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.
Best Answer
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.