![]() |
|||||
|
|||||
![]() |
|||||
![]() |
![]() |
An integral over an area A in the (x, y) plane is a special and easy to visualize
case of a surface integral. We can, in the results of the previous section, add
the condition that z(s, t) = c throughout our surface and we have an area integral.
In this case the direction normal to each element of area is always the k
direction. Thus we are interested only in the z or k component of w,
and that being the case can consider it to be a scalar field, that is we can set
wz = f(s, t). The integrand multiplied by the area element in this case becomes f(s, t)dA,
and if we have s = x and t = y, the area element dA is given by dxdy and this
becomes f(x, y)dxdy. The "cofactor" of wz in the determinant here is the two
by two determinant which is the z component of the two by two determinant of partial
derivatives of x and y with respect to s and t. You can then deduce that the expression f(x, y)dxdy can be written as f(x(s, t),y(s, t))dsdt J, where J is called the Jacobian of the transformation from variables x, y to s, t and J is given by the absolute value of the determinant of partial derivatives of x and y with respect to s and t. Of course we did not need to introduce the concept of a surface integral to
deduce this result. When infinitesimal changes ds and dt in s and t are made,
the resulting area in the xy plane is that of the parallelogram whose sides are
Exercises:24.2 What is the relationship between the Jacobian going from dxdy to dsdt, and that going the opposite way? 24.3 Explain this by examining the matrix product between the matrix above and the transpose (interchange rows and columns) of the one for going the other way. 24.4 Suppose x = uv and |
![]() |