Surface Integral Homework: Compute Int Int S F•n dS

[Saladsamurai]

Homework Statement

Let S be the part of the paraboloid z=1+x^2+y^2 lying above the rectangle
x between 0 and 1; y between -1 and 0 and oriented by the upward normal. Compute

\int\int_SF\cdot n\,dS where F=<xz, xy, yz>


So I have Parametrized the surface S as r(x,y,z)=<x, y, 1+x²+y²>

Then I have found dr\dx cross dr/dy =f

then I found F(r(x,y)) dot f

Now I need to integrate this over the domain of E but I am having trouble finding my bounds for x and y?

I need to project the paraboloid downward onto x-y plane right? This gives a curve, oh wait, the curve is just the equation of the paraboloid with z=0 right?


So the curve is 1+x²+y²=0

why does that not sit well with me?

 

[Saladsamurai]

Something is not right here. If my curve is 1+x²+y²=0

That means that x=sqrt[-1-x²] which is just stupid... wtf am I doing? 10 hours until the final and I'm effing up surface integrals

And did I compute the cross product wrong? Should it be dr/dy "cross" dr/dx instead?

 

[Saladsamurai]

I think I got it. If I project the paraboloid downwards, I get a circle of radius 1. So x=sqrt[1-y⁶] which is my UPPEr bound for x and y=sqrt[1-x⁶] which is my LOWER bound for y.

Yes?

Either way, the integral is retarded; F dot (dr/dy x dr/dx) is a huge mess. Converting to polar coordinates will help a little, but am I right to say that this is a long-a$$ integral?

 

[HallsofIvy]

The problem says " lying above the rectangle
x between 0 and 1; y between -1 and 0" Those are your limits of integration.

You should have
\int_{x=0}^1\int_{y= -1}^0 (2x^2- 2x^4- 3x^2y^2+ 2xy^2+ y^2- y^4)dydx
not all that hard, surely.