Change of Variables in Double and Triple Integrals

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Hello! Last lesson, you worked with a constraint that restricted which points were allowed. Here, the points stay the same, but you will describe them with different coordinates. A good choice of coordinates can turn a difficult integration region into a rectangle or a box—and can simplify the integrand too.

The catch is that a small rectangle in the new coordinates need not have the same area or volume in physical space. This lesson develops the Jacobian determinant as the correction factor, then uses it to change variables in double and triple integrals. The same principle will later matter whenever you integrate a spatial density, including a quantum probability density.

What a coordinate change does to area

Suppose a transformation $T$ describes points in the $x y$ -plane using coordinates $u, v$ :

T (u, v) = (x (u, v), y (u, v)) .

A small $uv$ -rectangle has sides $Δ u$ and $Δ v$ . Its image under $T$ may have curved edges, but on a sufficiently small scale it looks like a parallelogram. One side is approximately $\partial T / \partial u$ times $Δ u$ ; the other is approximately $\partial T / \partial v$ times $Δ v$ .

A rectangular grid in the $uv$-plane maps to a curved grid in the $xy$-plane. The Jacobian measures how the area of a tiny grid cell changes under the map.

The two derivative vectors are the columns of the Jacobian matrix you encountered in the chain-rule lesson:

D T = (x_{u} y_{u} x_{v} y_{v}) .

The area of the small image parallelogram is approximately the absolute value of this matrix’s determinant times $Δ u Δ v$ . Thus, if

J_{T} (u, v) = det D T = x_{u} y_{v} - x_{v} y_{u},

then the area element transforms as

d A = ∣ J_{T} (u, v)∣ d u d v .

A negative determinant means that the map reverses orientation; it does not mean that an area is negative. That is why an ordinary area or volume integral uses $∣ J_{T} ∣$ , not $J_{T}$ alone.

Watch Change of Variables & The Jacobian by Dr. Trefor Bazett for a visual account of this scaling. Its slanted-region example also shows how choosing coordinates to match the boundaries can make an integral easier.

Change of Variables & The Jacobian | Multi-variable Integration

Watch Dr. Trefor Bazett's Change of Variables & The Jacobian to connect a practical substitution with the geometry of small area elements.

In the substitution, follow how the slanted boundaries become constant-coordinate lines and how the transformation is written in the direction needed for integration. Then watch the area argument: the derivative vectors span the tiny transformed parallelogram. For ordinary area integrals, use the absolute value of its determinant, even when a calculation happens to give a positive determinant.

The double-integral formula in use

If $T$ maps a region $S$ in the $uv$ -plane once onto a region $R$ in the $x y$ -plane, the change-of-variables formula is

\iint_{R} f (x, y) d A = \iint_{S} f (x (u, v), y (u, v)) \frac{\partial ( x , y )}{\partial ( u , v )} d u d v .

There are three coordinated changes here: describe the region in $u, v$ , express the integrand in $u, v$ , and replace the area element using the Jacobian. Changing only two of the three gives the wrong integral.

For example, let $R$ be bounded by

0 \leq x + y \leq 2, 0 \leq x - y \leq 1,

and consider $\iint_{R} (x^{2} - y^{2}) d A$ . The boundary equations suggest setting $u = x + y$ and $v = x - y$ . The new region is simply the rectangle $0 \leq u \leq 2, 0 \leq v \leq 1$ . To compute the forward Jacobian—from the new coordinates to $x, y$ —solve for

x = \frac{u + v}{2}, y = \frac{u - v}{2} .

Consequently,

\frac{\partial ( x , y )}{\partial ( u , v )} = det (\frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{1}{2}) = - \frac{1}{2}, x^{2} - y^{2} = (x + y) (x - y) = uv .

The negative sign records an orientation reversal; the area factor is $1/2$ . The integral becomes

\iint_{R} (x^{2} - y^{2}) d A = \int_{0}^{2} \int_{0}^{1} \frac{uv}{2} d v d u = \frac{1}{2} .

Notice the direction of the determinant. We first defined $u, v$ in terms of $x, y$ , but the area factor needed here is $∣ \partial (x, y) / \partial (u, v)∣$ . Where the inverse exists, its determinant is the reciprocal of $\partial (u, v) / \partial (x, y)$ .

The other common reason to change coordinates is radial symmetry. For polar coordinates,

x = r cos θ, y = r sin θ, \frac{\partial ( x , y )}{\partial ( r , θ )} = det (cos θ sin θ - r sin θ r cos θ) = r .

Thus $d A = r d r d θ$ , for $r \geq 0$ . Over a disk of radius $a$ , the quantity $x^{2} + y^{2}$ becomes $r^{2}$ , giving

\iint_{x^{2} + y^{2} \leq a^{2}} (x^{2} + y^{2}) d A = \int_{0}^{2 π} \int_{0}^{a} r^{2} r d r d θ = \frac{π a ^{4}}{2} .

The extra $r$ is not part of the original integrand: it says that an angular interval of fixed width sweeps out a larger area farther from the origin.

From area scaling to volume scaling

The three-dimensional version replaces tiny coordinate boxes with tiny parallelepipeds. For

T (u, v, w) = (x (u, v, w), y (u, v, w), z (u, v, w)),

the three columns of $D T$ describe its three local edge directions. Their determinant gives signed volume scaling, so

∭_{D} f (x, y, z) d V = ∭_{G} f (T (u, v, w)) det x_{u} y_{u} z_{u} x_{v} y_{v} z_{v} x_{w} y_{w} z_{w} d u d v d w,

when $T$ maps the new-coordinate region $G$ once onto $D$ .

A rectangular box in $uvw$-space maps to a curved region in $xyz$-space. The absolute value of the $3\times3$ Jacobian determinant gives the local volume-scaling factor.

Read the indicated parts of OpenStax Calculus Volume 3, Section 5.7, to see the general triple-integral rule alongside the cylindrical and spherical cases. The displayed mathematics in those examples is worth following: each familiar coordinate-volume factor comes from a determinant, rather than being an unrelated rule to memorize.

5.7 Change of Variables in Multiple Integrals

Read OpenStax's discussion of triple integrals to connect the general Jacobian rule with cylindrical and spherical coordinates.

Under “Change of Variables for Triple Integrals,” read the setup and theorem, including the three-variable determinant. Then, in “Example 5.72: Obtaining Formulas in Triple Integrals for Cylindrical and Spherical Coordinates,” follow both coordinate derivations. Stop before Example 5.73; focus on which variable order each determinant uses and why its absolute value supplies the volume factor.

A general transformation need not be cylindrical or spherical. For instance, let $G$ be the unit box $0 \leq u, v, w \leq 1$ and set

x = 2 u + v, y = v, z = 3 w .

Its image $D$ is a slanted, stretched prism. The determinant is

det D T = det 200110003 = 6.

If $f (x, y, z) = z$ , then $f (T (u, v, w)) = 3 w$ , and

∭_{D} z d V = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} 3 w \cdot 6 d u d v d w = 9.

The constant $6$ accounts for the prism’s volume relative to the unit box; the $3 w$ accounts separately for the changing value of $z$ .

For cylindrical coordinates, $x = r cos θ, y = r sin θ, z = z$ . The third coordinate is unchanged, so the three-dimensional Jacobian has the same magnitude $r$ as the polar area Jacobian. Spherical coordinates, with $ϕ$ measured down from the positive $z$ -axis, use

x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ, z = ρ cos ϕ,

and have volume factor $ρ^{2} sin ϕ$ . A ball of radius $a$ therefore provides a radial example:

∭_{x^{2} + y^{2} + z^{2} \leq a^{2}} (x^{2} + y^{2} + z^{2}) d V = \int_{0}^{2 π} \int_{0}^{π} \int_{0}^{a} ρ^{2} ρ^{2} sin ϕ d ρ d ϕ d θ = \frac{4 π a ^{5}}{5} .

Here the first $ρ^{2}$ comes from the function being integrated; the second comes from the volume element. Keeping those roles distinct is particularly useful when integrating physical densities.

Choosing coordinates without losing the region

A reliable order of work is to identify the old region’s boundaries, choose variables that simplify them or the integrand, find the new region, and only then assemble the transformed integral. OpenStax condenses this into a short checklist.

5.7 Change of Variables in Multiple Integrals

Use OpenStax's change-of-variables strategy as a check on the order of your work.

Find “Problem-Solving Strategy,” then its “Change of Variables” subsection. Read the six steps, from sketching the original region through replacing the area element. As you read, check where the region, integrand, and Jacobian each enter the calculation.

Finally, check whether your coordinates cover the physical region once. A full polar or spherical angular interval meets itself at a seam, and the angular coordinate is undefined at the origin or along an axis. Those sets have zero area or volume in the integrals above, so they do not alter the results. A map that covers an entire positive-area region twice is different: using it without restricting or splitting the coordinate domain would double-count that region. The standard formula is simplest for a one-to-one smooth map with nonzero Jacobian in the region’s interior; coordinate singularities and overlapping images deserve an explicit check.

Looking ahead

A change of variables rewrites the region and the integrand, while $∣ det D T ∣$ corrects the area or volume element. Its geometric meaning is local: the derivative matrix turns tiny coordinate rectangles or boxes into parallelograms or parallelepipeds, and its determinant measures their scaling.

This completes the multivariable-calculus module. Next, you will work with fields in space, beginning with divergence, curl, and the Laplacian in Cartesian coordinates.