# Volume of rotation

a) the 3D object obtained by revolving the shaded area about y axis is a cone, having as basins a circle od radius r and with height h

b) the general equation of the line is

$y = m*x+n$

$m = (y-y0)/(x-x0) = (h-0)/(0- r) = – h/r$ is the slope

$n = h$ is the free term

$y = -h/r*x +h$ is the eq of the line

the infinitesimal area under the line is

$dA = y*dx = [(-h/r)*x +h]*dx$ (see the attached figure)

By revolving this infinitesimal area aroud y one gets the infinitesimal volume

$dV = (2*pi*x)*dA =2*pi*x*[(-h/r)*x+h]*dx$

The volume of the generated cone is simply

$V = int_0^r {2pi x[(-h/r)*x +h]*dx} = [-2pi (h/r)(x^3)/3 +2pi h(x^2)/2]_0^r =$

$=-2*pi h(r^3)/3 +pi h(r^2) = pi h(r^3)/3$

The volume of a cone having the basis radius r and height h is indeed

$V(cone) =(1/3)*pi h(r^2)$