Labware - MA35 Multivariable Calculus - Two Variable Calculus



Change of Variables


Integration in a non-Cartesian coordinate system requires an application of the Change of Variables Theorem.

Consider what would happen for a function of two variables. Up until now, we have integrated over two variables by adding up volumes of rectangular prisms with heights f(x, y) and rectangular bases of area dx * dy. If we decide to express x and y as functions of variables u and v, we need to find some new expression for the volume of the differential.

The height will still be the value of the function, expressed as f(x(u,v), y(u,v)).

The base of the volume will be a parallelogram. The two vectors representing the sides of this parallelogram are based on what happens when u and v change. When u changes by some amount du, x increases by xu*du and y increases by yu*du. The resulting vector is (xu * du, yu * du).

The other vector, which is based on what happens when v increases some amount dv, is (xv * dv, yv * dv).

To find the area of this base, take the cross product of the vectors, which is equal to

(xuyv - xvyu)dudv.

Finally, multiply by hieght to get the Change of Variables Theorem for double integrals:

∫∫Df(x, y)dxdy = ∫∫D*f(x(u, v), y(u, v))(xuyv - xvyu)dudv,

Where D is the domain in Cartesian coordinates and D* is the domain in the new coordinate system.

Note that the expression

(xuyv - xvyu)

is the determinant of the matrix with rows

(xu, xv), (yu, yv).



  • 1. Find the change of variables formula (i.e. form of ∫∫Df(x, y)dxdy = ∫∫D*f(x(u, v), y(u, v))(xuyv - xvyu)dudv) for each of the following coordinate systems:
    • x(u, v) = (u - v)√(2), y(u, v) = (u + v)√(2)
    • x(u, v) = ucos(v), y(u, v) = usin(v) (polar coordinates)
    • x(u, v) = u2 - v2, y(u, v) = 2uv
  • 2. Under what conditions will the cyan parallelogram shown in the "D" window have constant area for all (u, v)?