Q. 25 - Calculus

Question

In Exercises $25-40$, evaluate the integral

${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$

for the specified function $\mathbf{F}(x,y,z)$ and the given surface $S$. In each integral, $\mathbf{n}$ is the outwards-pointing normal vector.

$\mathbf{F}(x,y,z)=x{y}^2\mathbf{i}+y(z-3x)\mathbf{j}+4xyz\mathbf{k}$, and $S$ is the surface of the region $W$ bounded by the planes $y=0, y=z, z=3, x=0$, and $x=4$.

Step-by-Step Solution

Answer

The required integral is ${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS=3{\pi }^7$

Step 1

Consider the vector field below:

$\mathbf{F}(x,y,z)=4x^{3}yz\mathbf{i}+6x^2{y}^{2}z\mathbf{j}+6x^{2}y{z}^2\mathbf{k}$

The goal is to find the integral${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$  for the surface s, which is defined as follows:

The surface $S$ is the first-octant cube's surface with side length $\pi$ and one vertex at the origin, and $n$ is the normal vector heading outwards.

Step 2

To evaluate this integral, use the Divergence Theorem.

"Let $W$ be a bounded region in $R^3$ whose border $S$ is a smooth or piecewise-smooth closed oriented surface," says the Divergence Theorem. If an open region containing $W$ has a vector field $F(x,y,z)$, then

${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS={\iiint }_{W}div\mathbf{F}(x,y,z)dV ........ \left(1\right)$

where $n$ is the normal vector pointing outwards

Step 3

First, determine the vector field's divergence $\mathbf{F}(x,y,z)=4x^{3}yz\mathbf{i}+6x^2{y}^{2}z\mathbf{j}+6x^{2}y{z}^2\mathbf{k}$

A vector field's divergence $\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$ has the following definition:

$div F(x,y,z)=(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k).(F_{1}i+F_{2}j+F_{3}k)$

$=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}$

Then there's the vector field's divergence $\mathbf{F}(x,y,z)=4x^{3}yz\mathbf{i}+6x^2{y}^{2}z\mathbf{j}+6x^{2}y{z}^2\mathbf{k}$ will be,

$div\mathbf{F}(x,y,z)=\left(\frac{\partial }{\partial x}\mathbf{i}+\frac{\partial }{\partial y}\mathbf{j}+\frac{\partial }{\partial z}\mathbf{k}\right)·\left(F_1\mathbf{i}+F_2\mathbf{j}+F_3\mathbf{k}\right)$

$=\frac{\partial }{\partial x}\left(4x^{3}yz\right)+\frac{\partial }{\partial y}\left(6x^2{y}^{2}z\right)+\frac{\partial }{\partial z}\left(6x^{2}y{z}^2\right)$

$=12x^{2}yz+12x^{2}yz+12x^{2}yz$

$=36x^{2}yz$

$divF(x,y,z)=(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k).(4x^{3}yzi+6x^2{y}^{2}zj+6x^{2}y{z}^{2}k)$

$div\mathbf{F}(x,y,z)=\left(\frac{\partial }{\partial x}\mathbf{i}+\frac{\partial }{\partial y}\mathbf{j}+\frac{\partial }{\partial z}\mathbf{k}\right)·\left(F_1\mathbf{i}+F_2\mathbf{j}+F_3\mathbf{k}\right)$

Step 4

The region is defined by the surface $S$, which is the surface of the first-octant cube with a side length of$\mathrm{\pi }$ and one vertex at the origin.

The integration region will be here,

$R=\left\{\right(x,y,z)\mid 0\le x\le \pi ,0\le y\le \pi ,0\le z\le \pi \}$

Now evaluate the integral using the Divergence Theorem (1) ${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$ as follows:

$\iint F.n dS=\iint {\int }_{R}div F dV$

$={\int }_0^{\pi }{\int }_0^{\pi }{\int }_0^{\pi }\left(36x^{2}yz\right)dxdydz$

$={\int }_0^{\pi }{\int }_0^{\pi }\left[\right(36x^{2}yz\left)dx\right]dydz$

$={\int }_0^{\pi }{\int }_0^{\pi }{\left[12x^{3}yz\right]}_0^{\pi }dydz$

$={\int }_0^{\pi }{\int }_0^{\pi }\left(12{\pi }^{3}yz-12{\left(0\right)}^{3}yz\right)dydz$

$={\int }_0^{\pi }\left[{\int }_0^{\pi }12{\pi }^{3}yzdy\right]dz$

$={\int }_0^{\pi }{\left[6{\pi }^3{y}^{2}z\right]}_0^{\pi }dz$

$={\int }_0^{\pi }(6{\pi }^3{\left(\pi \right)}^{2}z-6{\pi }^3{\left(0\right)}^{2}z)dz$

$={\int }_0^{\pi }6{\pi }^{5}zdz$

$={\left[3{\pi }^5{z}^2\right]}_0^{\pi }$

$=3{\pi }^5{\pi }^2-3{\pi }^5{\left(0\right)}^2$

$=3{\pi }^7$