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Please help! sphereical cap

  • Thread starter bpuk
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  • #1
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ok so i have a homework question that goes around the lines of a perfect sphere is used as a measureing bowl with radius R is given, this has a hole in the top in which it is filled with water (volume of this discounted as it is relatively small), we are asked to find at what height (relative to the centre) marks should be made to show that there is p (mm) of liquid in the bowl??
i am really stuck on this someone please.
 

Answers and Replies

  • #2
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i dont know how to edit but i know in order to do find the volume i will have to model a cone within th experiment obtian the limits and then derive a function but i am for the life of me unable to comprehend on how to start it
 
  • #3
tiny-tim
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Welcome to PF!

Hi bpuk! Welcome to PF! :smile:

Divide the volume into horizontal slices each of thickness dz, find the volume of each slice (as a function of z), and then integrate. :smile:
 
  • #4
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Hi bpuk! Welcome to PF! :smile:

Divide the volume into horizontal slices each of thickness dz, find the volume of each slice (as a function of z), and then integrate. :smile:
so basically i will be working in terms of finding the integral (from the R to R-h) of the area of a cross section being pi*r^2??
 
  • #5
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Hi bpuk! Welcome to PF! :smile:

Divide the volume into horizontal slices each of thickness dz, find the volume of each slice (as a function of z), and then integrate. :smile:
or is it the int[sqrt(radius^2 - z^2)] between Radius and radius - height??
 
  • #6
HallsofIvy
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No, sqrt(radius^2- z^2) is just "r" itself for each "slice". Since the volume of each slice is
[itex]\pi r^2dz[/itex], you want [itex]\pi\int (radius^2- z^2)dz[/itex].
 
  • #7
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No, sqrt(radius^2- z^2) is just "r" itself for each "slice". Since the volume of each slice is
[itex]\pi r^2dz[/itex], you want [itex]\pi\int (radius^2- z^2)dz[/itex].
hmm,
but yeah thats what i thought but how do we then know the radius of each "slice" if we continued to work through the problem?

and were the Z limits ok?
 
  • #8
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using that [itex]
\pi\int (radius^2- z^2)dz
[/itex]
i found that intergrating it came out to be 1/2 ln (z+sqrt(r^2+x^2)r^2 + 1/2*x*sqrt(r^2+x^2)
 
  • #9
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No, sqrt(radius^2- z^2) is just "r" itself for each "slice". Since the volume of each slice is
[itex]\pi r^2dz[/itex], you want [itex]\pi\int (radius^2- z^2)dz[/itex].
and going on from this i found the integral of this is like [.5*ln(x+sqrt(r^2+x^2)r^2 + .5*x*sqrt(r^2+x^2)]
which seems a bit crazy.
 
  • #10
tiny-tim
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Hi bpuk! :smile:

I don't understand where your sqrt came from … there's no sqrt in the formula.

And is your x the same as your z, or something different? :confused:

Just integrate π(r^2 - z^2). :smile:
 
  • #11
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Hi bpuk! :smile:

I don't understand where your sqrt came from … there's no sqrt in the formula.

And is your x the same as your z, or something different? :confused:

Just integrate π(r^2 - z^2). :smile:
umm the sqrt came from the spherical formula cos you have to sqrt the side you have as in x^2+y^2+z^2=R^2.
and yesh i understand to integrate the eqn but in terms of the next step what is that lil r in reference to is that the main radius i know or the one of the smaller cut.

as in if i know go to integrate it i get pi*r^2*z - (z^3)/3*pi
do i then go my known limits of R and R-H here
and then what happens to the small r??
 
Last edited:
  • #12
tiny-tim
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Homework Helper
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umm the sqrt came from the spherical formula cos you have to sqrt the side you have as in x^2+y^2+z^2=R^2.
and yesh i understand to integrate the eqn but in terms of the next step what is that lil r in reference to is that the main radius i know or the one of the smaller cut.

as in if i know go to integrate it i get pi*r^2*z - (z^3)/3*pi
do i then go my known limits of R and R-H here
and then what happens to the small r??
(My r should have been R.)

Yes … calculate π[R^2*z - (z^3)/3] from -R to H. :smile:
 

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