Spherical coordinates in integrals ```Name: yendor Status: N/A Age: N/A Location: N/A Country: N/A Date: Around 1995 ``` Question: 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? Replies: 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- cal coordinates dV = (dRho) * (Rho * dPhi) * (Rho * sin(Phi) * dTheta). rcwinther Click here to return to the Mathematics Archives

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