Question

Could someone please help with this double polar integral please?
I have worked it through but can't figure out where/if I've gone wrong? I would just like to compare answers.

Answer 1

\[\int\limits_{\mathbb{R}}^{}\int\limits_{\mathbb{R}}^{}\frac{ -4 }{ (1+u^2+v^2)^4 } dudv\]
let:\[u=rcos \theta, v = rsin \theta \]
then we evaluate: \[\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi} \frac{ -4 }{ (1+r^2)^4 }r.dr.d \theta \]

Answer 2

Note that you can rewrite the double integral in polar coordinates as follows:
\[\begin{aligned} \int_{\Bbb{R}}\int_{\Bbb{R}} -\frac{4}{(1+u^2+v^2)^4}\,du\,dv \xrightarrow{\text{polar coords.}}{} &\phantom{=}\int_0^{2\pi}\int_0^{\infty} -\frac{4r}{(1+r^2)^4}\,dr\,d\theta\\ &= \left(\int_0^{\infty}-\frac{4r}{(1+r^2)^4}\,dr\right)\left(\int_0^{2\pi}\,d\theta\right)\end{aligned}\]
Now note for the integral with respect to r, you can make the substitution \(u=1+r^2\implies \,du=2r\,dr\) (don't forget to change your limits of integration as well!). Hence, we now see that
\[ \left(\int_0^{\infty}-\frac{4r}{(1+r^2)^4}\,dr\right)\left(\int_0^{2\pi}\,d\theta\right) \xrightarrow{u=1+r^2}{} \left(\int_1^{\infty}-\frac{2}{u^4}\,du\right)\left(\int_0^{2\pi}\,d\theta\right)\]
Can you take things from here? :-)

Answer 3

Okay using that I get 4pi/3?
The problem is I'm trying to find the Total Curvature of a surface, which is given by: \[\int\limits_{}^{}k(s) ds\]
where k(s) is the Gaussian Curvature, which is the original function I gave to integrate.
I have to prove that the total curvature is 4pi

Any idea where I'm going wrong? (ie getting 4pi/3 instead?)

Answer 4

What I showed above is the proper way to evaluate, so that means you probably set up the integral incorrectly... :-/

What surface were you given? Or did they explicitly give you the Gaussian curvature \(\large \kappa(s)\)?

Answer 5

wow thats a nasty question , what level maths u doing?

Answer 6

It's the Enneper Minimal surface I am evaluating:

I'm using the parametrisation: \[\sigma=\left(\begin{matrix}u-\frac{ u^3 }{ 3 }+uv^2 \\ -v-u^2v+\frac{ v^3 }{ 3 } \\ u^2-v^2 \end{matrix}\right)\]
From this I get:
Gaussian Curvature = \[K(u,v) = \frac{ -4 }{ (1+u^2+v^2)^4 }\]

I then have to prove that:\[\int\limits_{}^{}k(s) ds = 4\pi \]

But it's the fact that my Gaussian curvature is in terms of u,v but the total curv is in 's'

Answer 7

@ayeshaafzal221 I'm a third year university maths student

Answer 8

I just want to make sure you're familiar with the first and second fundamental forms before I proceed with my calculation (this is for a differential geometry class...right?)

Answer 9

Yeah I'm familiar with them. (Yep differential geometry! ) :-) (ps thankyou so much!)

Answer 10

It turns out that your Gaussian curvature is correct. I figured out where you went wrong. The integral \(\large\displaystyle \int\kappa(s)\,dS\) that you want to compute is actually a surface integral, not just any ordinary double integral. In particular, the formula you want to use is
\[\large \iint\limits_{\sigma}\kappa(s)\,dA = \int_{\Bbb{R}}\int_{\Bbb{R}} \kappa(\sigma(u,v))\|\sigma_u\times\sigma_v\|\,du\,dv.\]
Now, since
\[\large\sigma = \begin{pmatrix}u-\frac{1}{3}u^3+uv^2\\ -v-u^2v+\frac{1}{3}v^3\\ u^2-v^2\end{pmatrix}\]
it follows that
\[\large\sigma_u = \begin{pmatrix}1-u^2+v^2\\ -2uv\\ 2u\end{pmatrix}\quad\text{ and }\quad\sigma_v = \begin{pmatrix}2uv\\ -1-u^2+v^2\\ -2v\end{pmatrix}\]
Therefore
\[\large \begin{aligned}\sigma_u\times\sigma_v &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1-u^2+v^2 & -2uv & 2u\\ 2uv & -1-u^2+v^2 & -2v\end{vmatrix}\\ &= \begin{pmatrix} 2uv^2+2u+2u^3\\ 2v+2u^2v+2v^3\\v^4+2u^2v^2+u^4-1 \end{pmatrix}\\ &=\begin{pmatrix}2u(u^2+v^2+1)\\ 2v(u^2+v^2+1)\\ (u^2+v^2)^2-1\end{pmatrix}\\ &= \begin{pmatrix}2u(u^2+v^2+1)\\ 2v(u^2+v^2+1)\\ (u^2+v^2+1)(u^2+v^2-1) \end{pmatrix} \end{aligned}\]
Now we see that
\[\large \begin{aligned}\| \sigma_u\times\sigma_v \| &= (u^2+v^2+1)\sqrt{4u^2+4v^2+(u^2+v^2-1)^2 } \\ &=(u^2+v^2+1)\sqrt{(u^2+v^2-1)^2 + 4u^2+4v^2-4+4}\\ &= (u^2+v^2+1)\sqrt{(u^2+v^2-1)^2+2\cdot(2)(u^2+v^2-1) + 2^2}\\ &= (u^2+v^2-1)\sqrt{(u^2+v^2-1+2)^2}\\ &= (u^2+v^2+1)^2\end{aligned}\]
Therefore the total curvature is
\[\begin{aligned}\int_{\Bbb{R}}\int_{\Bbb{R}}-\frac{4}{(1+u^2+v^2)^4}\cdot\|\sigma_u\times\sigma_v\|\,du\,dv &= -\int_{\Bbb{R}}\int_{\Bbb{R}} \frac{4(1+u^2+v^2)^2}{(1+u^2+v^2)^4}\,du\,dv\\ &= -\int_{\Bbb{R}}\int_{\Bbb{R}} \frac{4}{(1+u^2+v^2)^2}\,du\,dv\end{aligned}\]
Now, making the conversion to polar gives us
\[\begin{aligned}-\int_{\Bbb{R}}\int_{\Bbb{R}} \frac{4}{(1+u^2+v^2)^2}\,du\,dv\xrightarrow{\text{polar coords.}}{} &\phantom{=} -\int_0^{2\pi}\int_0^{\infty}\frac{4r}{(1+r^2)^2}\,dr\,d\theta\\ &= \left(-\int_0^{\infty}\frac{4r}{(1+r^2)^2}\,dr\right) \left(\int_0^{2\pi}\,d\theta\right)\end{aligned}\]
Making the substitution \(\large u=1+r^2\)then turns the integral into
\[\begin{aligned} \left(-\int_0^{\infty}\frac{4r}{(1+r^2)^2}\,dr\right) \left(\int_0^{2\pi}\,d\theta\right)\xrightarrow{u=1+r^2}{} &\phantom{=}\left(-\int_1^{\infty}\frac{2}{u^2}\,du\right) \left(\int_0^{2\pi}\,d\theta\right) \\ &= \left.\left(\frac{2}{u}\right)\right|_1^{\infty}\cdot\left.\bigg(\theta\bigg) \right|_0^{2\pi} \\ &= -2\cdot 2\pi \\ &= -4\pi\end{aligned}\]
Therefore, the total curvature for the Enneper surface is \(-4\pi\), which, thankfully, matches with what's on Wikipedia: http://en.wikipedia.org/wiki/Enneper_surface

Phew...that was a lot of work. XD

I hope this makes sense!!

Answer 11

Oh my god you are an actual life saverI cannot thank you enough for spending the time to do that! I've been trying to work this out for days!

I'm going to go through it now and understand it

Again thankyou! :-)

Answer 12

@sarahusher may I invite you to the http://math.stackexchange.com/ website?

Answer 13

I've just joined it :-)

Answer 14

nice, see you in the chatroom soon :)