In the photo? Cut a short section as a cylinder.

The height is △h? ,? What is the radius of the bottom? (R^2-h^2)^0.5

Cylinder? Volume? △V=π*(R^2-h^2)*△h？

Hemisphere volume V=? ∫ π * (r 2-h 2) * DH (h value range? 0-R)

=∫π*R^2*dh-∫π*h^2*dh

=π(r^2*r-r^2*0)- 1/3π(r^3-0^3)？ (Calculation of definite integral)

=πR^3- 1/3*πR^3

=2/3*πR^3

The volume of the ball? V=2/3*πR^3*2？ =4/3*πR^3

High school knowledge answer:

Similar to the above, so use the same picture to explain.

Put the hemisphere from top to bottom? Points? M copies? (m tends to infinity), each copy? They are all small columns. What is the height of this cylinder? R/m

Let's see? N th? Situation,

Or use the ground radius of the n part of the triangle in the figure? =(R^2-(n/m*R)^2)^ 1/2

N th? What is the volume? V(n)=π*(R^2-(n/m*R)^2)*R/m

Simplify it:

V(n)=πR^3(m^2-n^2)/m^3

The volume of the hemisphere? =V( 1)+V(2)+V(3)+......+V(n)+...+V(m)

Namely. ∑ π r 3 (m 2-n 2)/m 3 (n is? 1 m)

∑πR^3(m^2-n^2)/m^3？ =πR^3*n/m-πR^3*/m^3∑n^2

Why don't you go and have a look? ∑n^2？ This total

12+22+32+...+n 2 = n (n+1) (2n+1)/6 (this formula can be derived by cubic difference formula).

After knowing this? Let's look at the above formula again.

So what? ∑πR^3(m^2-n^2)/m^3？ =? πR^3*n/m-πR^3*/m^3？ *n(n+ 1)(2n+ 1)/6？ (n=m)

=πr^3-πr^3*m(m+ 1)(2m+ 1)/6m^3

M tends to infinity

The volume of the hemisphere =? lim[πr^3-πr^3*m(m+ 1)(2m+ 1)/6m^3]？ (m→∞)

=πR^3-πR^3*lim？ [m(m+ 1)(2m+ 1)/6m^3]

(About this limit? lim？ [m(m+ 1)(2m+ 1)/6m^3]？ = 1/3,?

The limit is actually equal to? Coefficient ratio of the highest term)

So what? The volume of the hemisphere = π r 3-π r 3 *? 1/3=2/3πR^3

Sphere volume = 2/3 π r 3? *2? =4/3πR^3