Question

5. Setup (but do not evaluate) one integral (of any type) to find the flux of vector field F through surface S, where S s the unit cube given by 0 < x < 1,0 < y 1.0 < z 1,

Answer 1

$\text{Solution:}$

$\vec{\text{F}}=\triangledown f =\left \langle 3x^2+y,3y^2+x+z,3z^2+y \right \rangle$

By gauss's divergence theorem

divFdV

$\int \int _S\vec{\text{F}}\cdot \vec{n}\mathrm{d}S=\int \int \int _E\left \langle \frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right \rangle \cdot \left \langle 3x^2+y,3y^2+x+z,3z^2+y \right \rangle \mathrm{d}V$

$\int \int _S\vec{\text{F}}\cdot \vec{n}\mathrm{d}S=\int \int \int _E\left \left ( 6x+6y+6z \right )\mathrm{d}V$

$\int \int _S\vec{\text{F}}\cdot \vec{n}\mathrm{d}S=6\int \int \int _E\left \left ( x+y+z \right )\mathrm{d}V$

$\int \int _S\vec{\text{F}}\cdot \vec{n}\mathrm{d}S=6\int_0^1 \int_0^1 \int _0^1\left \left ( x+y+z \right )\mathrm{d}z\mathrm{d}y\mathrm{d}x$

$\int \int _S\vec{\text{F}}\cdot \vec{n}\mathrm{d}S=6\int_0^1 \int_0^1\left [ xz+yz+\frac{z^2}{2} \right ]_0^1\mathrm{d}y\mathrm{d}x$

$\int \int _S\vec{\text{F}}\cdot \vec{n}\mathrm{d}S=6\int_0^1 \int_0^1\left (x+y+\frac{1}{2} \right )\mathrm{d}y\mathrm{d}x$

$\int \int _S\vec{\text{F}}\cdot \vec{n}\mathrm{d}S=6\int_0^1 \left [ xy+\frac{y^2}{2}+\frac{y}{2} \right ]_0^1 \mathrm{d}x$

$\int \int _S\vec{\text{F}}\cdot \vec{n}\mathrm{d}S=6\int_0^1 \left ( x+1 \right ) \mathrm{d}x$

$\int \int _S\vec{\text{F}}\cdot \vec{n}\mathrm{d}S=6\left [ \frac{x^2}{2}+x \right ]_0^1$

$\int \int _S\vec{\text{F}}\cdot \vec{n}\mathrm{d}S=6*\frac{3}{2}=9$