# surface integral

F=x^2i+xyj+zk,and S is the part of the paraboloid z=x^2+y^2 below the plane z=1with upward orientation,find double integral of F over S

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Do you mean find the surface integral of F over S?

The surfave integral of F over S is defined as the double integral [over S]of {F 內積n}dS , where n is the (outward) normal of S. It can be derived to be the double integral [over D]of {F 內積[r_u外積r_v]}dA, if r(u,v) is a parametric representation of S, and D is the region of the projection of S on to uv-pane.

We need to find a parametric representation of the surface S first. A good choice for the paraboloid z=x^2+y^2 is r(u,v)=<u,v,u^2+v^2>. Thus r_u外積r_v=<-2u,-2v,1>, the integrand {F 內積[r_u外積r_v]} = (u^2+v^2)(1-2u). So we arrive at the double integral [over D]of {(u^2+v^2)(1-2u)}dA.

Notice that the region D is the unit disc u^2+v^2<=1. That means the best way to calculate the last integral is converting to polar coordinates: u=rcost, v=rsint; 0<=r<=1, 0<=t<=2pi. I leave it as an exercise to finish. The answer I got was pi/2.

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