# Evaluating a surface integral $\iint A.dS$where $A=y\hat i+2x\hat j-z\hat k$

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.

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.

• That's mean we don't need to consider the sketch..we just focus on the surface mentioned in the question?? Am I right??? @Doug M – emonhossain Sep 7 '18 at 20:16
• I always recommend a sketch. Nonetheless, you have just been asked to evaluate the one surface. – Doug M Sep 7 '18 at 20:20
• Oh! I got it.i really confused with the sketch and thinking that all surfaces should be consider.Thanks @Doug M – emonhossain Sep 7 '18 at 20:23
• There are times when you will want to consider whether it is easier to directly evaluate a surface, or to evaluate the remaining surfaces that enclose a volume. This is not one of those times. – Doug M Sep 7 '18 at 20:27