 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/1/a/solution

h_{r} h_{α} h_{β} = (sin α cos β, sin α sin β, cos α), = (r cos α cos β, r cos α sin β, - r sin α), = (- (1 + r sin α) sin β, (1 + r sin α) cos β, 0) .

1, r, 1 + r sin α

. Since

R < 1

, the last is positive throughout the body. Thus these circular torus coordinates are orthogonal wherever $r > 0$ ; at

r = 0

the cross-section angle degenerates, as in ordinary polar coordinates.

J = det \frac{\partial ( x , y , z )}{\partial ( r , α , β )} = h_{r} \cdot (h_{α} \times h_{β}) = r (1 + r sin α) .

Its sign is positive for

r > 0

, so the volume element is

d V = r (1 + r sin α) d r d α d β

. The coordinate degeneracy on the central circle has zero volume and does not obstruct volume integration.

 New to topics? Read the docs here!