Let $S$ be a one-sheeted hyperboloid $S = \{x^2+y^2-z^2=1, z\in(0,1)\}$. Evaluate $\int_S z dA$.
I tried solving this in two methods:
- By finding a parameterization: for each $z_0\in (0,1)$, the intersection of $S$ with the plane $z=z_0$ is the circle with radius $\sqrt{1+z_0^2}$. So a parameterization for the desired surface would be $r(\theta, z) = (\sqrt{1+z^2}\cos\theta, \sqrt{1+z^2}\sin\theta, z), \theta \in (0,2\pi), z\in(0,1)$. To evaluate $dA$:
$$ \partial _\theta r(\theta,z) = (-\sqrt{1+z^2}\sin\theta,\sqrt{1+z^2}\cos\theta,0)$$ $$\partial _zr(\theta,z) = (\frac{1}{\sqrt{1+z^2}}\cos\theta, \frac{1}{\sqrt{1+z^2}}\sin\theta, 1)$$
$$dA = \sqrt{\det
\left(\begin{matrix}
1+z^2 & 0 \\
0 & \frac{2+z^2}{1+z^2} \\
\end{matrix}\right)}=\sqrt{2+z^2}
$$
(We've defined $dA$ as the determinant of the Gram matrix of the partial derivatives of $r$).
And so we get: $$\int_SzdA = 2\pi\int_0^1z\sqrt{2+z^2}dz=\pi\int^3_2\sqrt{u}du=\frac{2}{3}\pi(\sqrt{27}-\sqrt{8})$$
- By defining $\gamma (t) = (t,\sqrt{t^2-1}), t\in(1,2)$ a curve in the $xz$-plane, rotating it around the $z$-axis such that $r(t,\theta) = (t\cos\theta, t\sin\theta, \sqrt{t^2-1})$ is a parameterization for the surface of revolution, so $dA = \frac{t^2}{\sqrt{t^2-1}}d\theta dt$ and then: $$\int_SzdA = 2\pi\int_1^2\sqrt{t^2-1}\frac{t^2}{\sqrt{t^2-1}}dt=2\pi\frac{t^3}{3}|^{t=2}_{t=1}=\frac{14\pi}{3}$$
which seems nicer! But I'm not sure where my error is. Would appericiate any help.
It seems like the errors in your original computation have been cleared up in the comments. Here is a slightly different way to proceed. Notice that for the surface
$$\vec{n} = \nabla(x^2+y^2-z^2) = 3(x,y,-z)$$
with corresponding unit vector
$$\hat{n} = \frac{(x,y,-z)}{\sqrt{x^2+y^2+z^2}} = \frac{(x,y,-z)}{\sqrt{2+z^2}}$$
and thus we can rewrite your integral as
$$\int_S z\:dS = \int_S (0,0,-\sqrt{2+z^2})\cdot \hat{n}dS$$
Therefore, we can apply divergence theorem to the vector field $(0,0,-\sqrt{2+z^2})$. Closing the sheet off with disks at $z=0$ and $z=1$, divergence theorem tells us
$$\int_E \frac{-z}{\sqrt{2+z^2}}\:dV = \int_S z\:dS + \int_{D_{\sqrt{2}}} -\sqrt{3}\:dS + \int_{D_1} \sqrt{2}\:dS$$
The integral on the left can be done w.r.t. to cylindrical coordinates
$$2\pi \int_0^1\int_0^{\sqrt{1+z^2}}\frac{-zr}{\sqrt{2+z^2}}\:dr\: dz = -\pi\int_0^1 \frac{z(1+z^2)}{\sqrt{2+z^2}}\:dz$$
$$=-\pi\left[\frac{1}{3}\sqrt{2+z^2}^3-\sqrt{2+z^2}\right]_0^1=-\frac{\pi\sqrt{2}}{3}$$
giving us the final answer
$$\int_Sz\:dS = -\frac{\pi\sqrt{2}}{3} +2\pi\sqrt{3} - \pi\sqrt{2} = \boxed{\pi\left(2\sqrt{3}-\frac{4\sqrt{2}}{3}\right)}$$
or equivalently $\frac{2\pi}{3}(\sqrt{27}-\sqrt{8})$