Evaluate $\int \int \vec A\hat n \,dS$, where$\vec A = 18z\hat i – 12\hat j + 3y\hat k$ and $S$ is that part of the plane $S$
$2x+3y+6z = 12$ which is located in the first octant.
The surface S and its projection R on the xy plane are shown in the figure below.
Here we are calculating the area of the triangle on the $x,y$ plane by that surface integral. Now if we calculate it in general method i.e. $\text{Area}= 1/2 \cdot (\text{base}) \cdot (\text{length})$, then we are getting $1/2 \cdot 6 \cdot 4 = 12$ Why we are not getting the same result?
[Math] Surface Integral problem. May be some misconception.
integrationsurface-integrals
Best Answer
No you are not calculating the area of the triangle on the $xy$ plane. If so, in the last line, the function that would be integrated would be $1$ and not $6-2x$.
You are calculating the flux of the vector field $\vec{A}$ across the surface, which is very different.