Question

So... Surface integrals...?

Answer 1

Good times.

Answer 2

are fun?

Answer 3

hahaha they are many, many things.... but I've got to say that "fun" doesn't exactly make my list :P

Answer 4

Consider \[\int\limits_{}^{}\int\limits_{s}^{} xdS\]
where S is part of the cylinder z=2-x^2 , bound by the planes x=0, y=0, y=4, z=0

Answer 5

Any ideas?? I'm not after a straight out answer, but some guidance would be nice :S

Answer 6

One second. The hardest part of these for me is figuring out the shape of the thing we're working with ;)

Answer 7

Same... I have quite a shotty looking scribble in front of me atm...

Answer 8

lol, one sec.

Answer 9

i tried drawing it, it turned out looking more like a dilapidated trogdor than anything else...i think i need to try again.

Answer 10

Ok, I got it.

Answer 11

That's not a cylinder, it's some kind of extruded parabola.

Answer 12

It's infinitely many parabolas of the form z = 2-x^2 extending along the y axis

Answer 13

Yeah it's a big tunnel.

Answer 14

Ok, so our integral will be:

\[\int_0^4\int_0^{\sqrt{2}} 2-x^2\ dxdy\]

Answer 15

Well, anyway, you have the limits for y (from 0 to 4) and z (from 0 to 2). Therefore, you can parametrize like this:\[\gamma(y, z) = (\sqrt{2-z},y,z)\quad y \in [0, 4], z \in [0, 2]\]

Answer 16

Oh, that's the volumn.. Sorry. Silly git I am

Answer 17

volume also

Answer 18

do you get the limits purely from the diagram..?

Answer 19

Yes

Answer 20

The area element is\[dA = \left\|\frac{\partial \gamma}{\partial y} \times \frac{\partial \gamma}{\partial z}\right\|dydz\]

Answer 21

We know that y is bounded from 0 to 4.
x is bounded at 0, and z is bounded at 0, so we know that x is also upper bounded at \(\sqrt{2}\)

Answer 22

I'd hardly bother with all that I think krebante. Just compute the line integral of the trace in the xz plane, and multiply it by 4.

Answer 23

Yes, but I'm trying to explain in a more general way since in most problems you can't do that :P.

Answer 24

true enough =)

Answer 25

\[z \ge 0 \implies 2-x^2 \ge 0\implies x \le \sqrt{2}\]