I'm reading Matthew's(adsbygoogle = window.adsbygoogle || []).push({}); Vector Calculus. In problem 2.8, I'm asked to find the surface integral for a surface which can be parametrized as [tex](x,y,x+y^2)[/tex]. I understand that to find a normal vector, I can cross product the x and y partial derivatives. When I do so, I get

[tex](1,0,1)\times(0,1,2y) = (-1, -2y, 1)[/tex]

The problem specifies that the normal has to have a negative z component. Not a complication -- reverse the signs: [tex](1,2y,-1)[/tex].

At this point, I thought, it's in the bag. Just normalize and work the double integral. BUT, when I checked my work up to that point with the answers in back, I saw that the solution DID NOT involve normalization at all. He just ran with (1, 2y, -1) as the unit normal vector.

Am I crazy in thinking this is wrong? Did I miss something? I just started, and I can't think that I'm already good enough at this to spot a major error by page 43. I keep thinking I've missed something, but the fact is, there's no way I can imagine anything of the form (1, whatever, -1) has length = 1. (Well, without throwing in a few imaginary numbers, but you know.)

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# Unit Normal Vector normal, but not normalized?

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