From Cartesian to spherical integral

In summary, the correct setup for the given problem is: $$\int_{3\pi/4}^\pi \int_0^{2\pi}\int_0^{\sqrt{2}}\frac{\sin\phi\;\text{d}\rho \text{d} \theta \text{d}\phi}{\rho^2}$$with limits for ##\rho## from 0 to ##\sqrt{2}##, ##\theta## from 0 to ##2\pi##, and ##\phi## from ##3\pi/4## to ##\pi##.
  • #1
Feodalherren
605
6

Homework Statement


coords.png



Homework Equations



[itex][/itex]

The Attempt at a Solution


Is this the correct setup?

[itex]\int^{\pi}_{\frac{3\pi}{4}}\int^{2\pi}_{0}\int^{\sqrt{2}}_{0}\frac{1}{\rho^{2}} rho^2 Sin\phi d\rho d\theta d\phi[/itex]

I gave up on itex. It was either that or my computer flying out the window. My apologies.
 
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  • #2
You came up with:
$$\int_{3\pi/4}^\pi \int_0^{2\pi}\int_0^{\sqrt{2}}\frac{\sin\phi\;\text{d}\rho \text{d} \theta \text{d}\phi}
{\rho^2}$$
(use the quote button on this post to see how I did that ;) )

Please show your working.
 
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  • #3
God, that looks complicated. I'll dedicate some time to actually learning how to write those integrals as nicely as you do after this week. I'm drowning in chemistry and math right now and haven't got a second to spare to learn itex right now :).

What I did was I drew a picture and then basically looked at it.
The limits for theta are obvious. The phi I got from setting x=0 on the upper limit and getting z=-x in the zy plane, this is 45 degree line, it's negative so 3pi / 4.

As for the rho limits, that's where I get less sure of what I'm doing.

I think that what I'm looking for is an ice cream cone-shape. Therefore the rho starts from 0 and ... Oh no wait I think I see it now. If I set the z-limits equal to each other I get the intercept which is r=1.
So then rho would go from 0 to 1. Correct?
 
  • #4
Feodalherren said:

Homework Statement


coords.png



Homework Equations



[itex][/itex]

The Attempt at a Solution


Is this the correct setup?

[itex]\int^{\pi}_{\frac{3\pi}{4}}\int^{2\pi}_{0}\int^{\sqrt{2}}_{0}\frac{1}{\rho^{2}} rho^2 Sin\phi d\rho d\theta d\phi[/itex]

I gave up on itex. It was either that or my computer flying out the window. My apologies.

That is correct except why not cancel the ##\rho^2## factors?


Feodalherren said:
As for the rho limits, that's where I get less sure of what I'm doing.

I think that what I'm looking for is an ice cream cone-shape. Therefore the rho starts from 0 and ... Oh no wait I think I see it now. If I set the z-limits equal to each other I get the intercept which is r=1.
So then rho would go from 0 to 1. Correct?

No, ##\rho## isn't the same as ##r##. Your original limits are correct.
 
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1. What is the Cartesian coordinate system?

The Cartesian coordinate system is a mathematical system used to describe the position of a point in space. It is named after mathematician and philosopher René Descartes, who developed the system in the 17th century. The system uses three perpendicular axes (x, y, and z) to determine the location of a point in three-dimensional space.

2. What is the spherical coordinate system?

The spherical coordinate system is another mathematical system used to describe the position of a point in three-dimensional space. It uses three parameters - radius, inclination, and azimuth - to specify the location of a point in terms of its distance from a fixed origin, its angle with respect to the positive z-axis, and its angle with respect to the positive x-axis, respectively.

3. How do you convert from Cartesian to spherical coordinates?

To convert from Cartesian to spherical coordinates, you can use the following equations:

r = √(x^2 + y^2 + z^2)

θ = arccos(z/r)

φ = arctan(y/x)

where r is the distance from the origin, θ is the inclination angle, and φ is the azimuth angle.

4. Why is the spherical coordinate system useful?

The spherical coordinate system is useful in many fields of science and engineering, such as physics, astronomy, and navigation. It allows for a more natural representation of objects and phenomena that have spherical symmetry, such as planets, stars, and electromagnetic fields. It also simplifies certain mathematical calculations and equations.

5. What are some real-world applications of converting from Cartesian to spherical coordinates?

One example of a real-world application is in astronomy, where the positions of celestial objects are often described using the spherical coordinate system. Another application is in navigation, where the spherical coordinate system is used to determine the position of a ship or aircraft in terms of latitude, longitude, and altitude. Additionally, converting from Cartesian to spherical coordinates can be useful in 3D computer graphics and simulations.

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