i know this is something easy but i cant figure it out right now. and by the way on what kinda solids does this apply?
2006-10-20 07:21:55 · 7 answers · asked by outofthisworld 2 in Science & Mathematics ➔ Mathematics
why does this hold only for some solids and not all?
2006-10-20 07:45:14 · update #1
This, in general, is not true.
For a sphere, this is true. Let:
V: Volume
A: Surface area
Then, for a sphere with radius r,
A = 4*pi*r^2
V = (4/3)*pi*r^3 = (r/3)*A
dV/dr = 3*(4/3)*pi*r^2 = 4*pi*r^2 = A
However, this is not true for an x-by-x cube, in which:
A = 6*x^2
V = x^3 = 2*x*A
dV/dr = 3*x^2 = A/2
Perhaps the relationship between V and A is more interesting than the relationship between dV/dr and A.
Now, if you are simply interested in the degree of r (x) in the sphere equations (cube equations), keep in mind that volume is a three dimensional measure so it will always have cubic units. Surface area is a two dimensional measure so it will always have square units. Since lengths and their units are buried in the parameter r (x), then the degree of r (x) will always be the same as the degree of the unit. Additionally, because a derivative measures the rate of change along a certain direction, it is necessary to divide the units by the units of that direction. When you are taking a derivative with respect to r, you are going to divide by one length unit. That's why the derivative of volume has the same degree as the surface area.
2006-10-20 07:46:06 · answer #1 · answered by Ted 4 · 2⤊ 0⤋
It's not the outer surface area. Imagine a slice thru the solid.
The slice is perpendicular to the X-axis.
The area exposed by the slice is the derivative.
If a cube is length A each slice has area A squared, each iteration volume is A squared times dx. Anti-derivative is A squared times x, x from 0 to A is A cubed.
2006-10-20 15:17:49 · answer #2 · answered by fiascogrande 2 · 0⤊ 0⤋
Alright everyone, I am going to settle this once and for all (hopefully). How do we calculate the volume, we take up all the areas (because area is infinitesimaly small volume) and then add them all up (integrate). Since the integral of the area is the volume, by the fundamental theorem of calculus (which Newton spent many years deriving), the derative of the volume must be the area.
2006-10-20 14:37:13 · answer #3 · answered by The Prince 6 · 2⤊ 1⤋
The antiderivative of a function, i.e. the integral, is the area of that function under the curve. The double integral of a surface represents the volume under the surface. So my best guess is that taking the derivative of the volume (the double integral) would result in a single integral...which is the area.
But, I am speculating.
2006-10-20 14:28:27 · answer #4 · answered by csulbalgebra 2 · 1⤊ 2⤋
volume is 3 dimensional-third degree
when you differentiate it you will get 2nd degree which is 2 dimensional
area is 2 dimensional
2006-10-20 14:25:38 · answer #5 · answered by raj 7 · 1⤊ 2⤋
this is not true.
cube V(x) = x^3
dV(x)/dx = 3x^2
the surface is 6x^2
2006-10-20 14:25:49 · answer #6 · answered by gjmb1960 7 · 2⤊ 0⤋
It's so Bush can drink and lead the country. -- Although friends don't let friends drink and lead the country.
2006-10-20 14:23:50 · answer #7 · answered by martino 5 · 0⤊ 4⤋