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Evaluate the surface integral. double integral_S x^2 z^2 dS, where S is the part of the cone z^2...

Question:

Evaluate the surface integral. {eq}\displaystyle \iint_S x^2 z^2\ dS {/eq}, where {eq}S {/eq} is the part of the cone {eq}z^2 = x^2 + y^2 {/eq} that lies between the planes {eq}\displaystyle z = 2\ \text{and}\ z = 4 {/eq}.

Integration:

A mathematical tool used to blend the data or function to form the original equation or function of the system is known as integration. It widely used in computational fluid dynamics to know the behaviour of the fluid flow.

Answer and Explanation:

Given Data:

  • The surface integral is: {eq}\int\int_Sx^2z^2dS \cdots\cdots\rm{(I)} {/eq}
  • {eq}{z^2} = {x^2} + {y^2} \cdots\cdots\rm{(II)} {/eq}
  • {eq}z = 2 {/eq}
  • {eq}z = 4 {/eq}


Rearrange the expression (II)

{eq}\begin{align*} {z^2} &= {x^2} + {y^2}\\ z &= \sqrt {{x^2} + {y^2}} \end{align*} {/eq}


Differentiate the above expression with respect to x

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


Differentiate the above expression with respect to y

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


The expression for surface arc of the curve is

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


Substitute and solve the above expression is

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


In polar coordinates the value of x and y will be

{eq}\begin{align*} x &= R\cos \theta \\ y &= R\sin \theta \\ dxdy &= RdRd\theta \end{align*} {/eq}


The domain will be

{eq}\begin{align*} 0 \le \theta \le 2\pi \\ 2 \le R \le 4 \end{align*} {/eq}

The radius of curvature in polar coordinate is:

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


Substitute and solve the expression (I) is

{eq}\begin{align*} S&=\int\int_Sx^2z^2dS\\ &= \int_0^{2\pi } {\int_2^4 {\sqrt 2 {R^5}{{\cos }^2}\theta dRd\theta } } \\ &= \sqrt 2 \int_0^{2\pi } {\left( {\dfrac{{1 + \cos 2\theta }}{2}} \right)d\theta } \int_2^4 {{R^5}} dR\\ &= \sqrt 2 \left[ {\dfrac{\theta }{2} + \dfrac{{\sin 2\theta }}{4}} \right]_0^{2\pi }\left[ {\dfrac{{{R^6}}}{6}} \right]_2^4\\ &= \sqrt 2 \left[ {\dfrac{{2\pi }}{2} + \dfrac{{\sin 2\left( {2\pi } \right)}}{4} - 0} \right]\left[ {\dfrac{{{{\left( 4 \right)}^6}}}{6} - \dfrac{{{2^6}}}{6}} \right]\\ &= \sqrt 2 \pi \left[ {672} \right] \end{align*} {/eq}


Thus the surface integral is {eq}\sqrt 2 \pi \left[ {672} \right]{/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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