91影视

Consider the vector field below:

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

The goal is to find the integral ${鈭�}_S\mathbf{F}(x,y,z)路\mathbf{n}dS$, where $S$is the sphere with radius 3 and centred at the origin, and$\mathbf{n}$ is the normal vector heading outwards.

To evaluate this integral, use the Divergence Theorem.

Divergence According to the theorem,

"Let $W$be a bounded region in ${\mathrm{鈩�}}^3$with a smooth or piecewise-smooth closed oriented surface as its boundary $S$. If a vector field $\mathbf{F}(x,y,z)$is defined on an open area containing $W$, then,

${鈭�}_S\mathbf{F}(x,y,z)路\mathbf{n}dS={鈭�}_{W}div\mathbf{F}(x,y,z)dV=$

Where $\mathbf{n}$is the unit normal vector inwards."

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

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

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

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

Then there's the vector field's divergence. $\mathbf{F}(x,y,z)=x^3\mathbf{i}+y^3\mathbf{j}+z^3\mathbf{k}$is going to be

$div\mathbf{F}(x,y,z)=\left(\frac{鈭�}{鈭�x}\mathbf{i}+\frac{鈭�}{鈭�y}\mathbf{j}+\frac{鈭�}{鈭�z}\mathbf{k}\right)路\left(x^3\mathbf{i}+y^3\mathbf{j}+z^3\mathbf{k}\right)$

$=\frac{鈭�}{鈭�x}\left(x^3\right)+\frac{鈭�}{鈭�y}\left(y^3\right)+\frac{鈭�}{鈭�z}\left(z^3\right)$

$=3x^2+3y^2+3z^2$

$=3\left(x^2+y^2+z^2\right)$.

Now evaluate the integral using the Divergence Theorem (1). ${鈭�}_s\mathbf{F}(x,y,z)路\mathbf{n}dS$in the following manner:

${鈭�}_S\mathbf{F}(x,y,z)路\mathbf{n}dS={鈭�}_{R}div\mathbf{F}(x,y,z)dV$

$={鈭�}_{R}3\left(x^2+y^2+z^2\right)dV$

$=3{鈭�}_R\left(x^2+y^2+z^2\right)dV$

The surface defines the region here as $S$, where, $S$is a three-dimensional sphere with its centre at the origin.

When represented in spherical coordinates, both the integrand and the region are considerably simplified. $(蚁,蠁,胃)$. Change the coordinates of this integral (2) to spherical and integrate it.

The zone of integration is defined as follows in spherical coordinates:

$R=\left\{\right(蚁,蠁,胃):0鈮�蚁鈮�3,0鈮�蠁鈮�蟺,0鈮�胃鈮�2蟺\}$

When expressed in spherical coordinates,

$x=蚁\sin 蠁\cos 胃,y=蚁\sin 蠁\sin 胃,z=蚁\cos 蠁$, and

$x^2+y^2+z^2={蚁}^2$, and

$dV={蚁}^2\sin 蠁d蚁d蠁d胃$

Then, as follows, assess the integral (2):

${鈭�}_S\mathbf{F}(x,y,z)路\mathbf{n}dS=3{鈭�}_R\left(x^2+y^2+z^2\right)dV$

$=3{鈭�}_0^{2蟺}{鈭�}_0^{蟺}{鈭�}_0^3\left({蚁}^2\right)\left({蚁}^2\sin 蠁d蚁d蠁d胃\right)$

$=3{鈭�}_0^{2蟺}{鈭�}_0^{蟺}{鈭�}_0^3{蚁}^4\sin 蠁d蚁d蠁d胃$

$=3{鈭�}_0^{2蟺}{鈭�}_0^{蟺}\sin 蠁\left[{鈭�}_0^3{蚁}^{4}d蚁\right]d蠁d胃$

$=3{鈭�}_0^{2蟺}{鈭�}_0^{蟺}\sin 蠁{\left[\frac{{蚁}^5}{5}\right]}_0^{3}d蠁d胃$

$=3{鈭�}_0^{2蟺}{鈭�}_0^{蟺}\sin 蠁\left[\frac{3^5}{5}-\frac{0^5}{5}\right]d蠁d胃$

$=3{鈭�}_0^{2蟺}{鈭�}_0^{蟺}\frac{243}{5}\sin 蠁d蠁d胃$

$=\frac{729}{5}{鈭�}_0^{2蟺}{鈭�}_0^{蟺}\sin 蠁d蠁d胃$

Reduce the last integral to the following:

${鈭�}_s\mathbf{F}(x,y,z)路\mathbf{n}dS=\frac{729}{5}{鈭�}_0^{2蟺}{鈭�}_0^{蟺}\sin 蠁d蠁d胃$

$=\frac{729}{5}{鈭�}_0^{2蟺}\left[{鈭�}_0^{蟺}\sin 蠁d蠁\right]d胃$

$=\frac{729}{5}{鈭�}_0^{2蟺}[-\cos 蠁{]}_0^{蟺}d胃$

$=\frac{729}{5}{鈭�}_0^{2蟺}\left[\right(-\cos 蟺)-(-\cos 0\left)\right]d胃$

$=\frac{729}{5}{鈭�}_0^{2蟺}\left[\right(-(-1))-(-1\left)\right]d胃$

$=\frac{729}{5}{鈭�}_0^{2蟺}2d胃$

$=\frac{729}{5}[2胃{]}_0^{2蟺}$

$=\frac{729}{5}[2路2蟺-2路0]$

$=\frac{2916蟺}{5}$

As a result, the necessary integral is ${鈭�}_S\mathbf{F}(x,y,z)路\mathbf{n}dS=\frac{2916蟺}{5}$.