By integration derived the formula for the volume of a hemisphere of radius r?

Answer 1

One way is to simply write it as the volume between the xy-plane and the surface with equation z=sqrt(r2-x2-y2), where -r <= x,y <= r (specifically, this gives the volume of a hemisphere with radius r centered at the origin and lying above the xy-plane.) We therefore have a double integral. We can take the outer one over y, and the inner one over x, both with limits -r and +r (The order is immaterial.) The integrals involved, however, are somewhat messy.

A much better technique is the disk method. The hemisphere may be regarded as an infinite number of circular disks piled on top of each other. Thus, with z ranging from 0 to r, the infinitesimally thick disk at a particular z-value has a radius of R(z) = sqrt(r2-z2) (which follows from the Pythagorean theorem).

The total volume is then found by integrating πR(z)2 dz from z = 0 to z = r, because this expression is the volume of a single disk (dz being the infinitesimal thickness, and R(z) being the radius.) The integrand simplifies to π(r2-z2); the integral is then π(r2z - z3/3). Evaluated at z = r, this gives π(r3-r3/3) = π(2r3/3), and at z = 0, gives 0. The volume of the hemisphere is therefore 2πr3/3, in agreement with established facts.

This can also be done readily with spherical coordinates. The surface of the hemisphere is given by ρ = r, with θ ranging from 0 to 2π and φ ranging from 0 to π/2. The Jacobian determinant is ρ2 sin φ. Integrating with respect to φ gives -ρ2 cos φ, with -(ρ2 cos π/2) = 0 and -(ρ2 cos 0) = -ρ2, giving ρ2 for the innermost integral. Integrating next with respect to ρ gives ρ3/3, with (r3/3 - 03/3) = r3/3. Finally, integrating with respect to θ gives θr3/3, with (2πr3/3) - (0r3/3) = 2πr3/3.