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Let S be the cube with vertices ( 1 , 1 , 1 ) , F =< x + y , y + 2 z , x y + z > Find ...

Question:

Let {eq}S {/eq} be the cube with vertices {eq}(\pm 1, \pm 1, \pm 1), \ \ \vec{F} = <x+y,y+2z,xy+z> {/eq}

Find {eq}\int \int_S \vec{F} \cdot d\vec{S} {/eq}

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.

Answer and Explanation:

Let {eq}S_1 {/eq} be the face of the cube in the plane {eq}x=-1 {/eq}, {eq}S_2 {/eq} the face of the cube in the plane {eq}x=1 {/eq}, {eq}S_3 {/eq} the face of the cube in the plane {eq}y=-1 {/eq}, {eq}S_4 {/eq} the face of the cube in the plane {eq}y=1 {/eq}, {eq}S_5 {/eq} the face of the cube in the plane {eq}z=-1 {/eq} and {eq}S_6 {/eq} the face of the cube in the plane {eq}z=1 {/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 x+y,y+2z,xy+z \right \rangle\cdot \langle -1, 0, 0\rangle\, dS_1\\ &= \iint\limits_{D_1} \left(-x-y\right)\, dA\\ &= \int^1_{-1}\int^1_{-1} \left(-(-1)-y\right)\, dy\, dz\\ &= \int^1_{-1}\int^1_{-1} \left(1-y\right)\, dy\, dz\\ &= \int^1_{-1}\left[y - \frac{1}{2}y^2\right]^1_{-1}\, dz\\ &= \int^1_{-1}\left(\frac{1}{2} - \frac{3}{2}\right)\, dz\\ &= 2 \int^1_{-1}dz\\ &= 2 \left[z\right]^1_{-1}\\ &= 2(1 -(-1))\\ &= 4 \end{align*} {/eq}

For {eq}S_2 {/eq} we have {eq}\vec{n_2} = \langle 1, 0, 0\rangle {/eq}, so {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 x+y,y+2z,xy+z \right \rangle\cdot \langle 1, 0, 0\rangle\, dS_1\\ &= \iint\limits_{D_2} (x+ y)\, dA\\ &= \int^1_{-1}\int^1_{-1}(1+ y)\, dy\, dz\\ &= \int^1_{-1}\left[y + \frac{1}{2}y^2\right]^1_{-1}\, dz\\ &= \int^1_{-1}\left(\frac{3}{2} + \frac{1}{2}\right)\, dz\\ &= 2\int^1_{-1}dz\\ &= 2\left[z\right]^1_{-1}\\ &= 2(1 -(-1))\\ &= 4 \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 x+y,y+2z,xy+z \right \rangle\cdot \langle 0, -1, 0\rangle\, dS_3\\ &= \iint\limits_{D_3} (-y-2z)\, dA\\ &= \int^1_{-1}\int^1_{-1}(1-2z)\, dx\, dz\\ &= \int^1_{-1}\left[x -2xz\right]^1_{-1}\, dz\\ &= \int^1_{-1}\left[\left(1-2z\right) - \left(-1 + 2z\right)\right]\, dz\\ &= \int^1_{-1}(2 -4z)\,dz\\ &= \left[2z - 2z^2\right]^1_{-1}\\ &= (2 - 2) - (-2 - 2)\\ &= 4 \end{align*} {/eq}

For {eq}S_4 {/eq} we have {eq}\vec{n_4} = \langle 0, 1, 0\rangle {/eq}, so {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 x+y,y+2z,xy+z \right \rangle\cdot \langle 0, 1, 0\rangle\, dS_4\\ &= \iint\limits_{D_4} (y+ 2z)\, dA\\ &= \int^1_{-1}\int^1_{-1}(1+ 2z)\, dx\, dz\\ &= \int^1_{-1}\left[x +2xz\right]^1_{-1}\, dz\\ &= \int^1_{-1}\left[\left(1+2z\right) - \left(-1 - 2z\right)\right]\, dz\\ &= \int^1_{-1}(2 +4z)\,dz\\ &= \left[2z + 2z^2\right]^1_{-1}\\ &= (2 + 2) - (-2 + 2)\\ &= 4 \end{align*} {/eq}

The unit normal vector to {eq}S_5 {/eq} is {eq}\vec{n_5} = \langle 0, 0, -1\rangle {/eq}, so we have {eq}d\vec{S_5} = \vec{n_5}\,dS_3 = \langle 0, 0, -1\rangle\, dS_5 {/eq}. Hence,

{eq}\begin{align*} \iint\limits_{S_5} \vec{F}\cdot d\vec{S_5} & = \iint\limits_{S_5}\vec{F}\cdot \vec{n_5}\, dS_5\\ & = \iint\limits_{S_5} \left \langle x+y,y+2z,xy+z \right \rangle\cdot \langle 0, 0, -1\rangle\, dS_5\\ &= \iint\limits_{D_5} (-xy - z)\, dA\\ &= \int^1_{-1}\int^1_{-1}(1-xy )\, dx\, dy\\ &= \int^1_{-1}\left[x -\frac{1}{2}x^2y \right]^1_{-1}\, dy\\ &= \int^1_{-1}\left[\left(1 -\frac{1}{2}y \right) - \left(-1-\frac{1}{2}y\right)\right]\, dy\\ &= 2\int^1_{-1}dy\\ &= 2\left[y\right]^1_{-1}\\ &= 2(1 - (-1))\\ &= 4 \end{align*} {/eq}

For {eq}S_6 {/eq} we have {eq}\vec{n_6} = \langle 0, 0, 1\rangle {/eq}, so {eq}d\vec{S_6} = \vec{n_6}\,dS_6 = \langle 0, 0, 1\rangle\, dS_6 {/eq}, and hence

{eq}\begin{align*} \iint\limits_{S_6} \vec{F}\cdot d\vec{S_6} & = \iint\limits_{S_6}\vec{F}\cdot \vec{n_6}\, dS_6\\ & = \iint\limits_{S_6} \left \langle x+y,y+2z,xy+z \right \rangle\cdot \langle 0, 0, 1\rangle\, dS_6\\ &= \iint\limits_{D_4} (xy+ z)\, dA\\ &= \int^1_{-1}\int^1_{-1} (xy+ 1)\, dx\, dy\\ &= \int^1_{-1}\left[\frac{1}{2}x^2y + x\right]^1_{-1}\, dy\\ &= \int^1_{-1}\left[\left(\frac{1}{2}y + 1\right) - \left(\frac{1}{2}y - 1\right)\right]\, dy\\ &= 2\int^1_{-1}dy\\ &= 2\left[y\right]^1_{-1}\\ &= 2(1 - (-1))\\ &= 4 \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}\\ &= 4 + 4 + 4 + 4 + 4 + 4\\ & =24 \end{align*} {/eq}


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