I don't understand how to proceed with an exercise. I will write down what I have done so far.
The exercise is:
Evaluate the following integral $$\iint_{\Sigma}\dfrac{1}{x^2+y^2}\ \text{d}\sigma $$
Where $\Sigma = \{(x, y, z): x^2+ y^2 = z^2+1,\ 1\leq z \leq 2 \}$
My attempt
I wrote
$$z = \sqrt{x^2+y^2-1} ~~~~~~~ \text{with} ~~~~~~~ 2 \leq x^2+y^2 \leq 5$$
Hence I fond a parametrisation curve like
$$\phi: \begin{cases} x = t \\ y = s \\ z = \sqrt{t^2+s^2-1} \end{cases} $$
Now I have to calculate the vector product of the partial derivatives gradients (I know I am expressing myself in a bad language, I apologise):
$$\frac{\partial \phi}{\partial t} = \left(1,\ 0,\ \dfrac{t}{\sqrt{t^2+s^2-1}}\right)$$
$$\frac{\partial \phi}{\partial s} = \left(0,\ 1,\ \dfrac{s}{\sqrt{t^2+s^2-1}}\right)$$
Hence
$$\frac{\partial \phi}{\partial t} \wedge \frac{\partial \phi}{\partial t} = \text{det}\begin{pmatrix} i & j & k \\ 1 & 0 & \dfrac{t}{\sqrt{t^2+s^2-1}} \\ 0 & 1 & \dfrac{s}{\sqrt{t^2+s^2-1}} \end{pmatrix} $$
Which lead me to
$$\bigg|\bigg| \frac{\partial \phi}{\partial t} \wedge \frac{\partial \phi}{\partial t}\bigg|\bigg| = \sqrt{\dfrac{2(t^2+s^2)-1}{t^2+s^2-1}} $$
Now I should evaluate the integral but I don't know how to proceed since I would get
$$\iint_{\Sigma} \dfrac{1}{s^2+t^2} \sqrt{\dfrac{2(t^2+s^2)-1}{t^2+s^2-1}}\ \text{d}\sigma\ \text{d}s$$
And I cannot continue...
I should get $3\pi$ as a result. Can anyone help me?
Thank you!
It is easy if you use the fact that $\Sigma$ have rotational symmetry around $z$ axis :
$$\iint_\Sigma\frac{d\sigma}{x^2+y^2}=\iint_\Sigma\frac{d\sigma}{z^2+1}=\int_1^2\frac{2\pi\sqrt{z^2+1}}{z^2+1}\sqrt{1+\bigg(\frac{d(\sqrt{z^2+1})}{dz}\bigg)^2}dz$$$$=2\pi\int_1^2\frac{\sqrt{2z^2+1}}{z^2+1}dz.$$
However, I don't think 3$\pi$ is the value of the integral. When I did numerical calculation of this integral, it gave me approx. $4.5595$, which clearly is not $3\pi$. Also Mathematica yields : $$2\pi\int_1^2\frac{\sqrt{2z^2+1}}{z^2+1}dz=\pi\bigg(2\sqrt{2}\Big(\arcsin(2\sqrt{2})-\arcsin(\sqrt{2})\Big)+\log((2+\sqrt{3})/5)\bigg)$$