by **Guest** » Thu May 06, 2021 9:16 pm

Do you know what "cylindrical coordinates" and "spherical coordinates" are?

Cylindrical coordinates have polar coordinates in the xy-plane, so a point (x, y) is given by the distance from the origin to (x, y), so that [tex]r= \sqrt{x^2+ y^2}[/tex], and the angle from the positive x-axis to the line from the origin to the line from the origin to (x, y), so [tex]\theta= arctan(y/x)[/tex], and the same z-axis as used in Cartesian coordinates. Going the other way, [tex]x= r cos(\theta)[/tex], [tex]y= r sin(\theta)[/tex], and, of course, z= z.

Spherical cooridnates designate the point (x, y, z) by [tex]\rho[/tex] the distance from the origin to (x, y, z) so [tex]\rho= \sqrt{x^2+ y^2+ z^2}[/tex], the angle [tex]\theta[/tex] from the x-axis to the line from the origin to the point (x, y, 0) in the xy-plane, so [tex]\theta= arctan(y/x)[/tex], and [tex]\phi[/tex], the angle from the positive z-axis to that line, so [tex]\phi= arctan(r/z)[/tex] (r is the same as in cylindrical coordinates. Going the other way, [tex]x= \rho cos(\theta) sin(\phi)[/tex], [tex]y= \rho sin(\theta) sin(\phi)[/tex] and [tex]z= \rho cos(\phi)[/tex].

Now, you try the other problems yourself!

The first question is about the Cartesian equation z= 2. That is the plane consisting of all point (x, y, 2), parallel to the xy-plane and 2 units above it. Since the "z" coordinate is the same for Cartesian and Cylindrical coordinates, the equation for Cylindrical coordinates is the same, z= 2. For spherical coordinates, [tex]z= \rho cos(\phi)= 2[/tex].

The second is about the Cartesian equation [tex]z= 5x^2+ 5y^2[/tex]. That is the same as [tex]z= 5(x^2+ y^2)= 5r^2[/tex]. In Cylindrical coordinates that is [tex]z= 5r^2[/tex]. It is particularly simple because, geometrically, the figure is a cylinder! In spherical coordinates [tex]z= \rho cos(\phi)= 5(\rho^2 cos^2(\theta)sin^2(\phi)+ \rho^2 sin^(\theta) sin^2(\phi))= 5\rho^2 sin^2(\phi)(cos^2(\theta)+ sin^2(\theta))= 5\rho^2 sin^2(\phi)[/tex]. [tex]\rho cos^(\phi)= 5\rho^2 sin^2(\phi)[/tex].[tex]\rho= \frac{cos(\phi)}{5 sin^2(\phi)}[/tex].