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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?

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  • $\begingroup$ If I am correct with what you mean by $dS$ then I did that and this is what I got. $\endgroup$ – Staki42 Feb 22 '18 at 21:18
  • $\begingroup$ Indeed, it is, I just wasn't following your work. $\endgroup$ – Doug M Feb 22 '18 at 21:20
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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

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

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  • $\begingroup$ Thanks, with these terrible looking integrals I am sometimes unsure if what I have done is correct. BTW: the inverse of the hyperbolic functions are called arsinh, arcos (because they measure area enclosed by the hyperbola and two rays from the origin). $\endgroup$ – Staki42 Feb 23 '18 at 8:53
  • $\begingroup$ @Staki42 In fact, they appear in many forms: as $\;\sinh^{-1}\;$ and etc. (i.stack.imgur.com/Jo52z.jpg), or as you say $\;\text{arshinh}\,x\;$ (en.wikipedia.org/wiki/Inverse_hyperbolic_functions), and also as $\;\text{argsinh}\,x\;$ and etc. (home.scarlet.be/math/hyp.htm#Inverse-hyperbolic-f) . The most usual, though, seems to be the first one mentioned above. $\endgroup$ – DonAntonio Feb 23 '18 at 9:48
  • $\begingroup$ Interesting! I only remember I was told to not write $\text{arcsinh}$ since it's not the arc length being measured :) $\endgroup$ – Staki42 Feb 23 '18 at 10:02
  • $\begingroup$ @Staki42 Whoever told you may have been a zealot of little points. Nobody will ever put in doubt what that means (as far as they know what hyperbolic functions are...). If that were a good reason, why then take the name hyperbolic sine, cosine and etc. if those are not trigonometric functions? Well, we want to stress the similarities between both kinds of functions. Mind you, I agree with the notation arsinh$\,(x)\;$ more...but I really don't pay that much attention to those little details $\endgroup$ – DonAntonio Feb 23 '18 at 10:11
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    $\begingroup$ thanks for your opinion, good point that they aren't actually trigonometric functions! $\endgroup$ – Staki42 Feb 23 '18 at 11:12

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