Use integrals to prove that the volume of a sphere of radius
R is equal to (4/3)(pi)R^3
well the really easy way is to assume that you know the area of a sphere is 4 pi r^2
integral dr (4 pi r^2)
= 4 pi r^3 / 3 = (4/3) pi r^3
Otherwise radius of plane parallel to x,y axis as a function of height above the center plane is
r = R cos theta
area of circular plane at height z= R sin theta = pi r^2 = pi R^2 cos^2 theta
so hemisphere above axis
integral dz pi R^2 cos^2 theta
but dz = d theta R cos theta
so
integral d theta R cos theta pi R^2 cos^2 theta
from theta = 0 to theta = pi/2
pi R^3 d theta cos^3 theta
integral of cos^3 = sin - (1/3) sin^3
so
pi R^3 [sin theta - (1/3 sin^3 theta]
at pi/2 - at 0
pi R^3 [ 2/3]
that is for the upper hemisphere, multiply by 2 to include the bottom half
pi R^3 [4/3]
To prove that the volume of a sphere of radius R is equal to (4/3)(π)R^3 using integrals, we can utilize the technique of triple integration.
To begin, let's consider a differential volume element, dV, within the sphere. This differential volume element can be expressed in spherical coordinates as dV = r^2sin(θ)drdθdϕ, where r is the radial distance, θ is the polar angle, and ϕ is the azimuthal angle.
Next, we need to determine the limits of integration for each variable. Since we are working with a sphere of radius R, we have the following limits: r ranges from 0 to R, θ ranges from 0 to π, and ϕ ranges from 0 to 2π.
Now, we can set up the triple integral to calculate the volume V of the sphere:
V = ∭dV
= ∭r^2sin(θ)drdθdϕ
= ∫[0 to 2π] ∫[0 to π] ∫[0 to R] r^2sin(θ)drdθdϕ
We will evaluate this integral in three steps, starting from the innermost integral:
∫[0 to R] r^2sin(θ)dr = (1/3)r^3sin(θ) evaluated from 0 to R
= (1/3)R^3sin(θ)
Next, we integrate with respect to θ:
∫[0 to π] (1/3)R^3sin(θ)dθ = -(1/3)R^3cos(θ) evaluated from 0 to π
= -(1/3)R^3(cos(π) - cos(0))
= -(1/3)R^3(-1 - 1)
= (2/3)R^3
Finally, we integrate with respect to ϕ:
∫[0 to 2π] (2/3)R^3dϕ = (2/3)R^3ϕ evaluated from 0 to 2π
= (2/3)R^3(2π - 0)
= (4/3)πR^3
Hence, we have shown that the volume V of a sphere of radius R is equal to (4/3)πR^3 using integrals.