I want to compute the surface area of a hyperboloid:
$$ x^2 + y^2 -(z-11)^2 = a$$
For $0 < z < 14$, where $a$ is some positive number such that the hyperboloid is of 1 sheet.
What I did was:
$$ x = r \cos(\theta), y = r \sin(\theta) \Rightarrow r(z) = \sqrt{a+(z-11)^2}$$
And hence, the perimeter of every "ring" of the hyperboloid is:
$$ 2 \pi r(z) $$
Which means that the surface area should be:
$$ \int_{0}^{14} 2\pi r(z)dz = \int_{0}^{14}2\pi\sqrt{a+(z-11)^2}dz$$
What my friends did was computing:
$$\left | N \right | = \left | \frac{\partial}{\partial z} (r \cos(\theta)) \times \frac{\partial}{\partial \theta} (r \sin(\theta)) \right | = \left | (-r\cos(\theta), r \sin(\theta),r \cdot \frac{\partial r}{\partial z} )) \right | = \sqrt{r^2 + (r \cdot \frac{\partial r}{\partial z})^2 }$$
And:
$$\frac{\partial r}{\partial z} = \frac{z-11}{\sqrt{(z-11)^2 + a}} $$ Which means that: $$ \left | N \right | = \sqrt{a+ 2 (z-11)^2}$$
And hence the surface area should be:
$$ \iint\limits_S 1 \cdot ds = \int_{0}^{2\pi}\int_{0}^{14}\sqrt{a + 2(z-11)^2}dzd\theta$$
Which naturally, yields a different result.
I tend to think that my own way, using $ \int 2\pi r(z)dz$ is wrong. But, It's striking me that the integrands of both ways are so similar, and I can't explain why or what's wrong with my own way.