Spherical coordinates angles of phi

In summary, spherical coordinates angles of phi are angles used in the spherical coordinate system to describe the location of a point in three-dimensional space. They differ from other coordinate systems in that they use a different set of coordinates and have a range of values from 0 to 180 degrees. These angles are commonly used in various real-world applications, and can also be converted to other coordinate systems using trigonometric functions.
  • #1
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im having trouble determining the angles of phi in spherical coordinates when asked to convert a triple integral into spherical, and find the limits of the phi integral. can anybody point out any hints/tips/tricks how this may be done??Please...i have an exam tomorrow and I am tryn to prepare for it...
 
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  • #2
mugzieee said:
im having trouble determining the angles of phi in spherical coordinates when asked to convert a triple integral into spherical, and find the limits of the phi integral. can anybody point out any hints/tips/tricks how this may be done??Please...i have an exam tomorrow and I am tryn to prepare for it...

well... phi is the angle measured from the positive z-axis. a solid sphere consists of points whose value of phi vary from 0 to pi.

cones, on the other hand, have points whose phi values vary from 0 to some specified or implied angle.

umm... this is what you're asking for? :confused:
 
  • #3
for example I am given a triple integral for the function w=z
outermost integral is:
the integral from -2 to 2, dx

middle integral is:
the integral from -sqrt(4-x^2) to sqrt(4-x^2), dy

and inner most integral is:
the integral from x^+y^2 to 4

they say to convert this to spherical.
i have a hard time finding the new limits of integration

the way i interpret this is to say that z=4, z=x^2+y^2
y=-sqrt(4-x^2) and y=sqrt(4-x^2)
and x=-2 and x=2.
then from here i use the conversion factors to get the limits?

man I am stumped..
 
  • #4
Okay, have you drawn the graph?
First draw a 2 dimensional graph and draw vertical lines at x= -2, x= 2.
Now y is between [tex]y=-\sqrt{4-x^2}[/tex] and [tex]y= /sqrt{4-x^2}[/tex] which you should recognise immediately as both giving [tex]y^2= 4- x^2[/tex] or [tex]x^2+y^2= 4[/tex], the circle centered at (0,0) with radius 2 (and so fitting nicely between x= -2 and x= 2).
The innermost integral has z between [tex]z= x^2+ y^2[/tex] and z= 4, a paraboloid and a horizontal plane.

Since everything is circularly symmetric, obviously [tex]\theta[/tex] runs from 0 to [tex]2\pi[/tex].
The paraboid [tex]z= x^2+y^2[/tex] is tangent to the xy-plane at (0,0) so [tex]\phi[/tex] starts at [tex]\frac{\pi}{2}[/tex].

The hard part (are you really required to do this in spherical coordinates? Cylindrical coordinates would be much easier.) is determining what [tex]\phi[/tex] is when we change from the paraboloid to the plane. [tex]z= x^2+ y^2[/tex] to z= 4 when, of course, [tex]x^2+ y^2= 4[/tex]. Looking along the x-axis (which we can do because of the symmetry), when z= 4 and x= 2. The line through (0,0,0) to (2, 0, 4) has slope 4/2= 2 so cot(φ)= 2 (remember that φ is measured from the z-axis).\
That means we need to break the integral into 2 parts, one with φ ranging from 0 to arccot(2), the other from arccot(2) to [tex]\frac{\pi}{2}[/tex]. Now we need to calculate ρ for each of those. For the first integral, ρ is measured along the line from (0,0,0) to the plane z= 4 along the line with slope cot(φ). Use cos(φ)= cos(arccot(2))= 4/ρ so that [tex]\rho= 2\sqrt{5}[/tex].
Doing the same for the paraboloid, φ from arccot(2) to [tex]\frac{\pi}{2}[/tex] is going to be harder.
 
  • #5
Hey guys. Always nice to draw a plot of what you're trying to integrate. As I see it, you're integrating f(x,y,z)=z over the volume enclosed inside the paraboloid and below the surface z=4 as shown below. :smile:

Edit: I tell you what, if I were attempting to convert this into spherical coordinates, I might try to flip it over and work from the flat surface UP to the paraboloid. This to me seems more amendable to spherical coordinates or am I making it more difficult?
 

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  • #6
I modified the definition of a volume element of a spherical partition in 3-D in order to work this problem in terms of spherical coordinates. The best way to envision this is to look at the standard definition of the partition and then just turn it upside down.

Rho then becomes the line segment from the point (0,0,4) to the point (x,y,z). Phi is the angle between rho and the z axis with phi=0 when rho is vertical and phi=pi/2 when rho is horizontal. We then have the following definitions:

[tex]x=\rho Sin[\phi]Cos[\theta][/tex]

[tex]y=\rho Sin[\phi]Sin[\theta][/tex]

[tex]z=4-\rho Cos[\phi][/tex]

I then expressed the paraboloid [itex]z=x^2+y^2[/tex] in terms of this new coordinate system:

[tex]\rho[\phi]=\frac{-Cos[\phi]+\sqrt{Cos^2(\phi)+16 Sin^2(\phi)}}{2Sin^2(\phi)}[/tex]


The integral in cartesian corrdinates is:

[tex]\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{x^2+y^2}^4 z dz dy dz[/tex]

The integral in this new spherical coordinate system is then:

[tex]\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\rho[\phi]} (4-\rho Cos(\phi)\ \rho^2 Sin(\phi)d\rho d\theta d\phi[/tex]

The value in both cases is:

[tex]\frac{64 \pi}{3}[/tex]
 
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1. What are spherical coordinates angles of phi?

Spherical coordinates angles of phi are angles used in the spherical coordinate system to describe the location of a point in three-dimensional space. Phi represents the angle measured from the positive z-axis to the point, and ranges from 0 to 180 degrees.

2. How are spherical coordinates angles of phi different from other coordinate systems?

Spherical coordinates angles of phi are different because they use a different set of coordinates to describe the location of a point in three-dimensional space. In spherical coordinates, the location of a point is given by the radial distance, polar angle (theta), and azimuthal angle (phi) instead of the traditional x, y, and z coordinates.

3. What is the range of values for spherical coordinates angles of phi?

The range of values for spherical coordinates angles of phi is from 0 to 180 degrees. This represents a half-circle in the spherical coordinate system, with 0 degrees pointing in the positive z-axis direction and 180 degrees pointing in the negative z-axis direction.

4. How are spherical coordinates angles of phi used in real-world applications?

Spherical coordinates angles of phi are used in many real-world applications, including astronomy, physics, and navigation. They are particularly useful for describing the location of objects in spherical or nearly spherical shapes, such as planets, stars, and satellites.

5. Can spherical coordinates angles of phi be converted to other coordinate systems?

Yes, spherical coordinates angles of phi can be converted to other coordinate systems such as Cartesian coordinates or cylindrical coordinates. This conversion involves using trigonometric functions to calculate the x, y, and z coordinates from the given radial distance, polar angle, and azimuthal angle.

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