The Student Room Group
Reply 1
Pie are squared.
It is a slight variation on πr2 \pi r^2 . In the normal spherical polar coordinate system where z=rcosθ z = r \cos \theta i.e theta is the polar angle between the radius vector and the z axis, then the cross sectional area at an angle theta is simple:

A=πz2=πr2cos2θ A = \pi z^2 = \pi r^2 \cos^2 \theta , r is the radius of the sphere.

Quick check: At θ=0,π2,π \theta = 0, \frac{\pi}{2}, \pi we get sensible results (namely πr2&0 \pi r^2\,\, \&\,\, 0 .)
0 div curl F
It is a slight variation on πr2 \pi r^2 . In the normal spherical polar coordinate system where z=rcosθ z = r \cos \theta i.e theta is the polar angle between the radius vector and the z axis, then the cross sectional area at an angle theta is simple:

A=πz2=πr2cos2θ A = \pi z^2 = \pi r^2 \cos^2 \theta , r is the radius of the sphere.

Quick check: At θ=0,π2,π \theta = 0, \frac{\pi}{2}, \pi we get sensible results (namely πr2&0 \pi r^2\,\, \&\,\, 0 .)

Given that the OP is asking for the area of a circle, do you reckon he knows what spherical polars (or even radians) are?

Here's the simplest way of thinking about it. Cut the top off your sphere, and pretend you're looking at it side-on. Then it'll basically look like a circle with a bit of the top replaced by a horizontal line. Draw in the radii of the sphere to the endpoints of this line, and try and work out the length of this line. Now, the length of that line is actually the diameter of the cross-sectional area... can you see why?
generalebriety
Given that the OP is asking for the area of a circle, do you reckon he knows what spherical polars (or even radians) are?

Here's the simplest way of thinking about it. Cut the top off your sphere, and pretend you're looking at it side-on. Then it'll basically look like a circle with a bit of the top replaced by a horizontal line. Draw in the radii of the sphere to the endpoints of this line, and try and work out the length of this line. Now, the length of that line is actually the diameter of the cross-sectional area... can you see why?



Well he just asked what the area was, not to derive it; I included that for free plus its only a simple right angled triangle.
Reply 5
The easiest way is thus:

Get yourself a can of paint, of known volume. Paint the sphere in its entirety, trying to keep an even coat thickness all around. Then measure the remaining volume of paint you have left. There will be losses on the brush, and the sides of the can, so take that into account when stating your accuracy. Remember, it's Science!

Happy Scienceing!
0 div curl F
Well he just asked what the area was, not to derive it; I included that for free plus its only a simple right angled triangle.

For all you know, this might be a GCSE student asking a past paper question, and a more helpful answer should then really have been "there isn't a standard formula, but you can derive one". It might not be a GCSE student, I dunno. The title does say "finding cross-sectional area"...
generalebriety
For all you know, this might be a GCSE student asking a past paper question, and a more helpful answer should then really have been "there isn't a standard formula, but you can derive one". It might not be a GCSE student, I dunno. The title does say "finding cross-sectional area"...



For all you know this might be a 1st year physics student doing a lab experiment and just needing to know the form of the equation. And the first post does say "What is the formula for calculating the cross-sectional area of a sphere?" which I supplied with the addition of a little background as to where I got it from, rather than just plucking it out of the air.

Molehill \to Mountain.
Reply 8
Well, both of you are right in your own way so stop the childish bickering.... Cross-sectional area, as explained in a physics class I once had, is the area of the shadow something would cast if light was shined from above. For a sphere, the shadow would take the form of a circle. Thus the answer for its cross-sectional area would be pi r^2. :biggrin: