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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.

(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 ?

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 ?

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