Question

Okay, I know I'm not crazy on this one.

@Kainui

(Sorry, but the textbook's been full of mistakes so far, and some are obvious, some make me feel like I'm going insane).

Answer 1

\[
\iint_S \sqrt{x^2+y^2+z^2}\;dx\wedge dy
\]For \(S:x^2+y^2=z^2,\;0\le z\le1\)

Answer 2

Changing to polar, with \(z=r\) yields, pretty clearly:
\[
\sqrt 2\iint_{[0,2\pi]\times [0,1]}d\theta\wedge dr \;r^2
\]

Answer 3

Which is:
\[
\frac{2\sqrt{2}}{3}\pi
\]Not the textbook's:
\[
\pi\sqrt{2}
\]

Answer 4

(The textbook seems to have missed the \(r\) from the change of variables. But I just want to make sure I'm not completely insane)

Answer 5

i understand

Answer 6

I think you need to bring in an extra r into your integral, but I'm kind of too hungry to think right now. brb.

Answer 7

hahahah know feeling cant sleep

Answer 8

An extra \(r\)? We have one from the root, and the second from the change of variables, which yields the \(r^2\).

I'm asking because this is the 3rd mistake in the book, at this point. Unless I'm completely missing something.

Answer 9

3rd mistake in the book *at the intro,* I should add...

Answer 10

I've not been properly introduced to differential forms. Does dx wedge dy just mean dA normal to the surface of the cone?

Answer 11

You can just treat it as \(dx\,dy\), as it's the area element.

Answer 12

Or... well, in this case \(d\theta\;dr\).

Answer 13

Yeah, doesn't dxdy=rdrdθ?

Answer 14

Yep, that's included in the integral above, I just didn't add it explicitly there.

Answer 15

("there" being in the post two above the quoted)

Answer 16

I am getting the exact same thing as you, however I feel like something is off in how we're considering the problem. Maybe it is just a typo.

Answer 17

I'm pretty sure it's just a typo (it seems the author forgot(?) to include the \(r\) from the change to simple polar), but it's like the third one in this intro, that's why I asked.

Thanks, though.

Answer 18

Yeah, sorry to hear about that book of yours haha.

Answer 19

Yeah >.>

Well, from looking ahead, the proofs are nice, at least.