Question

Evaluate the Gauss divergence theorem for F =x2 i+y2 j +z2 k where S is the surface of the cuboid
formed by the planes x=0, x=a, y=0, y=b, z=0 and z=c.
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Answer 1

Answer:

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Answer 2

Answer:

The evaluation of the Gauss divergence is $abc(a+b+c)$.

Step-by-step explanation:

Gauss Divergence theorem:-

The $surface\ integral$ of the $normal$ component of a $vector$ function $F$ over a closed surface $S$ is equal to the $volume$ integral of the divergence of $\vec F$ over the volume $V$ enclosed by the surface $S$.

Mathematically,

$\int \int\limits_S {\vec F\cdot \vec n} \, dS = \int \int \int\limits_V {\nabla \vec F} \, dV$ . . . . . (1)

Step 1 of 2

Given:-

$F=x^2i+y^2j+z^2k$, where $S$ : The surface of the $cuboid$ formed by the $planes$ $x=0$, $x=a$, $y=0$, $y=b$, $z=0$ and $z=c$.

We need to evaluate the gauss divergence theorem.

Calculate the value of $\nabla \vec F$ as follows:

$\nabla \vec F=\frac{\partial \vec F}{\partial x} +\frac{\partial \vec F}{\partial y} +\frac{\partial \vec F}{\partial z}$

$\nabla \vec F=\frac{\partial}{\partial x}(x^2+y^2+z^2) +\frac{\partial}{\partial y}(x^2+y^2+z^2) +\frac{\partial }{\partial z}(x^2+y^2+z^2)$

$\nabla \vec F=2x+2y+2z$

Step 2 of 2

Using the formula of the Gauss-divergence theorem, we have

$\int \int\limits_S {\vec F\cdot \vec n} \, dS = \int \int \int\limits_V {(2x+2y+2z)} \, dV$

                  $= \int^c_{z=0} \int^b _{y=0} \int^a _{x=0}{(2x+2y+2z)} \, dxdydz$

Integrate with respect to $x$ as follows:

$\int \int\limits_S {\vec F\cdot \vec n} \, dS= \int^c_{z=0} \int^b _{y=0} {(2\frac{x^2}{2} +2xy+2xz)^a_0} \, dydz$

Simplify as follows:

$\int \int\limits_S {\vec F\cdot \vec n} \, dS= \int^c_{z=0} \int^b _{y=0} {(x^2 +2xy+2xz)^a_0} \, dydz$

                  $= \int^c_{z=0} \int^b _{y=0} {(a^2 +2ay+2az-0^2-2(0)y-2(0)z)} \, dydz$

                  $= \int^c_{z=0} \int^b _{y=0} {(a^2 +2ay+2az)} \, dydz$

Now, integrate with respect to $y$ as follows:

$\int \int\limits_S {\vec F\cdot \vec n} \, dS= \int^c_{z=0} {(a^2y +2a\frac{y^2}{2} +2ayz)^b_0} \,dz$

                  $= \int^c_{z=0} {(a^2y +ay^2 +2ayz)^b_0} \,dz$

                  $= \int^c_{z=0} {(a^2(b) +a(b)^2 +2a(b)z-0)} \,dz$

                  $= \int^c_{z=0} {(a^2b +ab^2 +2abz)} \,dz$

Lastly, integrate with respect to $z$ as follows:

$\int \int\limits_S {\vec F\cdot \vec n} \, dS= {(a^2bz +ab^2z +2ab\frac{z^2}{2} )}^c_0$

                  $= {(a^2bz +ab^2z +abz^2 )^c_0$

                  $= a^2bc +ab^2c +abc^2-0$

                  $=abc(a+b+c)$

Final answer: The evaluation of the Gauss divergence is $abc(a+b+c)$.

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