Threaded Mode | Linear Mode

Slushba132 · 11/10/2011 07:01 PM

the volume of the sphere is calculated as
4/3*π*R^2

the volume of the bar is calculated as
π*(D/2)^2*L

how to find the volume of this cross section?

also I will probably need to figure out how to find the surface area of the spherical arc/curve/region that overlaps

cR@Zy!NgLi$h · 11/10/2011 08:34 PM

You will find the answer.  Then you will pass the test.  Then you will never, ever, EVER, need the answer EVER again.

ZiNgA BuRgA · 11/10/2011 09:28 PM

Volume of a generic sphere is:
V = ⁴/₃ π r³ (not squared)

Otherwise, wee can see that the intersection is a cut off portion of the sphere, the diameter of which is equal to a side depth of the bar.

Google shows me this: http://mathforum.org/dr.math/faq/formula...phere.html
Use the volume of the spherical cap.

I'm not quite sure how to isolate the height components in the formula though.

Slushba132 · 11/10/2011 10:33 PM

yeah that would be the problem... I assume there is probably a trig ratio for this somewhere

Slushba132 · 11/10/2011 10:41 PM

I've got it

[Image: sDJ8k.png]

you can use trig on the triangle
Pythagorean at the least

now to tackle the surface area issue

ZiNgA BuRgA · 11/10/2011 10:48 PM

Nah, you'd just have to grab the height component and substitute it into the volume formula I'd imagine.

Actually, I think my algebra's comming back to me:
(where r₁ is the radius of the base)
r = (h² + r₁²)/(2h)
2hr = h² + r₁²
h² - 2hr + r₁² = 0

h = (2r ± √4r² - 4r₁²) / 2
= r ± √r² - r₁²

In this case, r > h, so the ± is probably a -

Plug h into volume formula:
V = (Pi/6)(3r₁² + h²)h

EDIT: I'm not sure your thing exactly works cause you're introducing more variables?

Slushba132 · 11/10/2011 11:23 PM

Think I got it

Threaded Mode \| Linear Mode Computing Mathematical Cross Section From two Known Areas
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