Asked by Steven
on 28 Sep 2017

I have a set of data, fitted by polynomial of 8th order. I want to find the volume under this surface (due to rotation around y axis for example). I am not sure how to do it. Should I use dblquad or trapz? Since dblquad needs function handle and I do not have the function, just the interpolated data (I also can not use interp2 (as suggested in some answers), because it uses linear interpolation. Any idea how I can find the volume?

Thank you!

Answer by Teja Muppirala
on 28 Sep 2017

Edited by Teja Muppirala
on 28 Sep 2017

Accepted Answer

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; % y can be any arbitrary function of r. Put the function for your curve here (POLYVAL?).

a = 0; % Some limits of integration r = a to b

b = 1;

% Volume under surface of rotation:

vol = 2*pi*integral(@(r) y(r) .* r, a, b) % Put in the extra ".*r" for polar coords.

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 ]; % Again, just use a unit cone for testing purposes

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) % Again, you'll get 1.0472

Steven
on 29 Sep 2017

Torsten
on 29 Sep 2017

Teja is right - sorry for the confusion.

The formula says that the volume of the surface of revolution around the y-axis is obtained by integrating the lateral surface area A(x) of the cylinders which you get when you rotate a vertical line from the x-axis up to your polynomial function around the y-axis ( which is A(x) = 2*pi*x*P(x) ).

Best wishes

Torsten.

Steven
on 29 Sep 2017

Thanks Torsten

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