Warning, the protected names norm and trace have been redefined and unprotected

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1 22M:28 Spring 5 J. Simon An example of computing a triple integral with several coordinate systems > > with(plots):with(linalg): Warning, the name changecoords has been redefined Warning, the protected names norm and trace have been redefined and unprotected Consider the solid cone W with radius R and height H. Think of the cone as sitting on its vertex, with the vertex at the origin and the circular base paralell to the xy-plane at height H. The simplest equation for a circular cone is z=sqrt(x^2y^2) (note that intersects the xzplane in the graph z = x, which is what we want). But to have the cone attain radius = R at height z=h, we need the slope to be H/R, not 1. So we modify the equation to z = (H/R)*sqrt(x^2y^2). We can parametrize W by saying x = -R.. R y =-sqrt(r^2-x^2).. sqrt(r^2-x^2) z = (H/R)*sqrt(x^2y^2).. H These would be the limits of integration (in xyz-coordinates) for the integral of some function f(x,y,z) on W. Let's use R=1, H= > R:=1;H:=; R := 1 H := > DrawConeSurface:=plotd([x,y,(H/R)*sqrt(x^2y^2)], x=- R..R, y=-sqrt(r^2-x^2)..sqrt(r^2-x^2), color=red, style=patchnogrid): > DrawTopDisk:=plotd([x,y,H], x=-r..r, y=-sqrt(r^2- x^2)..sqrt(r^2-x^2), color=blue, style=wireframe, thickness=2, grid=[1,1]): > display({drawconesurface, DrawTopDisk}, scaling=constrained, tickmarks=[1,1,], axes=normal, lightmodel=light2, orientation=[-44,6]);

2 Example: f(x,y,z) = xyz > f:=xyz; f := x y z > TheIntegral:=Int(Int(Int(f, z=(h/r)*sqrt(x^2y^2).. H),y =-sqrt(r^2-x^2).. sqrt(r^2-x^2)), x = -R.. R); TheIntegral := x 2 d 1 x 2 1 x 2 y 2 x y z z dy dx > Step1:=Int(Int(int(f, z=(h/r)*sqrt(x^2y^2).. H),y =- sqrt(r^2-x^2).. sqrt(r^2-x^2)), x = -R.. R); Step1 := x 2 x d 1 x 2 y 2 y x 2 y x2 2 y2 y dx 1 x 2 > Step2:=Int(int(int(f, z=(h/r)*sqrt(x^2y^2).. H),y =- sqrt(r^2-x^2).. sqrt(r^2-x^2)), x = -R.. R);

3 Step2 := 1-1 x 1 x x x2 1 x 2 2 x ln ( ( x 1 ) ( x 1) 1) 2 x ln( ( x 1 ) ( x 1) 1) dx > Step:=int(int(int(f, z=(h/r)*sqrt(x^2y^2).. H),y =- sqrt(r^2-x^2).. sqrt(r^2-x^2)), x = -R.. R); Step := 1-1 x 1 x x x2 1 x 2 2 x ln ( ( x 1 ) ( x 1) 1) 2 x ln( ( x 1 ) ( x 1) 1) dx Notice Maple couldn't even calculate the antiderivative for Step. > evalf(step); ###################################################################### Let's see if the problem becomes easier if we use cylindrical coordinates. Now x = -R.. R y =-sqrt(r^2-x^2).. sqrt(r^2-x^2) z = (H/R)*sqrt(x^2y^2).. H becomes theta =..2*Pi r =..R z = (H/R)*sqrt(r^2).. H. and f(x,y,z) = xyz = r*cos(theta)r*sin(theta)z and we remember to multiply by the Jacobian determinant = r. > TheCylindricalIntegral:= Int(Int(Int((r*cos(theta)r*sin(theta)z)*r, z = (H/R)*sqrt(r^2)..H), r=..r), theta=..2*pi); TheCylindricalIntegral := 1 1 r 2 ( r cos( θ ) r sin( θ ) z ) r dz dr dθ > Step1Cyl:=Int(Int(int((r*cos(theta)r*sin(theta)z)*r, z = (H/R)*sqrt(r^2)..H), r=..r), theta=..2*pi);

4 Step1Cyl := 1 1 d d 2 r r2 ( r cos( θ ) r sin( θ )) r 1 r2 r θ > Step2Cyl:=Int(int(int((r*cos(theta)r*sin(theta)z)*r, z = (H/R)*sqrt(r^2)..H), r=..r), theta=..2*pi); Step2Cyl := sin( θ ) 425 cos( θ ) > StepCyl:=int(int(int((r*cos(theta)r*sin(theta)z)*r, z = (H/R)*sqrt(r^2)..H), r=..r), theta=..2*pi); StepCyl := 7225 π dθ > evalf(stepcyl); NOTICE: (except for a little epsilon error in the numerical integral, we DO get the same value as before. But this time the integral 7225*pi is something we could have calculated by hand. ######################################################################## ################## Now let's see what the integral looks like in spherical coordinates. ρ rho = distance from the origin (so rho >/= ) (using notation >= for "greater than or equal to") ϕ phi = angle of declination from the vertical ( </= phi </= Pi) θ theta = usual polar angle as in cylindrical coordinates (theta any values, but usually <= theta <= 2Pi) The transformation to write x,y,z in terms of rho, phi, theta is a little complicated. But here is how you can remember it. First express z and polar r in terms of rho and phi: z = rho*cos(phi) r = rho*sin(phi) z = ρ cos ϕ ϕ length ρ r = ρ cos (complement of ϕ) = ρ sin (ϕ) Then write x = r cos(theta), y = r sin(theta), z=z. This gives x = rho sin(phi) cos(theta), y = rho sin(phi) sin(theta), z = rho cos(phi).

5 So the transformation for spherical coordinates is x = r cos(theta) = rho sin(phi) cos(theta) y = r sin(theta) = rho sin(phi) sin(theta) z = rho cos(phi). In addition to using the above transformation (x = expression in rho, phi, theta; etc.) to rewrite the integrand f(x,y,z), we also need to compute the Jacobian determinant for the transformation. > jacobian([rho*sin(phi)*cos(theta), rho*sin(phi)*sin(theta), rho*cos(phi)], [rho, phi, theta]); sin( φ ) cos( θ ) ρ cos( φ ) cos( θ ) ρ sin( φ ) sin( θ ) sin( φ ) sin( θ ) ρ cos( φ ) sin( θ ) ρ sin( φ ) cos( θ ) cos( φ) ρ sin( φ) > det(%); sin( φ) cos( θ ) 2 ρ 2 sin( φ) sin( θ ) 2 ρ 2 ρ 2 cos( φ) 2 cos( θ ) 2 sin( φ) ρ 2 sin( φ ) sin( θ ) 2 cos( φ) 2 > simplify(%); sin( φ) ρ 2 So we get rho^2*sin(phi) as the stretching factor that we need to use to change integrals written in terms of x,y,z into integrals written in terms of spherical coordinates. To write our integral in spherical coordinates, we need to describe the domain, that is the solid cone. The boundary cone surface is just phi = constant angle. What angle? From the little trig picture above, we see that the angle phi that describes this cone has tan(phi) = R/H. So the cone angle (from the vertical to the cone) is arctan(r/h). Don't be frightened by the arctangent. Remember (R/H) is just a constant, so is arctan(r/h). What is the equation of the plane z=h in spherical coordinates? z = H is the plane. In spherical coordinates, z = rho*cos(phi). So the plane is rho*cos(phi) = H, i.e. rho = H/cos(phi). We are now ready to rewrite the integral. The integrand = xyz = rho*sin(phi)*cos(theta) rho*sin(phi)*sin(theta) rho*cos(phi). To keep the limits of integration clear in what we are about to write, let's introduce "g" to denote the integrand expressed in terms of rho, phi, theta. > g:=rho*sin(phi)*cos(theta) rho*sin(phi)*sin(theta) rho*cos(phi);

6 g := ρ sin( φ ) cos( θ ) ρ sin( φ ) sin( θ ) ρ cos( φ) We also are going to have to multiply g by the stretching factor rho^2*sin(phi). Let's call that whole new integrand G. > G:=rho^2*sin(phi)*g; G := ρ 2 sin( φ ) ( ρ sin( φ ) cos( θ ) ρ sin( φ ) sin( θ ) ρ cos( φ) ) > TheSphericalIntegral:=Int(Int(Int(G, rho=..h/cos(phi)), phi=..arctan(r/h)), theta=..2*pi); TheSphericalIntegral := arctan 1 1 cos( φ ) ρ 2 sin( φ ) ( ρ sin( φ ) cos( θ ) ρ sin( φ ) sin( θ ) ρ cos( φ) ) dρ dφ dθ > > Step1Spher:=Int(Int(int(G, rho=..h/cos(phi)), phi=..arctan(r/h)), theta=..2*pi); Step1Spher := arctan sin( φ ) ( sin( φ ) cos( θ ) sin( φ ) sin( θ ) cos( φ) ) dφ dθ cos( φ) 4 > Step2Spher:=Int(int(int(G, rho=..h/cos(phi)), phi=..arctan(r/h)), theta=..2*pi); Step2Spher := sin( θ ) 425 cos( θ ) > StepSpher:=int(int(int(G, rho=..h/cos(phi)), phi=..arctan(r/h)), theta=..2*pi); StepSpher := 7225 π THIS IS IDENTICAL TO WHAT WE GOT BEFORE USING CYLINDRICAL COORDINATES. dθ