If you have a function y = f(r), which it sounds like you do since you have a polynomial, then you can use integral.
y = @(r) 1-r;
a = 0;
b = 1;
vol = 2*pi*integral(@(r) y(r) .* r, a, b)
Here I used y = 1-r, so it is a cone, and I can verify that the answer I got (vol=1.047) was indeed equal to the analytical answer for the volume of a cone with unit height and radius (pi/3).
If you don't have a function, but just some discrete set of values r and y, then you can use INTERP1 along with INTEGRAL.
r = [ 0 0.2 0.4 0.6 0.8 1.0 ];
y = [ 1 0.8 0.6 0.4 0.2 0 ];
Yfun = @(p) interp1(r, y, p);
vol = 2*pi*integral(@(r) Yfun(r).*r, 0, 1)