Area of 2-dimensional surface

Question

I want to calculate the area of $$M = \left \{ (x,y,z) \in \mathbb R^3 : 1-z^2 = \sqrt{x^2 + y^2}, -1 \leq z \leq 1\right \}.$$ First parametrize $M$ a.e. via $$\Phi : ]-1, 1[ \times ]0, 2\pi[ \to M, \\ (z,t) \mapsto ((1-z^2) \cos t, (1-z^2) \sin t, z)$$ and we get $$A = 2\pi\int _{-1}^1 (1-z^2)\sqrt{1+4z^2}dz = 2\pi \int_{\text{arsinh(-2)}}^{\text{arsinh(2)}}(1-4\sinh(x))\cosh^2(x)dx$$ Now is this the correct way so far and how to proceed?

Answer 1

Your work looks correct (I had to do the parametrization...but I didn't check the last substitution fully), and the last integral doesn't look so terrible. First:

$$\int\cosh^2x\,dx=\frac14\int\left(e^{2x}+e^{-2x}+2\right)dx=\frac18\left(e^{2x}-e^{-2x}+4x\right)+C$$

and

$$\int\sinh x\cosh^2x\,dx=\frac13\cosh^3x+C$$

Finally, remember that

$$\text{arcsinh}\,x=\log\left(x+\sqrt{x^2+1}\right)\;,\;\;\cosh(\text{arcsinh}\,x)=\sqrt{x^2+1}$$