Compute the surface integral over the given oriented surface: \vec{F} = \left \langle 0, 0,...

Question:

Compute the surface integral over the given oriented surface:

{eq}\vec{F} = \left \langle 0, 0, e^{y+z} \right \rangle{/eq}, boundary of the cube {eq}0 \leq x,y,z \leq 5{/eq}, outward-pointing normal

Surface Integrals:

Let {eq}\vec{F} = \langle f(x, y, z), g(x, y, z), h(x, y, z)\rangle {/eq} be a continuous vector field defined on an oriented surface {eq}S {/eq} given by a vector equation {eq}\vec{r}(s, t) = \langle x(s, t), y(s, t), z(s, t)\rangle {/eq}, where {eq}a \leq s \leq b {/eq} and {eq}c \leq t\leq d {/eq}. Then the surface integral of {eq}\vec{F} {/eq} over {eq}S {/eq} is

{eq}\displaystyle \iint\limits_S \vec{F}\cdot d\vec{S} = \iint\limits_S\vec{F}\cdot \vec{n}\, dS {/eq},

where {eq}\vec{n} {/eq} is the unit normal vector given by {eq}\displaystyle\vec{n} = \frac{\vec{r}_s \times \vec{r}_t}{|\vec{r}_s \times \vec{r}_t|} {/eq}. Hence, we have

{eq}\displaystyle \iint\limits_S \vec{F}\cdot d\vec{S} = \iint\limits_D\vec{F}\cdot (\vec{r}_s \times \vec{r}_t)\, dA {/eq},

where {eq}D {/eq} is the parameter domain. If the surface {eq}S {/eq} is given by {eq}z= G(x, y) {/eq} we can consider {eq}S {/eq} as a parametric surface with parametric equations

{eq}x=x\quad \quad y= y \quad \quad z = G(x, y) {/eq}

and so

{eq}\displaystyle \vec{r}_x = \left\langle 1, 0, \frac{\partial z}{\partial x}\right\rangle \quad \quad \vec{r}_y = \left\langle 0, 1, \frac{\partial z}{\partial y}\right\rangle {/eq}.

Hence,

{eq}\displaystyle \vec{r}_x \times \vec{r}_y= \left\langle -\frac{\partial z}{\partial x}, -\frac{\partial z}{\partial y}, 1\right\rangle {/eq},

which means

{eq}\displaystyle \vec{F}\cdot (\vec{r}_x \times \vec{r}_y) = \left\langle f(x, y, z), g(x, y, z), h(x, y, z)\right\rangle \cdot \left\langle - \frac{\partial z}{\partial x}, - \frac{\partial z}{\partial y},1\right\rangle {/eq},

and therefore

{eq}\displaystyle \iint\limits_S\vec{F}\cdot d\mathbf{S} = \iint\limits_D\left(-f(x, y, z)\frac{\partial z}{\partial x} -g(x, y, z)\frac{\partial z}{\partial y} + h(x, y, z)\right)\, dA {/eq}.

This formula assumes the upward orientation of {eq}S {/eq}; for the downward orientation we have to multiply by {eq}-1 {/eq}.

Answer and Explanation:

Let {eq}S_1 {/eq} be the face of the cube in the plane {eq}x=0 {/eq}, {eq}S_2 {/eq} the face of the cube in the plane {eq}x=5 {/eq}, {eq}S_3 {/eq} the face of the cube in the plane {eq}y=0 {/eq}, {eq}S_4 {/eq} the face of the cube in the plane {eq}y=5 {/eq}, {eq}S_5 {/eq} the face of the cube in the plane {eq}z=0 {/eq} and {eq}S_6 {/eq} the face of the cube in the plane {eq}z=5 {/eq}. Note that the unit normal vector to {eq}S_1 {/eq} is {eq}\vec{n_1} = \langle -1, 0, 0\rangle {/eq}, so we have {eq}d\vec{S_1} = \vec{n_1}\,dS_1 = \langle -1, 0, 0\rangle\, dS_1 {/eq}. Hence,

{eq}\begin{align*} \iint\limits_{S_1} \vec{F}\cdot d\vec{S_1} & = \iint\limits_{S_1}\vec{F}\cdot \vec{n_1}\, dS_1\\ & = \iint\limits_{S_1} \left \langle 0, 0, e^{y+z} \right \rangle\cdot \langle -1, 0, 0\rangle\, dS_1\\ &= \iint\limits_{S_1} 0\, dS_1\\ &= 0 \end{align*} {/eq}

For {eq}S_2 {/eq} we have {eq}\vec{n_2} = \langle 1, 0, 0\rangle {/eq}, so we have {eq}d\vec{S_2} = \vec{n_2}\,dS_2 = \langle 1, 0, 0\rangle\, dS_2 {/eq}, and therefore

{eq}\begin{align*} \iint\limits_{S_2} \vec{F}\cdot d\vec{S_2} & = \iint\limits_{S_2}\vec{F}\cdot \vec{n_2}\, dS_2\\ & = \iint\limits_{S_2} \left \langle 0, 0, e^{y+z} \right \rangle\cdot \langle 1, 0, 0\rangle\, dS_1\\ &= \iint\limits_{S_2} 0\, dS_2\\ &= 0 \end{align*} {/eq}

The unit normal vector to {eq}S_3 {/eq} is {eq}\vec{n_3} = \langle 0, -1, 0\rangle {/eq}, so we have {eq}d\vec{S_3} = \vec{n_3}\,dS_3 = \langle 0, -1, 0\rangle\, dS_3 {/eq}. Hence,

{eq}\begin{align*} \iint\limits_{S_3} \vec{F}\cdot d\vec{S_3} & = \iint\limits_{S_3}\vec{F}\cdot \vec{n_3}\, dS_3\\ & = \iint\limits_{S_3} \left \langle 0, 0, e^{y+z} \right \rangle\cdot \langle 0, -1, 0\rangle\, dS_3\\ &= \iint\limits_{S_3} 0\, dS_3\\ &= 0 \end{align*} {/eq}

For {eq}S_4 {/eq} we have {eq}\vec{n_4} = \langle 0, 1, 0\rangle {/eq}, so we have {eq}d\vec{S_4} = \vec{n_4}\,dS_4 = \langle 0, 1, 0\rangle\, dS_4 {/eq}, and hence

{eq}\begin{align*} \iint\limits_{S_4} \vec{F}\cdot d\vec{S_4} & = \iint\limits_{S_4}\vec{F}\cdot \vec{n_4}\, dS_4\\ & = \iint\limits_{S_4} \left \langle 0, 0, e^{y+z} \right \rangle\cdot \langle 0, 1, 0\rangle\, dS_4\\ &= \iint\limits_{S_4} 0\, dS_4\\ &= 0 \end{align*} {/eq}

Since {eq}S_5 {/eq} has downward orientation, we have

{eq}\begin{align*} \iint\limits_{S_5}\vec{F}\cdot d\vec{S_5} & = -\iint\limits_{D_5}\left(-f(x, y,z)\frac{\partial z}{\partial x} -g(x, y, z)\frac{\partial z}{\partial y} + h(x, y, z)\right)\, dA\\ &= -\iint\limits_{D_5}\left(-(0)(0) -(0)(0) + e^{y+0}\right)\, dA\\ &=- \iint\limits_{D_5}e^{y}\, dA\\ &=- \int^5_0\int^5_0e^y\, dx\, dy\\ &=- \int^5_0\left[xe^y\right]^5_0\, dy\\ &=- \int^5_05e^y\, dy\\ &=- \left[5e^y\right]^5_0\\ &=-( 5e^5- 5)\\ &= 5(1-e^5) \end{align*} {/eq}

Finally, {eq}S_6 {/eq} has upward orientation, so we get

{eq}\begin{align*} \iint\limits_{S_6}\vec{F}\cdot d\vec{S_6} & = \iint\limits_{D_6}\left(-f(x, y,z)\frac{\partial z}{\partial x} -g(x, y, z)\frac{\partial z}{\partial y} + h(x, y, z)\right)\, dA\\ &= \iint\limits_{D_6}\left(-(0)(0) -(0)(0) + e^{y+5}\right)\, dA\\ &= \iint\limits_{D_6}e^{y + 5}\, dA\\ &= \int^5_0\int^5_0e^{y+ 5}\, dx\, dy\\ &= \int^5_0\left[xe^{y+ 5}\right]^5_0\, dy\\ &= \int^5_05e^{y+ 5}\, dy\\ &= \left[5e^{y+ 5}\right]^5_0\\ &=( 5e^{10}- 5e^5)\\ &= 5(e^{10} - e^5) \end{align*} {/eq}

Therefore,

{eq}\begin{align*} \iint\limits_{S}\vec{F}\cdot d\vec{S} &=\iint\limits_{S_1}\vec{F}\cdot d\vec{S_1} + \iint\limits_{S_2}\vec{F}\cdot d\vec{S_2} + \iint\limits_{S_3}\vec{F}\cdot d\vec{S_3} + \iint\limits_{S_4}\vec{F}\cdot d\vec{S_4} + \iint\limits_{S_5}\vec{F}\cdot d\vec{S_5} + \iint\limits_{S_6}\vec{F}\cdot d\vec{S_6}\\ &= 0 + 0 + 0 + 0 + 5(1-e^5) + 5(e^{10} - e^5)\\ & = 5 - 10 e^5 + 5e^{10} \end{align*} {/eq}


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