Question

Use the divergence theorem to find the outward flux of the vector field F(x,y,z)=2x2i+5y2j+3z2k across the boundary of the rectangular prism: 0≤x≤1,0≤y≤3,0≤z≤1.

Answer 1

Answer:

The answer is "120".

Step-by-step explanation:

Given values:

$F(x,y,z)=2x^2i+5y^2j+3z^2k$

differentiate the above value:

$div F =2 \frac{x^2i}{\partial x}+5 \frac{y^2j}{\partial y}+3 \frac{z^2k }{\partial z}$

        $= 4x+10y+6z$

$\ flu \ of \ x = \int \int div F dx$

              $= \int\limits^1_0 \int\limits^3_0 \int\limits^1_0 {(4x+10y+6z)} \, dx \, dy \, dz \\\\ = \int\limits^1_0 \int\limits^3_0 {(4xz+10yz+6z^2)}^{1}_{0} \, dx \, dy \\\\ = \int\limits^1_0 \int\limits^3_0 {(4x+10y+6)} \, dx \, dy \\\\ = \int\limits^1_0 {(4xy+10y^2+6y)}^3_{0} \, dx \\\\ = \int\limits^1_0 {(12x+90+18)}\, dx \\\\= {(12x^2+90x+18x)}^{1}_{0} \\\\= {(12+90+18)} \\\\=30+90\\\\= 120$