The Main Idea 

Just as [latex]u[/latex]-substitution simplifies single-variable integrals, changing variables in multiple integrals can transform difficult integration problems into manageable ones. You’ve already used this idea with polar, cylindrical, and spherical coordinates. This section develops the general framework for any change of variables.

The Core Idea

A transformation [latex]T[/latex] maps points from one coordinate system to another. For example, [latex]T: G \to R[/latex] transforms region [latex]G[/latex] in the [latex]uv[/latex]-plane to region [latex]R[/latex] in the [latex]xy[/latex]-plane using equations [latex]x = g(u,v)[/latex] and [latex]y = h(u,v)[/latex]. We write [latex]T(u,v) = (x,y)[/latex] to show that point [latex](u,v)[/latex] maps to point [latex](x,y)[/latex].

The goal is to convert an integral over region [latex]R[/latex] (which might be complicated) into an integral over region [latex]G[/latex] (which is simpler). This parallels how [latex]\int_2^3 x(x^2-4)^5 dx[/latex] becomes [latex]\int_0^5 \frac{1}{2}u^5 du[/latex] after substituting [latex]u = x^2 - 4[/latex].

One-to-One Transformations

A transformation is one-to-one if different input points always produce different output points. Formally, [latex]T(u_1, v_1) = T(u_2, v_2)[/latex] implies [latex](u_1, v_1) = (u_2, v_2)[/latex]. When [latex]T[/latex] is one-to-one, an inverse transformation [latex]T^{-1}[/latex] exists that maps back from [latex]R[/latex] to [latex]G[/latex].

For example, polar coordinates give a one-to-one transformation when [latex]0 \leq r \leq 1[/latex] and [latex]0 \leq \theta \leq \frac{\pi}{2}[/latex]. The inverse exists: [latex]r = \sqrt{x^2 + y^2}[/latex] and [latex]\theta = \tan^{-1}(\frac{y}{x})[/latex].

Domain and Range

Region [latex]G[/latex] is the domain of [latex]T[/latex]—the set of input points. Region [latex]R[/latex] is the range or image—the set of output points. We encounter two types of problems: finding the image [latex]R[/latex] given [latex]G[/latex] and [latex]T[/latex] (forward problem), or finding the pre-image [latex]G[/latex] given [latex]R[/latex] and [latex]T[/latex] (inverse problem).

Finding Images

To find where a region transforms, examine how its boundaries map. For each edge or curve bounding [latex]G[/latex], substitute the transformation equations to find the corresponding boundary in [latex]R[/latex]. Points inside [latex]G[/latex] map to points inside [latex]R[/latex].

The Main Idea 

The Jacobian measures how a transformation scales areas (or volumes in higher dimensions). When changing variables in an integral, you need to know not just the new coordinates, but also how much the transformation stretches or compresses area. The Jacobian provides this crucial scaling factor.

What is the Jacobian?

For a transformation [latex]T(u,v) = (x,y)[/latex] where [latex]x = g(u,v)[/latex] and [latex]y = h(u,v)[/latex], the Jacobian is:

[latex]J(u,v) = \dfrac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} = \dfrac{\partial x}{\partial u} \dfrac{\partial y}{\partial v} - \dfrac{\partial x}{\partial v} \dfrac{\partial y}{\partial u}[/latex]

This [latex]2 \times 2[/latex] determinant captures how the transformation changes areas. A small rectangle of area [latex]\Delta u \Delta v[/latex] in the [latex]uv[/latex]-plane transforms to a region of approximate area [latex]|J(u,v)| \Delta u \Delta v[/latex] in the [latex]xy[/latex]-plane.

Why the Jacobian Matters

When a transformation maps region [latex]G[/latex] to region [latex]R[/latex], it distorts the shape. A rectangle in [latex]G[/latex] might become a curved region in [latex]R[/latex]. The Jacobian tells you the area scaling factor at each point. For example, polar coordinates have Jacobian [latex]J(r,\theta) = r[/latex], which is why [latex]dA = r \, dr \, d\theta[/latex] includes that factor of [latex]r[/latex].

Computing the Jacobian

Set up the [latex]2 \times 2[/latex] matrix with partial derivatives: first row has [latex]x[/latex] derivatives with respect to [latex]u[/latex] and [latex]v[/latex], second row has [latex]y[/latex] derivatives. Compute the determinant using [latex]ad - bc[/latex] for matrix [latex]\begin{vmatrix} a & b \ c & d \end{vmatrix}[/latex].

Geometric Interpretation

The transformation maps the sides of a small rectangle (length [latex]\Delta u[/latex], width [latex]\Delta v[/latex]) to tangent vectors [latex]\Delta u \mathbf{r}_u[/latex] and [latex]\Delta v \mathbf{r}_v[/latex] in the [latex]xy[/latex]-plane. These vectors form a parallelogram, and its area equals [latex]|\mathbf{r}_u \times \mathbf{r}_v| \Delta u \Delta v[/latex]. The cross product magnitude equals [latex]|J(u,v)|[/latex].

The Main Idea 

Changing variables in double integrals transforms a difficult integral over region [latex]R[/latex] into a simpler integral over region [latex]S[/latex]. The key formula replaces [latex]dA[/latex] with [latex]|J(u,v)| , du , dv[/latex], where the Jacobian accounts for how the transformation scales areas.

The Change of Variables Formula

For transformation [latex]T(u,v) = (x,y)[/latex] where [latex]x = g(u,v)[/latex] and [latex]y = h(u,v)[/latex]:

[latex]\underset{R}{\iint} f(x,y) , dA = \underset{S}{\iint} f(g(u,v), h(u,v)) \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right| du \, dv[/latex]

This means: replace [latex]x[/latex] and [latex]y[/latex] in the integrand with their expressions in [latex]u[/latex] and [latex]v[/latex], multiply by the absolute value of the Jacobian, and integrate over the transformed region [latex]S[/latex] using [latex]du , dv[/latex].

Why This Works

A small rectangle in the [latex]uv[/latex]-plane with area [latex]\Delta u \Delta v[/latex] transforms into a curved region in the [latex]xy[/latex]-plane with area approximately [latex]|J(u,v)| \Delta u \Delta v[/latex]. The Jacobian captures this area scaling, ensuring the integral measures the correct total.

Choosing a Good Transformation

Look for patterns that suggest natural substitutions. The region’s boundary curves often reveal the transformation—if boundaries are lines like [latex]x - y = c[/latex] or curves like [latex]x^2 - y^2 = c[/latex], let [latex]u = x - y[/latex]. The integrand itself may also suggest coordinates (expressions like [latex]x^2 + y^2[/latex] suggest polar coordinates). Choose transformations that make either the region simpler (turning curves into straight lines) or the integrand simpler (or both).

Problem-Solving Strategy

1. Sketch region [latex]R[/latex] in the [latex]xy[/latex]-plane and identify boundary curves
2. Choose transformation [latex]x = g(u,v)[/latex], [latex]y = h(u,v)[/latex] that simplifies the region or integrand
3. Determine how boundaries transform to find the new region [latex]S[/latex] and limits in the [latex]uv[/latex]-plane
4. Compute the Jacobian [latex]J(u,v) = \dfrac{\partial(x,y)}{\partial(u,v)}[/latex]
5. Substitute the transformation into the integrand
6. Replace [latex]dA[/latex] with [latex]|J(u,v)| , du , dv[/latex] and evaluate over region [latex]S[/latex]

Why the Absolute Value?

The Jacobian can be negative if the transformation reverses orientation, but area must be positive. Using [latex]|J(u,v)|[/latex] ensures we measure area correctly.

The Main Idea 

The change of variables formula extends naturally from double integrals to triple integrals. The key difference is that the Jacobian becomes a [latex]3 \times 3[/latex] determinant that measures how the transformation scales volumes instead of areas. Cylindrical and spherical coordinates are just special cases of this general method.

The Change of Variables Formula

For transformation [latex]T(u,v,w) = (x,y,z)[/latex] where [latex]x = g(u,v,w)[/latex], [latex]y = h(u,v,w)[/latex], and [latex]z = k(u,v,w)[/latex]:

[latex]\underset{D}{\iiint} F(x,y,z) , dV = \underset{G}{\iiint} F(g(u,v,w), h(u,v,w), k(u,v,w)) \left|\dfrac{\partial(x,y,z)}{\partial(u,v,w)}\right| du \, dv \, dw[/latex]

This replaces [latex]dV = dx \, dy \, dz[/latex] with [latex]|J(u,v,w)| \, du \, dv \, dw[/latex], where the Jacobian accounts for volume scaling.

The Three-Dimensional Jacobian

The Jacobian is the [latex]3 \times 3[/latex] determinant:

[latex]J(u,v,w) = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w} \\ \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w} \end{vmatrix}[/latex]

You can arrange the determinant with rows for [latex]x[/latex], [latex]y[/latex], [latex]z[/latex] derivatives (shown above) or with columns for [latex]u[/latex], [latex]v[/latex], [latex]w[/latex] derivatives—both give the same result.

Connection to Standard Coordinates

Cylindrical coordinates ([latex]x = r\cos\theta[/latex], [latex]y = r\sin\theta[/latex], [latex]z = z[/latex]) have Jacobian [latex]J(r,\theta,z) = r[/latex], giving [latex]dV = r \, dr \, d\theta \, dz[/latex].

Spherical coordinates ([latex]x = \rho\sin\varphi\cos\theta[/latex], [latex]y = \rho\sin\varphi\sin\theta[/latex], [latex]z = \rho\cos\varphi[/latex]) have Jacobian [latex]J(\rho,\varphi,\theta) = \rho^2\sin\varphi[/latex], giving [latex]dV = \rho^2\sin\varphi \, d\rho \, d\varphi \, d\theta[/latex].

These familiar volume elements come directly from the general change of variables theorem.

Problem-Solving Strategy

1. Identify region [latex]D[/latex] in [latex]xyz[/latex]-space and its bounding surfaces
2. Choose transformation equations [latex]x = g(u,v,w)[/latex], [latex]y = h(u,v,w)[/latex], [latex]z = k(u,v,w)[/latex] that simplify the region or integrand
3. Determine the transformed region [latex]G[/latex] in [latex]uvw[/latex]-space by transforming each bounding surface
4. Compute the [latex]3 \times 3[/latex] Jacobian determinant
5. Substitute the transformation into the integrand
6. Replace [latex]dV[/latex] with [latex]|J(u,v,w)| \, du \, dv \, dw[/latex] and integrate over region [latex]G[/latex]