Triple integers (cylindrical method)

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Ibraheem
Ibraheem 2022-11-6
编辑： Torsten 2022-11-6
I am trying to write a code for a triple integral with function of r (radius) t (theta) z
xy/2 <= z <= (x^2+2y^2) 1/2
x = rcos theta
y = r sin theta
r^2sin(theta)*cos(theta) <= z <= r^2 cos^2 theta + 2*r^2 sin^2 theta
0<= r <= 1
0 <= theta <= pi/2
I tried this to get my answer
%%% computing the integral to find the actual volume of the region
% we know our theta (t) is bounded by 0 and pi/2
% we know the radius is bounded by 0 and 1
syms r t z
theta = t;
x = @(r,t) r*cos(t);
y = @(r,t) r*sin(t);
zmin = @(x,y) x*y/2;
zmax = @(x,y) sqrt (x.^2+2*y.^2);
rmin = 0;
rmax = x.^2 + y.^2 ==1;
% rmax is 1
% since we are finding the volume with triple integral the function would
% be 1
fun = @(z,r,t) r
volumee = integral3(fun,zmin,zmax,0,1,0,pi/2)
I am trying to get this

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采纳的回答

Torsten
Torsten 2022-11-6
编辑：Torsten 2022-11-6
Ran in:
lower_z = @(r,theta) 0.5*r.^2.*sin(theta).*cos(theta);
upper_z = @(r,theta) r.*sqrt(cos(theta).^2+2*sin(theta).^2);
fun = @(r,theta,z) r;
value = integral3(fun,0,1,0,pi/2,lower_z,upper_z)
value = 0.5742

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