# Consider the surface S formed by the graph of z = 20x^4y^3 over the unit square [0,1] times...

## Question:

Consider the surface {eq}S {/eq} formed by the graph of {eq}z = 20x^4y^3 {/eq} over the unit square {eq}[0,1] \times [0,1]. {/eq}. Orient {eq}S {/eq} by normals that have a positive z - component.Integrate the vector field {eq}F = 3x i - 5y j + 6z k {/eq} over {eq}S {/eq}

## Vector Surface Integrals over the Graph of a Function:

Let {eq}g(x,y) {/eq} be a function of two variables, and suppose that the surface {eq}S {/eq} is formed by the graph of {eq}z=g(x,y) {/eq} over some domain {eq}R {/eq} in the {eq}xy {/eq}-plane, oriented with upward-pointing normal. Let {eq}\vec{F} {/eq} be a vector field whose domain contains {eq}S {/eq}. Then the flux integral (that is, the vector surface integral) of {eq}\vec{F} {/eq} over {eq}S {/eq} is given by the following formula:

{eq}\displaystyle \iint_S \vec{F} \cdot \vec{dS}=\iint_R \vec{F}(x,y,g(x,y)) \cdot (-g_x\mathbf{i}-g_y\mathbf{j}+\mathbf{k}) \, dA \, . {/eq}

## Answer and Explanation:

Let {eq}g(x,y)=20x^4y^3 {/eq}, so the surface {eq}S {/eq} is the graph of {eq}g(x,y) {/eq} over the unit square in the {eq}xy {/eq}-plane. Differentiating {eq}g {/eq} gives:

{eq}\begin{align*} g_x&=80x^3y^3\\ g_y&=60x^4y^2 \, . \end{align*} {/eq}

By the formula for a flux integral over a graph, we therefore have:

{eq}\begin{align*} \iint_S \vec{F} \cdot \vec{dS}&=\iint_{[0,1]\times [0,1]} \vec{F}(x,y,g(x,y)) \cdot (-g_x\mathbf{i}-g_y\mathbf{j}+\mathbf{k}) \, dA\\ &=\iint_{[0,1] \times [0,1]} \vec{F}(x,y,20x^4y^3) \cdot (-80x^3y^3\mathbf{i}-60x^4y^2\mathbf{j}+\mathbf{k}) \, dA&&\text{(by the above expressions for }g\text{ and its derivatives)}\\ &=\iint_{[0,1] \times [0,1]} (3x\mathbf{i}-5y\mathbf{j}+6(20x^4y^3)\mathbf{k}) \cdot (-80x^3y^3\mathbf{i}-60x^4y^2\mathbf{j}+\mathbf{k}) \, dA&&\text{(by the given expression for }\vec{F}\text{)}\\ &=\iint_{[0,1] \times [0,1]} (3x\mathbf{i}-5y\mathbf{j}+120x^4y^3\mathbf{k}) \cdot (-80x^3y^3\mathbf{i}-60x^4y^2\mathbf{j}+\mathbf{k}) \, dA\\ &=\iint_{[0,1] \times [0,1]} \left[3x(-80x^3y^3)-5y(-60x^4y^2)+120x^4y^3(1)\right] \, dA&&\text{(evaluating the dot product)}\\ &=\iint_{[0,1] \times [0,1]} (-240x^4y^3+300x^4y^3+120x^4y^3) \, dA\\ &=\iint_{[0,1] \times [0,1]} 180x^4y^3 \, dA&&\text{(combining like terms)}\\ &=\int_{x=0}^1\int_{y=0}^1 180x^4y^3 \, dy \, dx&&\text{(writing the area integral as a Cartesian iterated integral)}\\ &=\int_{x=0}^1\Big(60x^4y^4\Big|_{y=0}^1 \, dx&&\text{(evaluating the }y\text{-integral)}\\ &=\int_{x=0}^1 60x^4 \, dx\\ &=\Big(12x^5\Big|_0^1&&\text{(evaluating the }x\text{-integral)}\\ &=12 \, . \end{align*} {/eq}

In summary, {eq}\boxed{\iint_S \vec{F} \cdot \vec{dS} = 12 \, .} {/eq}

#### Learn more about this topic:

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4