Question:Evaluate$\iint A.dS$ where $A=y\hat i+2x\hat j-z\hat k$ and S is the surface of the plane $2x+y=6$ in the first octant cut off by the plane $z=4$
My Approach:I roughly sketch and consider $5$ surfaces.
$S_1$ is the triangle in the plane $z = 0$
$S_2$ is the triangle in the plane $z = 4$
$S_3$ is the rectangle in the plane $x = 0$
$S_4$ is the rectangle in the plane $y = 0$
$S_5$ is the plane $2x+y=6$.
The normal vectors to these respective surfaces are $(0,0,-1), (0,0,1), (-1,0,0),(0,-1,0), (2,1,0) $ respectively.Then i evaluate the surfaces $S_1, S_2 ,S_3,S_4,S_5$ And Add them together.But my answer is incorrect.the solution provided by the book is correct.
I think my approaching is incorrect.Please explain me how to do this in right way.
Thanks in advance.
It sounds like you are trying to apply the divergence theorem defining 5 surfaces that enclose the volume, but that sounds like more work than evaluating it directly.
The vector normal to the surface is $(1,2,0)$ We will say: $y = 6 - 2x$
$\int\int (6-2x,2x,-z)\cdot(1,2,0) \ dx\ dz$
Limits $0\le x\le 3, 0\le z\le 4$
$\int_0^4\int_0^3 6+4x \ dx\ dz\\ 4(3x+2x^2)|_0^3 = 108$
Since you suggest evaluating all 5 surfaces...
By the divergence theorem we should find that:
$108 + \int_0^4\int_0^6 -y \ dy\ dz + \int_0^4\int_0^3 -2x \ dx\ dz + \int_0^3\int_0^{6-2x} -4 \ dy\ dx = \iiint -1 dV$
But that isn't really what the question has asked.