Change of Variables in Multiple Integrals

When evaluating integrals, making an appropriate substitution can transform a difficult integral into a simpler one. You’ve already seen this technique in single-variable calculus with [latex]u[/latex]-substitution.

Consider the integral [latex]\int_2^3 x(x^2 - 4)^5 dx[/latex]. We can substitute [latex]u = g(x) = x^2 - 4[/latex]. Then [latex]du = 2x , dx[/latex], so [latex]x , dx = \frac{1}{2} du[/latex]. The limits change to [latex]u = g(2) = 2^2 - 4 = 0[/latex] and [latex]u = g(3) = 9 - 4 = 5[/latex]. The integral becomes [latex]\int_0^5 \frac{1}{2} u^5 du[/latex], which is much simpler to evaluate.

You’ve also used this idea when transforming double integrals from rectangular to polar coordinates, or triple integrals from rectangular to cylindrical or spherical coordinates. In each case, the goal is to make computations simpler through an appropriate change of variables.

For double integrals, when we substitute [latex]x = h(r, \theta) = r\cos\theta[/latex], [latex]y = g(r, \theta) = r\sin\theta[/latex], and [latex]dA = dx , dy = r , dr , d\theta[/latex], we get:

[latex]\iint_R f(x,y) dA = \iint_S f(r\cos\theta, r\sin\theta) r , dr , d\theta[/latex]

where region [latex]R[/latex] in the [latex]xy[/latex]-plane is replaced by region [latex]S[/latex] in polar coordinates.

The function we use to change variables is called a transformation or mapping. This section develops the general theory for changing variables in multiple integrals.

Planar Transformations

A planar transformation [latex]T[/latex] transforms a region [latex]G[/latex] in one plane into a region [latex]R[/latex] in another plane through a change of variables. Both [latex]G[/latex] and [latex]R[/latex] are subsets of [latex]\mathbb{R}^2[/latex].

Figure 1 shows region [latex]G[/latex] in the [latex]uv[/latex]-plane transformed into region [latex]R[/latex] in the [latex]xy[/latex]-plane by the change of variables [latex]x = g(u,v)[/latex] and [latex]y = h(u,v)[/latex]. We can also write this as [latex]x = x(u,v)[/latex] and [latex]y = y(u,v)[/latex].

We typically assume that each transformation function has continuous first partial derivatives. This means [latex]g_u[/latex], [latex]g_v[/latex], [latex]h_u[/latex], and [latex]h_v[/latex] all exist and are continuous. The reason for this requirement will become clear as we develop the theory.

One-to-One Transformations

A transformation [latex]T: G \to R[/latex] defined as [latex]T(u,v) = (x,y)[/latex] is one-to-one if no two different points in [latex]G[/latex] map to the same point in [latex]R[/latex].

To prove that [latex]T[/latex] is one-to-one, we assume [latex]T(u_1, v_1) = T(u_2, v_2)[/latex] and show that this implies [latex](u_1, v_1) = (u_2, v_2)[/latex].

When transformation [latex]T[/latex] is one-to-one on domain [latex]G[/latex], the inverse transformation [latex]T^{-1}[/latex] exists with domain [latex]R[/latex] such that [latex]T^{-1} \circ T[/latex] and [latex]T \circ T^{-1}[/latex] are identity functions.

Domain, Range, and Image

In Figure 1, the mapping [latex]T(u,v) = (x,y)[/latex] relates [latex]x[/latex] and [latex]y[/latex] to [latex]u[/latex] and [latex]v[/latex] through the equations [latex]x = g(u,v)[/latex] and [latex]y = h(u,v)[/latex].

Region [latex]G[/latex] is the domain of [latex]T[/latex], while region [latex]R[/latex] is the range of [latex]T[/latex], also called the image of [latex]G[/latex] under transformation [latex]T[/latex].

Suppose a transformation [latex]T[/latex] is defined as [latex]{T}{({r},{\theta})} = {({x},{y})}[/latex] where [latex]{x} = {r}{\cos}{\theta}, {y} = {r}{\sin}{\theta}[/latex]. Find the image of the polar rectangle [latex]{G} = {\left \{{({r},{\theta})}{\mid}{0} \leq {r} \leq {{1},{0}} \leq {\theta} \leq {{\pi}/{2}} \right \}}[/latex] in the [latex]{r}{\theta}[/latex]-plane to a region [latex]R[/latex] in the [latex]xy[/latex]-plane. Show that [latex]T[/latex] is a one-to-one transformation in [latex]G[/latex] and find [latex]{{T}^{-1}}{({x},{y})}[/latex].

Show Solution

Since [latex]r[/latex] varies from 0 to 1 in the [latex]{r}{\theta}[/latex]-plane, we have a circular disc of radius 0 to 1 in the [latex]xy[/latex]-plane. Because [latex]{\theta}[/latex] varies from 0 to [latex]{\pi}/{2}[/latex] in the [latex]{r}{\theta}[/latex]-plane, we end up getting a quarter circle of radius 1 in the first quadrant of the [latex]xy[/latex]-plane (Figure 2). Hence [latex]R[/latex] is a quarter circle bounded by [latex]x^{2}+y^{2}=1[/latex] in the first quadrant.

In order to show that [latex]T[/latex] is a one-to-one transformation, assume [latex]{T}{({r_1},{{\theta}_1})} = {T}{({r_2},{{\theta}_2})}[/latex] and show as a consequence that [latex]{({r_1},{{\theta}_1})} = {({r_2},{{\theta}_2})}[/latex]. In this case, we have

[latex]\begin{aligned} {T}{({r_1},{{\theta}_1})} & = {T}{({r_2},{{\theta}_2})} \\ {({x_1},{y_1})} & = {({x_2},{y_2})} \\ {({r_1}{\cos}{{\theta}_{1}},{r_1}{\sin}{{\theta}_{1}})} & = {({r_2}{\cos}{{\theta}_{2}},{r_2}{\sin}{{\theta}_{2}})} \\ {r_1}{\cos}{{\theta}_{1}} & = {r_2}{\cos}{{\theta}_{2}} \\ {r_1}{\sin}{{\theta}_{1}} & = {r_2}{\sin}{{\theta}_{2}} \end{aligned}[/latex]

Dividing, we obtain

[latex]\begin{aligned} {\frac{{r_1}{\cos}{{\theta}_{1}}}{{r_1}{\sin}{{\theta}_{1}}}} & = {{\frac{{r_2}{\cos}{{\theta}_{2}}}{{r_2}{\sin}{{\theta}_{2}}}}} \\ {\frac{{\cos}{{\theta}_{1}}}{{\sin}{{\theta}_{1}}}} & = {\frac{{\cos}{{\theta}_{2}}}{{\sin}{{\theta}_{2}}}} \\ {\tan}{{\theta}_{1}} & = {\tan}{{\theta}_{2}} \\ {{\theta}_{1}} & = {{\theta}_{2}} \end{aligned}[/latex]

since the tangent function is one-one function in the interval [latex]{0} \leq {\theta} \leq {{\pi}/{2}}[/latex]. Also, since [latex]{0} < {r} \leq {1}[/latex], we have [latex]{{r}_{1}} = {{r}_{2}}, {{\theta}_{1}} = {{\theta}_{2}}[/latex]. Therefore, [latex]{({{r}_{1}}, {{\theta}_{1}})} = {({{r}_{2}}, {{\theta}_{2}})}[/latex] and [latex]T[/latex] is a one-to-one transformation from [latex]G[/latex] into [latex]R[/latex].

To find [latex]{{T}^{-1}}{({x},{y})}[/latex] solve for [latex]{r},{\theta}[/latex] in terms of [latex]x, y[/latex]. We already know that [latex]{{r}^{2}} = {{x}^{2}} + {{y}^{2}}[/latex] and [latex]{\tan}{\theta} = {\frac{y}{x}}[/latex]. Thus [latex]{{T}^{-1}}{({x},{y})} = {({r},{\theta})}[/latex] is defined as [latex]{r} = {\sqrt{{x^{2}}+{y^{2}}}}[/latex] and [latex]{\theta} = {{\tan}^{-1}}{({\frac{y}{x}})}[/latex].

Let the transformation [latex]T[/latex] be defined by [latex]T(u, v)=(x, y)[/latex] where [latex]x=u^{2}+v^{2}[/latex] and [latex]y=uv[/latex]. Find the image of the triangle in the [latex]uv[/latex]-plane with vertices [latex](0, 0)[/latex], [latex](0, 1)[/latex], and [latex](1, 1)[/latex].

Show Solution

The triangle and its image are shown in Figure 3. To understand how the sides of the triangle transform, call the side that joins [latex](0, 0)[/latex] and [latex](0, 1)[/latex] side [latex]A[/latex], the side that joins [latex](0, 0)[/latex] and [latex](1, 1)[/latex] side [latex]B[/latex], and the side that joins [latex](1, 1)[/latex] and [latex](0, 1)[/latex] side [latex]C[/latex].

For the side [latex]A[/latex]: [latex]u=0[/latex], [latex]{0} \leq {v} \leq {1}[/latex] transforms to [latex]x=-v^{2}[/latex], [latex]y=0[/latex] so this is the side [latex]{A}^{\prime}[/latex] that joins [latex](-1, 0)[/latex] and [latex](0, 0)[/latex].

For the side [latex]B[/latex]: [latex]u=v[/latex], [latex]{0} \leq {u} \leq {1}[/latex] transforms to [latex]x=0[/latex], [latex]y=u^{2}[/latex] so this is the side [latex]{B}^{\prime}[/latex] that joins [latex](0, 0)[/latex] and [latex](0, 1)[/latex].

For the side [latex]C[/latex]: [latex]{0} \leq {u} \leq {1}[/latex], [latex]v=1[/latex] transforms to [latex]x=u^{2}-1[/latex], [latex]y=u[/latex] (hence [latex]x=y^{2}-1[/latex]) so this is the side [latex]{C}{\prime}[/latex] that makes the upper half of the parabolic arc joining [latex](-1, 0)[/latex] and [latex](0, 1)[/latex].

All the points in the entire region of the triangle in the [latex]uv[/latex]-plane are mapped inside the parabolic region in the [latex]xy[/latex]-plane.