# Evaluate the surface integral \iint_S F \cdot n dS, where \textbf F(x, y, z) = 4xy \textbf i +...

## Question:

Evaluate the surface integral {eq}\displaystyle \iint_S F \cdot n dS {/eq}, where {eq}\textbf F(x, y, z) = 4xy \textbf i + x^2\textbf j - 4yz\textbf k {/eq} and {eq}S {/eq} is the surface {eq}z = -3xe^y, 0 \leq x \leq 4, 0 \leq y \leq 2, {/eq} with upward orientation.

## Integration:

A mechanical quantity that represents the summation of the differential of the equation to form the basic equation is known as integration. It used in physics applications to solve the system equation.

Given Data:

• The surface integral is: {eq}\int {\int_S {F \cdot ndS} } {/eq}
• The function of surface integral is: {eq}F\left( {x,y,z} \right) = 4xyi + {x^2}j - 4yzk {/eq}
• {eq}z = - 3x{e^y} {/eq}
• {eq}0 \le x \le 4 {/eq}
• {eq}0 \le y \le 2 {/eq}

The diversion of function of surface integral is

{eq}\begin{align*} \nabla \cdot F\left( {x,y,z} \right) &= \left( {\dfrac{d}{{dx}},\dfrac{d}{{dy}},\dfrac{d}{{dz}}} \right) \cdot \left( {4xy,{x^2},4yz} \right)\\ &= 4y + 0 + 4y\\ &= 8y \end{align*} {/eq}

The expression for surface area is

{eq}\begin{align*} dS &= \left( {\dfrac{d}{{dx}},\dfrac{d}{{dy}},\dfrac{d}{{dz}}} \right)\left( { - 3x{e^y}} \right)dA\\ &= - 3{e^y} - 3x{e^y} \end{align*} {/eq}

Substitute the value and solve the surface integral with respective limit is

{eq}\begin{align*} \int {\int_S {F \cdot ndS} } &= \int_0^2 {\int_0^4 {8y\left( { - 3{e^y} - 3x{e^y}} \right)dxdy} } \\ &= \int_0^2 {\int_0^4 {\left( { - 24y{e^y} - 24yx{e^y}} \right)dxdy} } \\ &= - 24\int_0^2 {\left[ {xy{e^y} + y{e^y}\dfrac{{{x^2}}}{2}} \right]_0^4dy} \\ &= - 24\int_0^2 {\left[ {\left( {4 - 0} \right)y{e^y} + y{e^y}\dfrac{{{{\left( 4 \right)}^2} - 0}}{2}} \right]dy} \\ &= - 24\int_0^2 {\left[ {4y{e^y} + 8y{e^y}} \right]} dy\\ &= - 24\int_0^2 {\left[ { 12y{e^y}} \right]} dy\\ &= -288\int_0^2 {y{e^y}} dy \end{align*} {/eq}

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

{eq}\begin{align*} \int {\int_S {F \cdot ndS} } &= 96\int_0^2 {y{e^y}} dy\\ &= -288\left[ {y{e^y}} \right]_0^2 - \dfrac{{dy}}{{dy}}\int_0^2 {{e^y}dy} \\ &= -288\left[ {\left( 2 \right){e^2} - 0} \right] - \left[ {{e^y}} \right]_0^2\\ &= -288\left[ {2{e^2}} \right] - \left[ {{e^2} - {e^0}} \right]\\ &= -288\left( {{e^2} + 1} \right) \end{align*} {/eq}

Thus the surface integral is {eq}-288\left( {{e^2} + 1} \right) {/eq}

Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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