# A surface integral

1. Mar 17, 2013

Hi to all,

1. The problem statement, all variables and given/known data
Evaluate the surface integral of the vector F=xi+yj+zk over that portion of the surface x=xy+1
which covers the square 0≤x≤1 , 0≤y≤1 in the xy plane

2. Relevant equations
∫∫F.ndσ
n=∇g/|∇g|

maybe transformation to the volume integral

3. The attempt at a solution

g(x,y,z)=xy+1-z
n=∇g/|∇g|=(yi+xj-k)/√(y2+x2+1)

Plugging into integral,
i finally got

∫∫((xy-1)/(√(y2+x2+1))) dxdy

both x and y are from 0 to 1.
But i could not take this integral without help of a computer.

Since this is from a book's ordinary question i don't think it needs such a treatment.

I think i could transform the surface integral into a volume integral but there is not a well defined volume that can be used.

So, i stuck at this point.

Thanks.

2. Mar 17, 2013

ok, drawing the region, transforming into volume integral and subtracting three additional surface integrals; i finally found 3/4 with only pencil and paper.

thanks anyway.

3. Mar 18, 2013

### HallsofIvy

Staff Emeritus
From what you write below, you mean z= xy+ 1.

Do you mean using some version of Stoke's theorem? Those all require a closed surface which you do not have here.

You are confusing the integral of a scalar valued function with the integral of a vector valued function. What you are doing uses some parts of both!

Instead of the surface area integral you should use the vector surface differential.
Some texts write that as $$\vec{u}dS$$ where u is a unit vector perpendicular to the surface, say the gradient, that you calculated for dS. But that require multiplying by the length of grad s in dS and diving by it in u. And they will cancel anyway.

Instead, write the surface as the xy+1- z= 0 or, even simpler, xy- z= -1, which are just different ways of writing the same surface. Now, grad(xy- z)= <y, x, -1> and the "vector differential of surface area is $d\vec{S}= <y, x, -1>dx dy$ (because only x and y appear in that normal vector, it is simplest to choose x and y as parameters).

Now, the vector function to be integrated is $\vec{F}= xi+ yj+ zk= xi+ yj+ (xy+ 1)k$ on this surface where I have written z in terms of the parameters x and y.

So $\vec{F}\cdot d\vec{S}= <x, y, (xy+ 1)>\cdot <y, x, -1>dxdy= xy+ xy- xy- 1= xy- 1$

The integral you want is $\int_{x=0}^1\int_{y= 0}^1 xy- 1 dydx$

But, you did not specify an orientation for the surface here! You cannot do a vector surface integral without specifying an orientation for the surface. I, since I use <y, x, -1> as normal vector, am assuming downward (the z component is negative) orientation. If the problem specifies an upward or orientation, use <-y, -x, 1> which would just multiply the final result by -1.