91影视

Consider the vector field below:

$\mathbf{F}(x,y,z)=e^{z}x\sin y\mathbf{i}+e^z\cos y\mathbf{j}+e^x{\tan }^{-1}y\mathbf{k}$

The goal is to find the integral localid="1650788950659" ${鈭�}_S\mathbf{F}(x,y,z)路\mathbf{n}dS$, where localid="1650788956627" $S$is the surface of the arealocalid="1650788967744" $W$ that falls within the unit sphere and above the plane localid="1650788976198" $z=\frac{\sqrt{2}}{2}$, and localid="1650788983969" $n$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 border $S$If a vector field $\mathbf{F}(x,y,z)$is defined on an open area containing$W$, then,

localid="1650789022863" ${鈭�}_S\mathbf{F}(x,y,z)路\mathbf{n}dS={鈭�}_{W^{\text{'}}}div\mathbf{F}(x,y,z)dV=$

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

First, determine the vector field's divergence.$\mathbf{F}(x,y,z)=e^{2}x\sin y\mathbf{i}+e^z\cos y\mathbf{j}+e^x{\tan }^{-1}y\mathbf{k}$

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

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

Then, the divergence of the vector field $\mathbf{F}=e^{z}x\sin y\mathbf{i}+e^z\cos y\mathbf{j}+e^x{\tan }^{-1}y\mathbf{k}$will be,

$div\mathbf{F}(x,y,z)=\left(\frac{鈭�}{鈭�x}\mathbf{i}+\frac{鈭�}{鈭�y}\mathbf{j}+\frac{鈭�}{鈭�z}\mathbf{k}\right)路\left(e^{z}x\sin y\mathbf{i}+e^z\cos y\mathbf{j}+e^x{\tan }^{-1}y\mathbf{k}\right)$

$=0$.

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

localid="1650797449478" ${鈭�}_S\mathbf{F}路\mathbf{n}dS={鈭�}_{R}div\mathbf{F}dV$localid="1650797458922" $={鈭�}_{R}0dV$

localid="1650797465402" $=0$

The necessary integral is ${鈭�}_S\mathbf{F}(x,y,z)路\mathbf{n}dS=0$.