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

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

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Answer 1 · 2015-02-16T18:15:03

Hello !

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

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

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

enter image source here

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

enter image source here

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

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

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