Area Differentials Via Geometry   

The Jacobian determinant for polar coordinates can be obtained both computationally and geometrically. Computationally, we notice that x = rcos( q) and y = rsin( q) implies that

¶( x,y) ¶( r,q) =  ¶x ¶r  ¶y ¶q -  ¶x ¶q  ¶y ¶r = cos( q) rcos( q) -rsin(q) sin( q) = rcos2( q) +rsin2( q) = r

Thus, the area differential is dA =  r drdq.

However, we can also obtain the result geometrically. In particular, suppose that at a point P, the polar distance changes from r to r+dr for some small dr > 0 and suppose that the polar angle changes from q to q+dq for some small angle dq. Then the region covered is practically the same as a small rectangle with height dr and width ds, which is the distance from P to Q due to a change from q to q+dq along a circle of radius r. If an arc subtends an angle q of a circle of radius r, then the length of the arc is s = rq. Thus, small changes ds and dq with r constant satisfy ds = rdq, and 

dA = drds = rdrdq

Indeed, if a coordinate transformation is sufficiently smooth, then images of straight lines are ''practically straight'' over short distances. Thus, if du and dv are small, then the rectangle with width p = á du,0 ñ and height q = á0,dv ñ in the uv-plane is practically the same as the parallelogram implied by the vectors w = J( u,v) p and z = J( u,v) q in the xy-plane. The area differential dA is the area of the parallelogram implied by w = J( u,v) p and z = J( u,v) q .

Let's look at an example. The area differential of the transformation T( u,v) = á u²-v²,2uv ñ is

dA = ê ê  ¶( x,y) ¶( u,v) ê ê dudv = ( 4u2+4v2) dudv

as was shown in example 4. Let's use the ideas discussed above to btain this same result geometrically.

To begin with, the Jacobian of T( u,v) = áu²-v²,2uv ñ is

J( u,v) = é ê ë xu xv yu yv ù ú û = é ê ë 2u -2v 2v 2u ù ú û

Since p = á du,0 ñ and height q = á 0,dv ñ , the vectors w = J(u,v) p and z = J( u,v) q are

w = é ê ë 2u -2v 2v 2u ù ú û é ê ë du 0 ù ú û = é ê ë 2udu 2vdu ù ú û         and        z = é ê ë 2u -2v 2v 2u ù ú û é ê ë 0 dv ù ú û = é ê ë -2vdv 2udv ù ú û

That is, w = á 2udu,2vdu ñ and z = á -2vdu,2udu ñ .

Moreover, w·z = 0, so that w and z are orthogonal. That is, a small rectangle in the uv-plane is mapped to a region that is practically a rectangle with sides w and z in the xy-plane. The lengths of w = á 2udu,2vdu ñ and z = á -2vdu,2udu ñ are

|| w|| = 4u2du2+4v2du2        and        || z|| = 4v2dv2+4u2dv2

The area differential is the area of the rectangle with sides w and z in the xy-plane, which means that

dA = || w||   || z|| = ( 4u2+4v2) du2      (4v2+4u2) dv2 = ( 4u2+4v2)2 du2dv2 = ( 4u2+4v2) dudv

This matches the computational result in example 4.