P
pLuvia
Guest
did any of your teachers ever show you how the Volume Formulae came to what they are?
-
Looking for HSC notes and resources? Check out our Notes & Resources page
P
pLuvia
Guest
my tutor today showed me how to get the volume formulae for cone, cylinder and sphere. I have to try and find out how to get the formula for an area of a sphere 
, it quite interesting how it originated lol
through integration
, it quite interesting how it originated lolthrough integration
Dumsum
has a large Member;
We were shown in 3u... but it's just the 'slicing' method from 4u.
For any function y = f(x), take the area from a to b and rotate around the x-axis. By taking very thin slices perpendicular to the x-axis, you end up with a series of "circles" (which are more like cylinders with next to no height). So the volume of these is pi.r^2.δx (δx is the width of the slice, just 'a very small distance'). Now r happens to be y (the distance from the x-axis to the function...sorry that I don't have a graph) and by adding up all the slices you get the volume of the solid.
δV = pi.y^2.δx
V = Sum{a->b} lim{δx->0} pi.y^2.δx
this limiting sum is defined as an integral (in the same way as the area under a curve is defined)
V = Int{a->b} pi.y^2.dx
Similarly for rotation around y-axis.
Would have made more sense with a graph, sorry.
For any function y = f(x), take the area from a to b and rotate around the x-axis. By taking very thin slices perpendicular to the x-axis, you end up with a series of "circles" (which are more like cylinders with next to no height). So the volume of these is pi.r^2.δx (δx is the width of the slice, just 'a very small distance'). Now r happens to be y (the distance from the x-axis to the function...sorry that I don't have a graph) and by adding up all the slices you get the volume of the solid.
δV = pi.y^2.δx
V = Sum{a->b} lim{δx->0} pi.y^2.δx
this limiting sum is defined as an integral (in the same way as the area under a curve is defined)
V = Int{a->b} pi.y^2.dx
Similarly for rotation around y-axis.
Would have made more sense with a graph, sorry.
gman03
Active Member
- Joined
- Feb 7, 2004
- Messages
- 1,283
- Gender
- Male
- HSC
- 2003
In second year maths language:
in spherical coordinates the points on the sphere (with fixed radius R) can be represented by (R, theta, phi), where theta = [0..pi] the angle from positive z-axis, phi = [0..2pi] the angle around the z-axis from the positive x-axis.
Then area is simply the double integral of unit dA = [R * dtheta] * [R * sin(theta) * dphi]. The reason for all these factors is due to the change of basis from cartesian to spherical and the metric associted with it...
so Area
= integral over phi { integral over theta { dA } }
= integral over phi { integral over theta { R^2 * sin(theta) * dphi * dtheta } }
= R^2 * integral over theta { sin(theta) * dtheta } * integral over phi { dphi }
= R^2 * [ -cos(theta) from 0 to pi] * [ phi from 0 to 2pi ]
= R^2 * [ 1 - -1] * [2pi]
= 4 pi R^2
Similar for the suface area of a cone but uses cylindrical coordinates i think.
For the volume on sphere, we use dV = dr * [r * dtheta] * [r * sin(theta) * dphi] and do V = integral over r = [0..R] { integral over phi { integral over theta { dV } } }
These should confuse some people.
in spherical coordinates the points on the sphere (with fixed radius R) can be represented by (R, theta, phi), where theta = [0..pi] the angle from positive z-axis, phi = [0..2pi] the angle around the z-axis from the positive x-axis.
Then area is simply the double integral of unit dA = [R * dtheta] * [R * sin(theta) * dphi]. The reason for all these factors is due to the change of basis from cartesian to spherical and the metric associted with it...
so Area
= integral over phi { integral over theta { dA } }
= integral over phi { integral over theta { R^2 * sin(theta) * dphi * dtheta } }
= R^2 * integral over theta { sin(theta) * dtheta } * integral over phi { dphi }
= R^2 * [ -cos(theta) from 0 to pi] * [ phi from 0 to 2pi ]
= R^2 * [ 1 - -1] * [2pi]
= 4 pi R^2
Similar for the suface area of a cone but uses cylindrical coordinates i think.
For the volume on sphere, we use dV = dr * [r * dtheta] * [r * sin(theta) * dphi] and do V = integral over r = [0..R] { integral over phi { integral over theta { dV } } }
These should confuse some people.
ahhah, strangely enough i wasn't lost through that. I have some basic concept as to what a double integral is. At the time I don't claim to know how exactly they work, I could compute one, but not sure as to EXACTLY how to set it up. I am sure there are a lot of people doing the HSC who could follow precisely what you had. Nevertheless, that was just mean. But funny.