Integration operations

If you are computing a change of variables for a new coordinate system (i.e., in a triple integral), Mathematica has some built in functions that are useful for that as well:

JacobianDeterminant[Spherical[ρ, ϕ, θ]]

ρ^2 Sin[ϕ]

JacobianDeterminant[Cylindrical]

Rr

As usual, be sure you know what the default variables of your coordinate system are before using this:

JacobianDeterminant[Cylindrical[r, θ, z]]

r

SetCoordinates[Cartesian[x, y, z]]

Cartesian[x, y, z]

Also, if you are doing a path integral, you will need to be able to find the "arc length factor", which is the factor you have to introduce to convert an integral ds into an integral dt (or whatever your independent variable is).  Of course, since we know that ds=|Overscript[c, ]'(t)|dt, this "arc length factor" is just |Overscript[c, ]'(t)|.  Now, you could certainly compute that yourself, but Mathematica has a built-in function for this:

ArcLengthFactor[{2Sin[t], 2Cos[t], t}, t]//FullSimplify

5^(1/2)

Now, even though we haven't really discussed this possibility in class, there isn't any reason you can't have curves defined in other coordinate systems.  So, for example, you could define a parametric curve in cylindrical coordinates as:  r=1, θ=t, z=t or, in vector form:

curveCylindrical = {2, t, t}

{2, t, t}

Now, if you just apply the ArcLengthFactor directly to this curve, you get:

ArcLengthFactor[{2, t, t}, t]

2^(1/2)

Notice, that this is actually the same curve I defined above in Cartesian coordinates, a helix with a radius of 2 about the z axis (moving at the same "speed").  Therefore, they ought to have the same factors.  The way to fix this, of course, is to tell Mathematica that your curve is actually specified in cylindrical coordinates (the above command assumed Cartesian):

ArcLengthFactor[{2, t, t}, t, Cylindrical]

5^(1/2)

If you are curious about the actual formula:

ArcLengthFactor[{x[t], y[t], z[t]}, t, Cartesian]

(x^′[t]^2 + y^′[t]^2 + z^′[t]^2)^(1/2)

ArcLengthFactor[{r[t], θ[t], z[t]}, t, Cylindrical]

(r^′[t]^2 + z^′[t]^2 + r[t]^2 θ^′[t]^2)^(1/2)


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