How to calculate for the volume taken up by a partially filled sphere?

[mesa]

So here is something a little different than the last couple formulas I needed, this one has to calculate for the volume taken up by a partially filled sphere.

The actual volume I need is at a depth of 1/2r but it would be better (and more fun) to build a formula that can be used for any given depth up to value r.

Any Thoughts?

 

[berkeman]

So here is something a little different than the last couple formulas I needed, this one has to calculate for the volume taken up by a partially filled sphere.

The actual volume I need is at a depth of 1/2r but it would be better (and more fun) to build a formula that can be used for any given depth up to value r.

Any Thoughts?

Have you had integral calculus? That is a pretty standard problem to solve with an integral. You add up the small volumes of lots of horizontal disks centered on the vertical axis. You know the horizontal radius of each disk, and the vertical height of each disk is dz tall. You integrate up the volumes of all of the very thin (dz vertical thickness) disks using an integral.

 

[mesa]

integral calculus?

That's cheating ;)

 

[mesa]

Well, here is what I came up with so far. I decided to start with the specific volume (r/2) that I needed and go from there.

I drew out a "sphere" and decided to calculate the radius for the circle that represents the top of the volume using some trig. r×cos60=r/2 so that gave me the angle to radius~b on the sphere and gave a right triangle with the short side being the radius~b I was looking for, so used sin60 to get:

r~b = sin60 × r~a or
r~b = .866 × r~a

From here I decided to cut this thing up into cylinders of thickness determined by incremental angles of cos as a function of the radius and then add the stack up:

∏×(r~a×sin 60)^2×[r~a×cos60-r~a×cos59)] + ∏(r~a×sin59)^2×[r~a×cos59-r~a×cos58)] + ...

Now I need to find a function for the summation to sin0 although, to increase volume accuracy, it would be better to have the formula incorporate much thinner cylinders using a shorter function of height based off of cos. I know I don't need trig to represent the height but I think it will make it easier to come up with a formula using similar terms.

Any thoughts?

 

[mesa]

Okay, so I was told that the way I went about doing this is similar to calc so I tossed it out and tried something else. I think I have it, sort of...

So I started out with trying to find the area of a partially filled circle and then try to add the 3rd dimension to that formula.

I ended up getting (∏r^2)/2 - r^2[2∏(θ/360) + (sinθ)(cosθ)]
θ is a function of r based on the height of the volume the angle of the radius from the center to the top of the volume off of the ceneter horizontal diameter.

I then divided the formula for the volume of a sphere by the area of a circle and got 4/3r and multiplied the above forumal by that and it seems to work, Neat!

Now here comes the sort of lol. I need to convert θ into a function of radius (not a big deal) and this formula can only calculate for volumes under 1/2 the total filled volume (this might be tough).

Any thoughts?

 

[Point Conception]

Have you had integral calculus? That is a pretty standard problem to solve with an integral. You add up the small volumes of lots of horizontal disks centered on the vertical axis. You know the horizontal radius of each disk, and the vertical height of each disk is dz tall. You integrate up the volumes of all of the very thin (dz vertical thickness) disks using an integral.

What would that integral be ?

 

[mathwonk]

the derivative of the volume function is the area of the top of the portion of volume. hence the volume function is the antiderivative of that area function. you only need pythagoras to compute that area.

 

[mesa]

the derivative of the volume function is the area of the top of the portion of volume. hence the volume function is the antiderivative of that area function. you only need pythagoras to compute that area.

Can you show me how that works?

 

[mesa]

Need to figure out the Area taken up by a partially filled sphere

Okay, so now I need an equation that can figure out the surface area of a sphere that is taken up by whatever volume is filled. I believe I have the correct formula now but need a check on it. I got an area of 18.3 m^2 for a sphere with a diameter of 20m filled to 5m.

So how far off am I lol? :)

 

[Vorde]

Well I got something completely different when I did it, but I can't help without knowing your process.

Would you mind walking us through your solution?

 

[HallsofIvy]

Okay, so now I need an equation that can figure out the surface area of a sphere that is taken up by whatever volume is filled. I believe I have the correct formula now but need a check on it. I got an area of 18.3 m^2 for a sphere with a diameter of 20m filled to 5m.

So how far off am I lol? :)

What do you mean by this? Are you talking about the surface area of the part of the sphere from the bottom up to 5 m above the bottom? I get far more than that. The entire surface area of a sphere of radius 20 m is 4\pi (400)= 5026. Your answer would be only 18.3/5026= 0.3% of the whole sphere. That's much too small.

 

[coolul007]

According to CRC, V=(1/3)(Pi)(h^2)(3R - h) = (1/6)(Pi)h(3a^2 + h^2) where R = Radius of sphere, a = radius of circle of partial fill, h = (depth of substance in sphere)perpendicular distance from center of circle to sphere surface.

S = 2(Pi)Rh = (Pi)Dh = (Pi)p^2, where D = diameter of sphere, and p = square root of(a^2 + h^2)

 

[mesa]

What do you mean by this? Are you talking about the surface area of the part of the sphere from the bottom up to 5 m above the bottom? I get far more than that. The entire surface area of a sphere of radius 20 m is 4\pi (400)= 5026. Your answer would be only 18.3/5026= 0.3% of the whole sphere. That's much too small.

Hah hah, yup I goofed this one pretty good, the formula should at least kind of work I thought, let me recrunch and see what the heck I did wrong :)

 

[mesa]

Well I got something completely different when I did it, but I can't help without knowing your process.

Would you mind walking us through your solution?

I basically took an equation I had for figuring the volume of a partially fill sphere at r=11 and then subtracted from the same formula r=10 leaving a shell of a partial sphere volume 1 m thick so divided by 1m and got something probably wrong but hopefully ball park.

Your answer would be only 18.3/5026= 0.3% of the whole sphere. That's much too small.

Okay I see what I did wrong, I forgot to add the third dimension to the equation, whoops! With that the answer from this formula spits out 386.5m^2 which is about 7.5% so it still seems low but a good deal better than .3% :)

According to CRC, V=(1/3)(Pi)(h^2)(3R - h) = (1/6)(Pi)h(3a^2 + h^2) where R = Radius of sphere, a = radius of circle of partial fill, h = (depth of substance in sphere)perpendicular distance from center of circle to sphere surface.

S = 2(Pi)Rh = (Pi)Dh = (Pi)p^2, where D = diameter of sphere, and p = square root of(a^2 + h^2)

Those CRC books have everything lol. Okay this should work as a check, thanks for the info!

 

[Vorde]

Yeah your original attempt doesn't work (although I can see why it made sense at the time), using the other method will probably solve your question. I did it with calculus, but you shouldn't need it.

 

[mesa]

Yeah your original attempt doesn't work (although I can see why it made sense at the time), using the other method will probably solve your question. I did it with calculus, but you shouldn't need it.

Dang it, I keep getting ≈ 385 m^2, kind of close but not really. I think I'm going to scrap what I've been trying and go with a different method.

What are your thoughts on this?

 

[Vorde]

Dang it, I keep getting ≈ 385 m^2, kind of close but not really. I think I'm going to scrap what I've been trying and go with a different method.

What are your thoughts on this?

Are you comfortable with integral calculus? If not someone else will have to help you, I'm sure there is a general formula for a partially-filled sphere but I don't know it.

 

[mesa]

Are you comfortable with integral calculus? If not someone else will have to help you, I'm sure there is a general formula for a partially-filled sphere but I don't know it.

I figured out the 'volume' formula for a partially filled sphere, but am now trying to get the 'area' taken up by that volume.

The 'area' thread was mistakenly combined with this 'volume' thread because one of the admins thought I was posting the same thread twice (the titles are worded nearly identically so I can see where the confusion came from).

I wrote to the PF admins to get it fixed and re-separate the two threads since they are completely different questions and an area question does not belong in a volume thread.

 

[mesa]

Are you comfortable with integral calculus?

I'm actually headed out to go sign up for my first calculus class for the summer session, very excited about it!

 

[Vorde]

It's very fun, and I recommend it highly. As for this problem, my solution still works perfectly (with a small modification), but seeing as you don't know calculus I would wait for someone else to give you a general solution.

 

[mesa]

It's very fun, and I recommend it highly. As for this problem, my solution still works perfectly (with a small modification), but seeing as you don't know calculus I would wait for someone else to give you a general solution.

Yeah, it should be great even though it is a summer class it is the only one I am taking so I can really dig into it. This geometry stuff is fun but can be difficult to do with just algebra and trig lol.

So what was your final answer?