- #1

bpuk

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i am really stuck on this someone please.

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- Thread starter bpuk
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- #1

bpuk

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i am really stuck on this someone please.

- #2

bpuk

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- #3

tiny-tim

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Hi bpuk! Welcome to PF!

Divide the volume into horizontal slices each of thickness dz, find the volume of each slice (as a function of z), and then integrate.

- #4

bpuk

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Hi bpuk! Welcome to PF!

Divide the volume into horizontal slices each of thickness dz, find the volume of each slice (as a function of z), and then integrate.

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

bpuk

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Hi bpuk! Welcome to PF!

Divide the volume into horizontal slices each of thickness dz, find the volume of each slice (as a function of z), and then integrate.

or is it the int[sqrt(radius^2 - z^2)] between Radius and radius - height??

- #6

HallsofIvy

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[itex]\pi r^2dz[/itex], you want [itex]\pi\int (radius^2- z^2)dz[/itex].

- #7

bpuk

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[itex]\pi r^2dz[/itex], you want [itex]\pi\int (radius^2- z^2)dz[/itex].

hmm,

but yeah that's 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

bpuk

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\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

bpuk

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[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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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?

Just integrate π(r^2 - z^2).

- #11

bpuk

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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?

Just integrate π(r^2 - z^2).

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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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.

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