Spherical coordinates in integrals
Date: Around 1995
I was interested in how the psin(phi) comes about in the conversion from 3
space to spherical coordinates with integrals.
v=IntIntIntpsin(phi)dv . . . just curious, could someone help?
In rectangular coordinates the product (dx)*(dy)*(dz) represents dV, the
infinitesimal volume element, since you are integrating over a volume. When
you convert to spherical coordinates Rho, Theta, Phi (with Phi the angle
between the radius vector and the z-axis), the volume element dV is not
(dRho) * (dTheta) * (dPhi), it is
(Rho^2) * (sin(Phi)) * (dRho) * (dTheta) * (dPhi).
In both cases,the infinitesimal volume dV is generated by choosing an arbitrary point in
3-space and increasing each of the coordinates by an infinitesimal amount.
For rectangular coordinates, it is easy -- you form a box with sides dx, dy,
and dz. It is not so clear with spherical coordinates. Try to draw
yourself a picture of this dV, or look for one in a calculus text (there
should be such a picture in virtually every calc text). The solid thus
generated has some curved sides, but in the infinitesimal limit this is of
no consequence. Increasing Rho by dRho gives straight "sides" of length
dRho; increasing Phi by dPhi gives "sides" of length approximately
Rho * dPhi and increasing Theta by dTheta gives "sides" of length approxi-
mately Rho * sin(Phi) * dTheta. Here is where the sin(Phi) comes in: Theta
is measured in the xy-plane, so in computing the length of the "side"
generated by increasing Theta, we need to use not the radius vector, but
rather its projection in the xy-plane (again, a picture greatly helps to see
this). The length of this projection is Rho * sin(Phi). Hence, in spheri-
dV = (dRho) * (Rho * dPhi) * (Rho * sin(Phi) * dTheta).
Click here to return to the Mathematics Archives
Update: June 2012