Respuesta :
To find:
(a) Equation for the sphere of radius 5 centered at the origin in cylindrical coordinates
(b) Equation for a cylinder of radius 1 centered at the origin and running parallel to the z-axis in spherical coordinates
Solution:
(a) The equation of a sphere with center at (a, b, c) & having a radius 'p' is given in cartesian coordinates as:
[tex](x-a)^{2}+(y-b)^{2}+(z-c)^{2}=p^{2}[/tex]
Here, it is given that the center of the sphere is at origin, i.e., at (0,0,0) & radius of the sphere is 5. That is, here we have,
[tex]a=b=c=0,p=5[/tex]
That is, the equation of the sphere in cartesian coordinates is,
[tex](x-0)^{2}+(y-0)^{2}+(z-0)^{2}=5^{2}[/tex]
[tex]\Rightarrow x^{2}+y^{2}+z^{2}=25[/tex]
Now, the cylindrical coordinate system is represented by [tex](r, \theta,z)[/tex]
The relation between cartesian and cylindrical coordinates is given by,
[tex]x=rcos\theta,y=rsin\theta,z=z[/tex]
[tex]r^{2}=x^{2}+y^{2},tan\theta=\frac{y}{x},z=z[/tex]
Thus, the obtained equation of the sphere in cartesian coordinates can be rewritten in cylindrical coordinates as,
[tex]r^{2}+z^{2}=25[/tex]
This is the required equation of the given sphere in cylindrical coordinates.
(b) A cylinder is defined by the circle that gives the top and bottom faces or alternatively, the cross section, & it's axis. A cylinder running parallel to the z-axis has an axis that is parallel to the z-axis. The equation of such a cylinder is given by the equation of the circle of cross-section with the assumption that a point in 3 dimension lying on the cylinder has 'x' & 'y' values satisfying the equation of the circle & that 'z' can be any value.
That is, in cartesian coordinates, the equation of a cylinder running parallel to the z-axis having radius 'p' with center at (a, b) is given by,
[tex](x-a)^{2}+(y-b)^{2}=p^{2}[/tex]
Here, it is given that the center is at origin & radius is 1. That is, here, we have, [tex]a=b=0,p=1[/tex]. Then the equation of the cylinder in cartesian coordinates is,
[tex]x^{2}+y^{2}=1[/tex]
Now, the spherical coordinate system is represented by [tex](\rho,\theta,\phi)[/tex]
The relation between cartesian and spherical coordinates is given by,
[tex]x=\rho sin\phi cos\theta,y=\rho sin\phi sin\theta, z= \rho cos\phi[/tex]
Thus, the equation of the cylinder can be rewritten in spherical coordinates as,
[tex](\rho sin\phi cos\theta)^{2}+(\rho sin\phi sin\theta)^{2}=1[/tex]
[tex]\Rightarrow \rho^{2} sin^{2}\phi cos^{2}\theta+\rho^{2} sin^{2}\phi sin^{2}\theta=1[/tex]
[tex]\Rightarrow \rho^{2} sin^{2}\phi (cos^{2}\theta+sin^{2}\theta)=1[/tex]
[tex]\Rightarrow \rho^{2} sin^{2}\phi=1[/tex] (As [tex]sin^{2}\theta+cos^{2}\theta=1[/tex])
Note that [tex]\rho[/tex] represents the distance of a point from the origin, which is always positive. [tex]\phi[/tex] represents the angle made by the line segment joining the point with z-axis. The range of [tex]\phi[/tex] is given as [tex]0\leq \phi\leq \pi[/tex]. We know that in this range the sine function is positive. Thus, we can say that [tex]sin\phi[/tex] is always positive.
Thus, we can square root both sides and only consider the positive root as,
[tex]\Rightarrow \rho sin\phi=1[/tex]
This is the required equation of the cylinder in spherical coordinates.
Final answer:
(a) The equation of the given sphere in cylindrical coordinates is [tex]r^{2}+z^{2}=25[/tex]
(b) The equation of the given cylinder in spherical coordinates is [tex]\rho sin\phi=1[/tex]
