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Evaluate \iint_{s}\overrightarrow{F} \cdot d\overrightarrow{S}, where \overrightarrow{F}(x, y,...

Question:

Evaluate {eq}\iint_{s}\overrightarrow{F} \cdot d\overrightarrow{S}{/eq}, where {eq}\overrightarrow{F}(x, y, z)= \left \langle x^2,xy,z \right \rangle {/eq} and {eq}S{/eq} is the part of the paraboloid {eq}z=x^2+y^2{/eq} below the plane {eq}z=1{/eq}, with upward orientation.

Integration:

A mechanical quantity that represents the merge of the derivative of the function or equation in combine form in the respective domain to analyse the system is known as integration. It is part of the calculus analysis.

Answer and Explanation:

GIVEN DATA:

  • The function is: {eq}\mathop F\limits^ \to \left( {x,y,z} \right) = \left\langle {{x^2},xy,z} \right\rangle {/eq}
  • The equation of parabola is: {eq}z = {x^2} + {y^2} {/eq}


The function is written as

{eq}\mathop F\limits^ \to = xi + yj + zk {/eq}


Substitute the value in above expression

{eq}\mathop F\limits^ \to = {x^2}i + xyj + \left( {{x^2} + {y^2}} \right)k {/eq}


Differentiate the equation of paraboloid with respect to x

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


Differentiate the equation of paraboloid with respect to y

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


The S is part of the paraboloid {eq}z = {x^2} + {y^2} {/eq} below the plane {eq}z = 1 {/eq} with upward orientated

The expression for integral {eq}\int {\int_S {\mathop F\limits^ \to \cdot \mathop {dS}\limits^ \to } } {/eq} is

{eq}\int {\int_S {\mathop F\limits^ \to \cdot \mathop {dS}\limits^ \to } } = \int {\int_S {\left( { - {x^2}\left( {\dfrac{{dz}}{{dx}}} \right) - xy\left( {\dfrac{{dz}}{{dy}}} \right) + \left( {{x^2} + {y^2}} \right)} \right)dA} } {/eq}


Substitute the value in above expression

{eq}\begin{align*} \int {\int_S {\mathop F\limits^ \to \cdot \mathop {dS}\limits^ \to } } &= \int {\int_S {\left( { - {x^2}\left( {2x} \right) - xy\left( {2y} \right) + \left( {{x^2} + {y^2}} \right)} \right)} } dA\\ &= \int {\int_S {\left( { - 2{x^3} - 2x{y^2} + \left( {{x^2} + {y^2}} \right)} \right)} } dA\\ &= \int {\int_S {\left( {1 - 2x} \right)\left( {{x^2} + {y^2}} \right)dA} } \cdots\cdots\rm{(I)} \end{align*} {/eq}


The expression for radius is

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


The radius by compare above expression with equation of paraboloid is {eq}R = 1 {/eq}


Convert the integral in polar form

The coordinate of x in polar form is

{eq}x = R\cos \theta {/eq}


The domain of the integral is

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


Substitute the value in expression (I) to form polar integral

{eq}\begin{align*} \int {\int_S {\mathop F\limits^ \to \cdot \mathop {dS}\limits^ \to } } &= \int_0^{2\pi } {\int_0^1 {\left( {1 - 2R\cos \theta } \right){R^2}} \left( R \right)dRd\theta } \\ &= \int_0^{2\pi } {\int_0^1 {\left( {{R^3} - 2{R^4}\cos \theta } \right)dRd\theta } } \\ &= \int_0^{2\pi } {\left[ {\dfrac{{{R^4}}}{4} - \dfrac{{2{R^5}\cos \theta }}{5}} \right]_0^1} d\theta \\ &= \int_0^{2\pi } {\left[ {\left( {\dfrac{{{{\left( 1 \right)}^4}}}{4} - \dfrac{{2{{\left( 1 \right)}^5}\cos \theta }}{5}} \right) - 0} \right]} d\theta \\ &= \int_0^{2\pi } {\left[ {\left( {\dfrac{1}{4} - \dfrac{{2\cos \theta }}{5}} \right)} \right]} d\theta \end{align*} {/eq}


Integrate the above expression with respect t to {eq}\theta {/eq}

{eq}\begin{align*} \int {\int_S {\mathop F\limits^ \to \cdot \mathop {dS}\limits^ \to } } &= \int_0^{2\pi } {\left[ {\left( {\dfrac{1}{4} - \dfrac{{2\cos \theta }}{5}} \right)} \right]} d\theta \\ &= \left[ {\dfrac{\theta }{4} - \dfrac{{2\sin \theta }}{5}} \right]_0^{2\pi }\\ &= \left[ {\left( {\dfrac{{2\pi }}{4} - \dfrac{{2\sin \left( {2\pi } \right)}}{5}} \right) - 0} \right]\\ &= \dfrac{\pi }{2} \end{align*} {/eq}


Thus the value of integral {eq}\int {\int_S {\mathop F\limits^ \to \cdot \mathop {dS}\limits^ \to } } {/eq} is {eq}\dfrac{\pi }{2} {/eq}


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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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