How to find volume under fitted data?

Steven (view profile)

on 28 Sep 2017
Latest activity Commented on by Steven

Steven (view profile)

on 29 Sep 2017
Accepted Answer by Teja Muppirala

Teja Muppirala (view profile)

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!

KSSV

KSSV (view profile)

on 28 Sep 2017
YOu want a area or volume? I think it is area.....
Steven

Steven (view profile)

on 28 Sep 2017
No. I want the volume resulted from the revolution of this plot (say, half of it) around y axis which gives a spherical shape. Thank you
KSSV

KSSV (view profile)

on 28 Sep 2017
Go ahead with trapz.

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Answer by Teja Muppirala

Teja Muppirala (view profile)

on 28 Sep 2017
Edited by Teja Muppirala

Teja Muppirala (view profile)

on 28 Sep 2017

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

Steven (view profile)

on 29 Sep 2017
Thank you guys. Since I want the rotation of the right-half of the plot around y axis, I would use Teja's first formula. Although I don't understand that .*r multiplication in the formula for polar coordinates. So in this case, we are integrating the area of a cylinder over height? Thanks
Torsten

Torsten (view profile)

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