How do you sketch #f(x,y)=arcsin(x^2+y^2-2)#?

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Answer 1 · 2015-02-16T18:15:03

Hello !

Let S be the surface of equation #z = \text{arcsin}(x^2+y^2-2)#.

S is a surface of revolution because #z = F(r)# where #r=sqrt{x^2+y^2}#. Here, #F(r) = \text{arcsin}(r^2-2)#.

First, you study the curve of equation #z = \text{arcsin}(x^2-2)# : you get

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Second, you rotate this curve around (0z) axis and you get the surface S :

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Remark that #f# exists only on the domain defined by #1\leq x^2+y^2\leq 3# : it's a disk of radius #sqrt{3}# with an circular hole of radius 1.