# Verify the Divergence by evaluating \int_s \int F \cdot V ds as a surface integral and as a...

## Question:

Verify the Divergence by evaluating

{eq}\displaystyle\iint_S \mathbf F \cdot \mathbf N \: dS {/eq}

as a surface integral and as a triple integral

{eq}\mathbf F(x, y, z) = (zx - y) \: \mathbf{i} - (zy - z) \: \mathbf{j} + z\: \mathbf{k} {/eq}

{eq}S {/eq}: surface bounded by the plane {eq}2x + 4y + 2z = 12 {/eq} and the coordinate planes.

## Divergence:

The value of the divergence at a point in a vector field is a measure of the net flow that crosses a surface centered at that point. Therefore, if the divergence is zero then we can affirm that the outflow and inbound flow are canceled.

## Answer and Explanation:

The Divergence Theorem states: {eq}\iint_S \vec F \cdot \hat{n} \, dS= \iiint_D \nabla \cdot F \, dV {/eq}

Part 1.

{eq}I=\iiint_D \nabla \cdot F \, dV {/eq}

Calculate the divergence of the vector field F

{eq}\nabla \cdot F= \frac {\partial P} {\partial x} + \frac {\partial Q} {\partial y} + \frac {\partial R} {\partial z}\\ \vec F(x,y,z)=(zx-y) \vec {i} - (zy-z) \vec {j} +z \vec \\ \nabla \cdot F =z-z+1\\ \nabla \cdot F = 1\\ {/eq}

Calculate the triple integral

{eq}I= \int_D \nabla \cdot F \, dV\\ I=\int_{0}^{6} \int_{0}^{3} \int_{0}^{6-x-2y} 1 \, dzdydx\\ I=\int_{0}^{6} \int_{0}^{3} \left. z \right|_{0}^{6-x-2y} \, dydx\\ I=\int_{0}^{6} \int_{0}^{3} 6-x-2y \, dydx\\ I=\int_{0}^{6} \left. 6y-xy-y^2 \right|_{0}^{3} \, dx\\ I=\int_{0}^{6} 9-3x \, dx\\ I= \left. 9x-\frac{3x^2}{2} \right|_{0}^{6}\\ I=0\\ {/eq}

Part 2

{eq}\iint_S \vec F \cdot \hat{n} \, dS {/eq}

The the cube bounded by the planes six planes

{eq}S_1: x = 0 \,\, S_2:y = 0 \,\, S_3:z = 0 \,\, S_4:z =6-x-2y {/eq}

Calculate the surface integral in{eq}S_1: x = 0 {/eq}

{eq}\iint_S \vec F \cdot \hat{n} \, dS \\ \hat{n} = -i\\ \vec F(x,y,z)=(zx-y) \vec {i} - (zy-z) \vec {j} +z \vec \\ \vec F \cdot \hat{n} = y\\ I_{ S_1}=\iint_S \vec F \cdot \hat{n} \, dS \\ I_{ S_1}= \int_{0}^{3} \int_{0}^{6 -2y} y \, dzdy\\ I_{ S_1}= \int_{0}^{3} \left. zy \right|_{0}^{6 -2y} \, dy\\ I_{ S_1}= \int_{0}^{3} 6y-2y^2 \, dx\\ I_{ S_1}= \left. 3y^2-\frac{3y^3}{3} \right|_{0}^{3} \, \\ I_{ S_1}=0\\ {/eq}

Calculate the surface integral in {eq}S_2: y = 0 {/eq}

{eq}\iint_S \vec F \cdot \hat{n} \, dS \\ \hat{n} = -j\\ \vec F(x,y,z)=(zx-y) \vec {i} - (zy-z) \vec {j} +z \vec \\ \vec F \cdot \hat{n} = z\\ I_{ S_2}=\iint_S \vec F \cdot \hat{n} \, dS \\ I_{ S_2}= \int_{0}^{6} \int_{0}^{6 -z} z \, dxdz\\ I_{ S_2}= \int_{0}^{6} \left. xz \right|_{0}^{6 -z} \, dy\\ I_{ S_2}= \int_{0}^{6} 6z-z^2 \, dx\\ I_{ S_2}= \left. 3z^2-\frac{z^3}{3} \right|_{0}^{6} \, \\ I_{ S_2}=36\\ {/eq}

Calculate the surface integral in {eq}S_3: z = 0 {/eq}

{eq}\iint_S \vec F \cdot \hat{n} \, dS \\ \hat{n} = -k\\ \vec F(x,y,z)=(zx-y) \vec {i} - (zy-z) \vec {j} +z \vec \\ \vec F \cdot \hat{n} = -z=0\\ I_{S_3}=\iint_S \vec F \cdot \hat{n} \, dS =0\\ {/eq}

Calculate the surface integral in {eq}S_4: z = 6-x-2y {/eq}

{eq}\iint_S \vec F \cdot \hat{n} \, dS \\ \hat{n} = k\\ \vec F(x,y,z)=(zx-y) \vec {i} - (zy-z) \vec {j} +z \vec \\ \vec F \cdot \hat{n} = z\\ I_{ S_4}=\iint_S \vec F \cdot \hat{n} \, dS \\ I_{ S_4}= \int_{0}^{3} \int_{0}^{6 -2y} 6-x-2y \, dxdy\\ I_{ S_4}= \int_{0}^{3} \left. 6x-\frac{x^2}{2}-2xy \right|_{0}^{6 -2y} \, dy\\ I_{ S_4}= \int_{0}^{3} 36-12y-2(3-y)^2-2y(6-2y) \, dy\\ I_{ S_4}= \int_{0}^{3} 36-12y-2(9-2y+y^2)-12y+4y^2 \, dy\\ I_{ S_4}= \int_{0}^{3} 36-12y-18+4y-2y^2-12y+4y^2 \, dy\\ I_{ S_4}= \int_{0}^{3} 18-20y+2y^2 \, dy\\ I_{ S_4}= \left. 18y-10y^2+\frac{2y^3}{3} \right|_{0}^{3} \\ I_{ S_4}= -36\\ {/eq}

Calculate sum of the integral surfaces

{eq}I=0+0+36-36\\ I=0 {/eq}

The two sides of the equation give the same result, Divergence Theorem is proven