Gravitational Force Integral

Tazerfish · Jul 2, 2016

irst of all an apology : I was very uncertain where to put this. .I am doing this for fun. It isn't really homework so I don't care about any specifics or numbers.Additionally, I couldn't really follow the template with my question.
I also wasn't sure how difficult this problem really is.
Sorry, if this is the wrong place or phrased the wrong way .

I am confused about multiple integrals.Specifically, when integrating over angles.

Calculate the gravitational force on an object sitting in the middle of a hemispherical "planet".
(By middle, I mean middle of the flat surface)

The thing is I "know" the solution. I just don't understand it.
You integrate the downward force from "shells"
Solution:## dF=\int_0^{2\pi} d\phi \int_0^\frac{\pi}{2} d\theta r^2 sin(\theta) dr \rho \frac{Gm}{r^2} cos(\theta) ##
I don't know why this equation looks like it does...
Does ## \int_0^{2\pi} d\phi \int_0^\frac{\pi}{2} d\theta r^2 dr ## give us some infinitesimal volumes ?
Or rather, how to interpret the above formula?(Or the part about the volume in the first equation if I did mess something up)
Why is there a sin and a cos in the formula?
I suppose it isn't really that hard and I just don't "see" the thought behind it.
I would be glad if someone answered
PS: How to make the Integrals big in tex?
EDIT: I think I get some part of it now.
The integrals of the angles together with the r^2 produce the shells.
The theta integral makes a curve on the surface of the hemisphere.And the phi integral rotates it around to make it an area.The last integral with dr integrates the shells into volumes ... right ?
The cos is there because the component of the force downward is ##cos (\theta) F##
And the sin is there because the rings the phi integral would produce without the theta integral would have the radius ##r_0 * sin(\theta)=r_{of the ring}##

I think I understood it now.
But i dont think anyone else will by reading my rambling ....
Is there a way o delete my post ?

Charles Link · Jul 2, 2016

Tazerfish said: ↑

irst of all an apology : I was very uncertain where to put this. .I am doing this for fun. It isn't really homework so I don't care about any specifics or numbers.Additionally, I couldn't really follow the template with my question.
I also wasn't sure how difficult this problem really is.
Sorry, if this is the wrong place or phrased the wrong way .

I am confused about multiple integrals.Specifically, when integrating over angles.

Calculate the gravitational force on an object sitting in the middle of a hemispherical "planet".
(By middle, I mean middle of the flat surface)

The thing is I "know" the solution. I just don't understand it.
You integrate the downward force from "shells"
Solution:## dF=\int_0^{2\pi} d\phi \int_0^\frac{\pi}{2} d\theta r^2 sin(\theta) dr \rho \frac{Gm}{r^2} cos(\theta) ##
I don't know why this equation looks like it does...
Does ## \int_0^{2\pi} d\phi \int_0^\frac{\pi}{2} d\theta r^2 dr ## give us some infinitesimal volumes ?
Or rather, how to interpret the above formula?(Or the part about the volume in the first equation if I did mess something up)
Why is there a sin and a cos in the formula?
I suppose it isn't really that hard and I just don't "see" the thought behind it.
I would be glad if someone answered
PS: How to make the Integrals big in tex?
EDIT: I think I get some part of it now.
The integrals of the angles together with the r^2 produce the shells.
The theta integral makes a curve on the surface of the hemisphere.And the phi integral rotates it around to make it an area.The last integral with dr integrates the shells into volumes ... right ?
The cos is there because the component of the force downward is ##cos (\theta) F##
And the sin is there because the rings the phi integral would produce without the theta integral would have the radius ##r_0 * sin(\theta)=r_{of the ring}##

I think I understood it now.
But i dont think anyone else will by reading my rambling ....
Is there a way o delete my post ?

It's an interesting question. The integral solution you give is in spherical coordinates and the infinitesimal volume ## dv=r^2 \sin{\theta} \ d \theta \ d \phi \ dr ## . It really is not doing it in any kind of shell, but there is a z-axis symmetry. The ## \cos{\theta} ## gives the z-component of the force, since by symmetry any radial components will cancel. ## \rho ## is the density so that ## \rho dv ## gives the elemental mass in volume ## dv ##. The remaining part is just the inverse square law with the universal gravitational constant ## G ##. And you also need to integrate the ## dr ## from ## 0 ## to ## R ##.

Tazerfish · Jul 3, 2016

Charles Link said: ↑

It's an interesting question. The integral solution you give is in spherical coordinates and the infinitesimal volume ## dv=r^2 \sin{\theta} \ d \theta \ d \phi \ dr ## . It really is not doing it in any kind of shell, but there is a z-axis symmetry. The ## \cos{\theta} ## gives the z-component of the force, since by symmetry any radial components will cancel. ## \rho ## is the density so that ## \rho dv ## gives the elemental mass in volume ## dv ##. The remaining part is just the inverse square law with the universal gravitational constant ## G ##. And you also need to integrate the ## dr ## from ## 0 ## to ## R ##.

Well in some sense there are shells...
## \int_0^{2\pi} r sin(\theta) d\phi ## gives us rings. And summing up the rings along the surface with the same r away from the "middle" gives us the shells (or rather hemispherical areas)
## \int_0^{\frac{\pi}{2}}(\int_0^{2\pi} r sin(\theta) d\phi)r d\theta ## and if you integrate that over dr you get the volume.
Is it correct now ?

Charles Link · Jul 3, 2016

Tazerfish said: ↑

Well in some sense there are shells...
## \int_0^{2\pi} r sin(\theta) d\phi ## gives us rings. And summing up the rings along the surface with the same r away from the "middle" gives us the shells (or rather hemispherical areas)
## \int_0^{\frac{\pi}{2}}(\int_0^{2\pi} r sin(\theta) d\phi)r d\theta ## and if you integrate that over dr you get the volume.
Is it correct now ?

The ## r^2 ## factors cancel (numerator and denominator). Also, integrate over ## dr ## from ## 0 ## to ## R ## to get ## R ##. If ## \rho=\rho(r,\theta) ## (you need ## \phi ## symmetry), the integral would still give the correct answer. Spherical coordinate integrals with ## \phi ## symmetry will give rings, etc., but that feature is of secondary importance. You don't need to look for rings or shells in using the spherical coordinates. (In doing integrals where you rotate a curve about an axis and compute the volume, there is a "shell" method, but going this route really isn't necessary in spherical coordinates.) You can simply mechanically do the integrals and most often, the triple integral easily separates into the product of 3 integrals.

Tazerfish · Jul 3, 2016

Charles Link said: ↑

Spherical coordinate integrals with ϕ \phi symmetry will give rings, etc., but that feature is of secondary importance.

I am fairly new to calculus and I really like to know what I am doing.This picture about rings and shells makes me understand why the formula looks like that.
Otherwise, how would you know that what you are integrating in the first formula is truly the gravitational force downward ?
How would you derive this formula if you just

Charles Link said: ↑

simply mechanically do the integrals

?

Charles Link · Jul 3, 2016

Tazerfish said: ↑

I am fairly new to calculus and I really like to know what I am doing.This picture about rings and shells makes me understand why the formula looks like that.
Otherwise, how would you know that what you are integrating in the first formula is truly the gravitational force downward ?
How would you derive this formula if you just
?

Your approach is a good one. When I first learned calculus, I also would analyze the steps very carefully in evaluating double and triple integrals, especially in picking the limits on the double integrals, etc. After doing quite a number of them, they have became very routine, but working them more carefully from first principles is a very healthy approach. As a college student I also would not take any shortcuts until it became a well-proven path.

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