Question

Evaluate the surface integral:\[\int\int_S x^2z^2 dS\]S is part of the cone \(z^2 =x^2+y^2\) that lies between planes \(z=1\) and \(z=3\)

Answer 1

How do I start this... do I find the intersection first?

Answer 2

\[ndS=\left( \frac{ \delta r }{ \delta u } \times \frac{ \delta r }{ \delta v }\right) dudv\]

Answer 3

you can plug in the two z values

Answer 4

it will give two circles of radius 1 and sqrt(3)

Answer 5

so you could use polar coordinates to integrate this

Answer 6

you first need to find the normal, since the integral is with respect to the surface.
you could parametrize the cone as r(x,y)=(x,y,sqrt(x^2+y^2))
then, you use the first formula that i posted, which is actually just to derive r with respect to x and y and take the cross product (since as you know this will give the normal vector perpendicular to both derivatives)

Answer 7

so this formula is not necessary?:\[\int\limits\int\limits_Df(x,y,g(x,y)\sqrt{\left(\frac{\partial z}{\partial x}\right)^2 +\left(\frac{\partial z}{\partial y}\right)^2+1}\]

Answer 8

Ohh...

Answer 9

\[I=\int\limits_{0}^{2 \pi} \int\limits_{r=1}^{r=\sqrt(3)} (r^2\cos^2(\theta))(r^2 )rdrd\theta \left( \frac{ \delta r(x,y) }{ \delta x } \times \frac{ \delta r(x,y) }{ \delta y } \right)\]

Answer 10

\[z^2 =x^2 +y^2 \implies z=\sqrt{x^2+y^2}\]

Answer 11

well it is the same thing, since at some point you will get dr/dx=(1,0,something) and dr/dy=(0,1,something), which will simplify to your formula above. but the cross product is a more general version , since it works for any parametrization

Answer 12

Oh, I wasn't really familiar with that notation.

Answer 13

ok, i though that you knew the Jacobian transformation. Then you could just use your formula, its simpler

Answer 14

its just that if you had something like \[\xi (s,t) = (t^2,s^3,te^s)\], then the formula wont work since it relies on the fact that the two first parameters are linear (i.e. have degree 0)

Answer 15

*degree 1

Answer 16

I would parametrize the cone in terms of \(r\) and \(\theta\).

Answer 17

(rcost,rsint,r)?

Answer 18

\[
x=r\cos\theta,y=r\sin\theta \implies z^2=(r\cos\theta)^2+(r\sin\theta)^2\implies z=r
\]

Answer 19

yeah, it ends up being the same.

Answer 20

I understood that part, just not this part: \[\left( \frac{ \delta r(x,y) }{ \delta x } \times \frac{ \delta r(x,y) }{ \delta y } \right)\]

Answer 21

It's the cross product of the partial derivatives\[
|\langle\cos\theta,\sin\theta,1\rangle \times \langle -r\sin\theta, r\cos\theta, 0\rangle|
\]

Answer 22

You know how to do the cross product, and you know how to find the magnitude of a vector.

Answer 23

Oh, I understand now.\[u=r ~ , ~ v = \theta\]\[\vec r_u = \langle \cos(\theta) ~ ,~\sin(\theta)~,~1\rangle\]\[\vec r_v = \langle -r\sin\theta ~ , ~ r\cos(\theta) ~ , ~ 0\rangle\]\[\vec n = \langle -r\cos(\theta)~,~ -r\sin(\theta)~,~ r\rangle\]\[|\vec n| = \sqrt{(-r\cos\theta)^2 +(-r\sin\theta)^2+r^2} = \sqrt{r^2\cos^2\theta +r^2\sin^2\theta +r^2}\] Am i doing some algebra mistake..

Answer 24

You are correct so far

Answer 25

\[\sqrt{r^2(1)+r^2} = \sqrt{2r^2} = r\sqrt{2}\]

Answer 26

So we just plug that in to our formula now..

Answer 27

Oh, whoops. I'm confusing everything. \[\sqrt{2}\int_0^{2\pi}\int_1^3 (r^2\cos^2\theta)(r^2) rdrd\theta\]