Computing the integral of a certain surface

Question

I am given the surface $$S:=\{(x,y,z)\in\mathbb{R}^3|\sqrt{x^2+y^2}=2\cosh 2z,z\in[0,1]\}$$ They ask me to find a parametrization for this surface of revolution. This corresponds to the following $$\varphi(u,v)=(2\cosh(2u)\cos(v),2\cosh(2u)\sin(v),u)$$ with $(u,v)\in[0,1]\times[0,2\pi]$. With this, they ask me to compute the integral over the 2-form $\omega=(z^2+2z)dx\wedge dy+(2x+2yz-z)dx\wedge dz$ if our surface is oriented by the normal vector $\nu(2,0,0)=(-1,0,0)$. Since $\varphi(0,0)=(2,0,0)$, $\varphi_u(0,0)=(0,0,1)$ and $\varphi_v(0,0)=(0,2,0)$, we have that $\nu(0,0)=\varphi_u(0,0)\wedge\varphi_v(0,0)=(-2,0,0)$ and thus we have chosen the correct orientation. Now, $$\int_{S}\omega=\int_{[0,1]\times[0,2\pi]}\varphi^{*}((z^2+2z)dx\wedge dy+(2x+2yz-z)dx\wedge dz)$$ To compute the pullback we do the following $$(u^2+2u)d(2\cosh(2u)\cos(v))\wedge d(2\cosh(2u)\sin(v))+(4\cosh(2u)\cos(v)+4u\cosh(2u)\sin(v)-u)$$ $$d(2\cosh(2u)\cos(v))\wedge du$$ $$= 4(u^2+2u)(2\sinh(2u)\cos(v)du-\sin(v)\cosh(2u)dv)\wedge (2\cosh(2u)\cos(v)dv+4\sin(v)\sinh(2u)du)+2(4\cosh(2u)\cos(v)+4u\cosh(2u)\sin(v)-u)(2\sinh(2u)\cos(v)du-\sin(v)\cosh(2u)dv)\wedge du$$ $$=4(u^2+2u)(4\sinh(2u)\cosh(2u)\cos(v)^2 du\wedge dv + 4\sinh(2u)\cosh(2u)\sin(v)^2 du\wedge dv)+2\sin(v)\cosh(2u)(4\cosh(2u)\cos(v)+4u\cosh(2u)\sin(v)-u)du\wedge dv$$ $$=16(u^2+2u)\sinh(2u)\cosh(2u)du\wedge dv+2\sin(v)\cosh(2u)(4\cosh(2u)\cos(v)+4u\cosh(2u)\sin(v)-u)du\wedge dv$$

But this seemes a super long process, plus I don't really know if I will be able to integrate the expression. Is there a simpler way? Like using Stokes theorem or so?

Answer 1

I will try to answer my own question using brute-force, i.e. integrating as usual. We have, $$4(u^2+2u) d(\cosh(2u)\cos(v))\wedge d(\cosh(2u)\sin(v))+(8\cosh(2u)\cos(v)+8u\cosh(2u)\sin(v)-2u) d(\cosh(2u)\cos(v))\wedge du$$ $$=4(u^2+2u)(2\sinh(2u)\cos(v)du-\sin(v)\cosh(2u)dv)\wedge (2\sinh(2u)\sin(v)du+\cos(v)\cosh(2u)dv)+(8\cosh(2u)\cos(v)+8u\cosh(2u)\sin(v)-2u)(-\sin(v)\cosh(2u)dv)\wedge du$$ $$=4(u^2+2u)(2\sinh(2u)\cosh(2u)\cos(v)^2du\wedge dv+2\sinh(2u)\cosh(2u)\sin(v)^2du\wedge dv)+\sin(v)\cosh(2u)(8\cosh(2u)\cos(v)+8u\cosh(2u)\sin(v)-2u)du\wedge dv$$ $$=8(u^2+2u)\sinh(2u)\cosh(2u)du\wedge dv+8\sin(v)\cos(v)\cosh(2u)^2du\wedge dv+8u\cosh(2u)^2\sin(v)^2du\wedge dv-2u\sin(v)\cosh(2u)du\wedge dv$$

Now we have to compute $$\int_{S}\omega=\int_{0}^{1}\int_{0}^{2\pi}(\star)$$ where $(\star)$ corresponds to the last expression. This can be split like the following, $$\int_{0}^{1}\int_{0}^{2\pi}8(u^2+2u)\sinh(2u)\cosh(2u)dvdu = \frac{\pi}{4}(-1-8\sinh(4)+25\cosh(4))$$ $$\int_{0}^{1}\int_{0}^{2\pi} 8\sin(v)\cos(v)\cosh(2u)^2dvdu = 0$$ $$\int_{0}^{1}\int_{0}^{2\pi} 8u\cosh(2u)^2\sin(v)^2dvdu = -\frac{\pi}{4}(-9-4\sinh(4)+\cosh(4))$$ $$\int_{0}^{1}\int_{0}^{2\pi} -2u\sin(v)\cosh(2u)dvdu = 0$$

So, our result is $\frac{\pi}{4}(-1-8\sinh(4)+25\cosh(4))-\frac{\pi}{4}(-9-4\sinh(4)+\cosh(4))=2\pi+\frac{7\pi}{2e^{4}}+\frac{5\pi e^{4}}{2}$

Answer 2

I think what @NinadMunshi was getting at was to see S as the boundary of the volume of revolution. (I don't know why they accused you of being disrespectful! - I thought their final comment sounded disrespectful if anything was.) So $$\int_V\mathrm{d}\omega = \int_S\omega + \text{contributions from top and bottom discs}$$ I'm too lazy to do the actual computations, so will leave them to you (though beware of the signs) ...

Answer 3

I didn't check all your developments, because it is a super long process, as you said, but here is a quicker way to carry such a computation.

The first step invokes Stokes' theorem in order to transform the surface integral over $S$ into a volume integral over $V$, with $S = \partial V$. One has thus $I := \int_S\omega = \int_V\mathrm{d}\omega$, with $\mathrm{d}\omega = 2\, \mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z$, which is nothing else than twice the volume form.

Now, we must choose a parametrization for $V$. The one you gave is based on the cylindrical coordinates with a varying radius; however, it can be simplified a little bit further by recasting it as $$ V = \{(r\cos\theta,r\sin\theta,z) \in \mathbb{R}^3 \,|\, r\in[0,2\cosh(2z)], \theta\in[0,2\pi], z\in[0,1]\}. $$ You already know the volume form expressed in cylindrical coordinates, so that $\mathrm{d}\omega = 2\,\mathrm{dvol}_3 = 2r\, \mathrm{d}r \wedge \mathrm{d}\theta \wedge \mathrm{d}z$ and $$ \begin{array}{rcl} I &=&\displaystyle 2\int_{[0,2\cosh(2z)] \times [0,2\pi] \times [0,1]} r\, \mathrm{d}r \wedge \mathrm{d}\theta \wedge \mathrm{d}z \\ &=&\displaystyle 8\pi\int_0^1 \cosh^2(2z) \,\mathrm{d}z \\ &=&\displaystyle 4\pi\left.\left(z+\frac{1}{4}\sinh(4z)\right)\right|_0^1 \\ &=&\displaystyle \pi(4+\sinh(4)) \end{array} $$ with the help of the identity $\cosh^2\left(\frac{z}{2}\right) = \frac{1}{2}(1+\cosh z)$.