^{\frac{1}{2}} {\dfrac{1}{{\sqrt {1 - {R^2}} }}RdR} \cdots\cdots\rm{(II)} \end{align*} {/eq}


Let {eq}1 - {R^2} = V \cdots\cdots\rm{(III)} {/eq}


Differentiate the above expression

{eq}\begin{align*} \dfrac{{d\left( {1 - {R^2}} \right)}}{{dR}} &= \dfrac{{dV}}{{dR}}\\ - 2R &= \dfrac{{dV}}{{dR}}\\ dR &= - \dfrac{{dV}}{{2R}} \end{align*} {/eq}


Substitute the value and solve the expression (III)_at {eq}R = \dfrac{{\sqrt 3 }}{2} {/eq}

{eq}\begin{align*} 1 - {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} &= V\\ V &= 1 - \dfrac{3}{4}\\ &= \dfrac{1}{4} \end{align*} {/eq}


Substitute the value and solve the expression (III)_at {eq}R = \dfrac{1}{2} {/eq}

{eq}\begin{align*} 1 - {\left( {\dfrac{1}{2}} \right)^2} &= V\\ V &= 1 - \dfrac{1}{4}\\ &= \dfrac{3}{4} \end{align*} {/eq}


The limit of the variable {eq}R {/eq} is

{eq}\dfrac{1}{4} \le R \le \dfrac{3}{4} {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} \int {\int_S {F \cdot dS} } &= 2\pi \int_{\frac{1}{4}}^{\frac{3}{4}} {\dfrac{1}{{\sqrt V }}R\left( { - \dfrac{{dV}}{{2R}}} \right)} \\ &= - \pi \int_{\frac{1}{4}}^{\frac{3}{4}} {\dfrac{1}{{\sqrt V }}dV} \\ &= - 2\pi \left[ {{{\left( V \right)}^{\dfrac{1}{2}}}} \right]_{\dfrac{1}{4}}^{\dfrac{3}{4}}\\ &= - 2\pi \left[ {{{\left( {\dfrac{3}{4}} \right)}^{\dfrac{1}{2}}} - {{\left( {\dfrac{1}{4}} \right)}^{\dfrac{1}{2}}}} \right]\\ &= - 2\pi \left[ {\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2}} \right]\\ &= \pi \left[ {1 - \sqrt 3 } \right]\\ &= - 1.732\pi \end{align*} {/eq}


Thus the surface integral is {eq}- 1.732\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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