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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. enter image description here 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?

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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.

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