Question

Identify the surface of the z² = 4 + 4r² by converting them into equations in the Cartesian form. Show the complete solutions.

Answer 1

The surface is given in cylindrical coordinates $(r, \theta, z),$ and the conversion formula:

$x=r cos \theta,$

$y=r sin \theta,$

$z=z.$

Since $x^2+y^2=r^2,$ we can convert the equation $z^2 = 4 + 4r^2$ into Cartesian form:

$z^2=4+4(x^2+y^2),$

$-4x^2-4y^2+z^2=4.$

By dividing the equation by 4 we obtain the equation of the hyperboloid of two sheets:

$-x^2-y^2+\frac{z^2}{4}=1,$

$-\frac{x^2}{1^2}-\frac{y^2}{1^2}+\frac{z^2}{2^2}=1.$