Math Is Fun Forum

Identity

Member

Registered: 2007-04-18

Posts: 934

Surface integral over a plane

I have a plane in the positive octant, with axes intercepts (0,0,a), (a,0,0), (0,2a,0), and I need to integrate

over it.

Do I to complete all these steps?

1. Find two vectors on the plane
2. Compute their cross product to find the normal
3. Use the normal to find a cartesian equation for the plane
4. Find the parametric representation of the plane
5. Take partial derivatives with respect to the variables
6. Make the partial derivatives unit vectors
7. Take their cross product
8. Dot this with v
9. Integrate over the allowed values of the parameters.

It seems incredible tedious for a simple plane...

Is there a faster (but systematic) way to do this, or is this the best there is? Thanks

Last edited by Identity (2010-07-24 01:05:24)

rzaidan

Member

Registered: 2009-08-13

Posts: 59

Re: Surface integral over a plane

1)find two vectors(a,0,-a) , (-a,2a,0)
2)find their cross product which is the normal n=(2a² , a² ,2a²)  to the plane.
3) compute the surface integral=∫∫v.n ds=∫∫(yk).n (dxdy/|n.k|) 
=∫∫v.n ds=∫∫2a²(dxdy/2a²=∫∫dxdy=area of triangle in the xy-plane with vertices(0,0) , ( a, 0), (0,2a)=(½) (a)(2a)=a²
I hope I am right
Best Wishes
Riad Zaidan