Jacobians and Change of Variables

A change of variables rewrites a multiple integral in new variables u and v to make the region or the integrand simpler. The Jacobian is the determinant that tells how much the substitution stretches area. You replace dA with |∂(x, y)/∂(u, v)| du dv and integrate over the new region.

Before this lesson: Double Integrals over General Regions, Double Integrals in Polar Coordinates, Partial Derivatives and Determinant of a Matrix

Why integrals need a stretch factor

In single-variable u-substitution, dx = g′(u) du. The factor g′(u) says how much a small piece of the u-axis is stretched when it is mapped to the x-axis. In two variables, a transformation x = x(u, v), y = y(u, v) maps a small rectangle in the uv-plane to a small, almost parallelogram-shaped patch in the xy-plane.

The area of that patch is the area of the rectangle, du dv, times a stretch factor. That factor is the absolute value of the Jacobian. You already met two Jacobians: r in polar and cylindrical coordinates, and ρ² sin φ in spherical coordinates.

Jacobian and change of variables formula

\dfrac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[2mm] \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}

\iint_R f(x, y)\,dA = \iint_S f\big(x(u, v), y(u, v)\big)\left|\dfrac{\partial(x, y)}{\partial(u, v)}\right|du\,dv

Here S is the region in the uv-plane that the transformation maps onto R. Use the absolute value: the Jacobian can be negative, but area cannot.

1. Choose u and v from the boundary curves or the integrand. Lines such as x + y = c suggest u = x + y.
2. Solve for x and y in terms of u and v (or use the shortcut in Related methods).
3. Compute the Jacobian ∂(x, y)/∂(u, v) and take its absolute value.
4. Find the new region S by rewriting each boundary of R in u and v.
5. Rewrite the integrand, multiply by |J| and integrate over S.

Jacobian change of variables examples

Example 1: The Jacobian of polar coordinates

Find the Jacobian of x = r cos θ, y = r sin θ.

\dfrac{\partial x}{\partial r} = \cos\theta, \quad \dfrac{\partial x}{\partial \theta} = -r\sin\theta

\dfrac{\partial y}{\partial r} = \sin\theta, \quad \dfrac{\partial y}{\partial \theta} = r\cos\theta

\dfrac{\partial(x, y)}{\partial(r, \theta)} = \cos\theta \cdot r\cos\theta - (-r\sin\theta)\sin\theta

= r\cos^2\theta + r\sin^2\theta

= r

Answer: the Jacobian is r. That is why dA = r dr dθ in double integrals in polar coordinates.

Example 2: A parallelogram region

Evaluate \iint_R (x + y)^2\,dA, where R is the parallelogram bounded by x + y = 0, x + y = 2, x − 2y = 0 and x − 2y = 3.

u = x + y, \quad v = x - 2y

0 \le u \le 2, \quad 0 \le v \le 3

u - v = 3y

y = \dfrac{u - v}{3}

x = u - y = \dfrac{2u + v}{3}

\dfrac{\partial(x, y)}{\partial(u, v)} = \dfrac{2}{3}\cdot\Big(-\dfrac{1}{3}\Big) - \dfrac{1}{3}\cdot\dfrac{1}{3}

= -\dfrac{3}{9} = -\dfrac{1}{3}

|J| = \dfrac{1}{3}

\iint_R (x + y)^2\,dA = \int_0^3 \int_0^2 u^2 \cdot \dfrac{1}{3}\,du\,dv

\int_0^2 \dfrac{u^2}{3}\,du = \Big[\dfrac{u^3}{9}\Big]_0^2 = \dfrac{8}{9}

\int_0^3 \dfrac{8}{9}\,dv = \dfrac{8}{3}

Answer: 8/3. In x and y, this region would need three separate integrals.

Example 3: Area of an ellipse

Find the area of the ellipse x²/9 + y²/4 ≤ 1.

x = 3u, \quad y = 2v

u^2 + v^2 \le 1

\dfrac{\partial(x, y)}{\partial(u, v)} = 3 \cdot 2 - 0 \cdot 0 = 6

\text{Area} = \iint_S 6\,du\,dv

= 6 \cdot \pi(1)^2

= 6\pi

Answer: 6π ≈ 18.85. In general, an ellipse with semi-axes a and b has area πab.

Example 4: A region between hyperbolas

Evaluate \iint_R xy\,dA, where R is the region in the first quadrant between the hyperbolas xy = 1 and xy = 4 and the lines y = x and y = 4x.

u = xy, \quad v = \dfrac{y}{x}

\dfrac{\partial(u, v)}{\partial(x, y)} = y \cdot \dfrac{1}{x} - x \cdot \Big(-\dfrac{y}{x^2}\Big)

= \dfrac{2y}{x} = 2v

\dfrac{\partial(x, y)}{\partial(u, v)} = \dfrac{1}{2v}

\iint_R xy\,dA = \int_1^4 \int_1^4 u \cdot \dfrac{1}{2v}\,du\,dv

\int_1^4 u\,du = \dfrac{16 - 1}{2} = \dfrac{15}{2}

\int_1^4 \dfrac{1}{2v}\,dv = \dfrac{1}{2}\ln 4 = \ln 2

\dfrac{15}{2}\ln 2 \approx 5.199

Answer: (15/2) ln 2 ≈ 5.20.

Example 5: Water in an oval pond

An oval pond is 30 m long and 20 m wide, so its edge is the ellipse x²/225 + y²/100 = 1. Its depth in meters is d = 2(1 − x²/225 − y²/100), so it is 2 m deep in the middle. How much water does it hold?

V = \iint_R 2\Big(1 - \dfrac{x^2}{225} - \dfrac{y^2}{100}\Big)\,dA

x = 15u, \quad y = 10v

\dfrac{\partial(x, y)}{\partial(u, v)} = 15 \cdot 10 = 150

V = \iint_{u^2 + v^2 \le 1} 2(1 - u^2 - v^2) \cdot 150\,du\,dv

= 300\int_0^{2\pi} \int_0^1 (1 - r^2)\,r\,dr\,d\theta

\int_0^1 (r - r^3)\,dr = \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{1}{4}

V = 300 \cdot 2\pi \cdot \dfrac{1}{4}

= 150\pi \approx 471.2

Answer: about 471 m³ of water. As a check, the surface area is π(15)(10) = 150π m², so the average depth is 1 m, half of the 2 m center depth.

Common mistakes

Related methods

The Jacobian of the inverse transformation

If it is hard to solve for x and y, find u and v as functions of x and y and use ∂(x, y)/∂(u, v) = 1 ÷ ∂(u, v)/∂(x, y). Example 4 used this shortcut. In Example 2, ∂(u, v)/∂(x, y) = (1)(−2) − (1)(1) = −3, and its reciprocal is −1/3, as found above.

Jacobians for triple integrals

For x = x(u, v, w), y = y(u, v, w), z = z(u, v, w), the Jacobian is the 3 × 3 determinant of all nine partial derivatives, and dV = |∂(x, y, z)/∂(u, v, w)| du dv dw. For spherical coordinates this determinant is ρ² sin φ.

Practice problems

1. Find the Jacobian of x = 3u − v, y = u + 2v.
2. Find the Jacobian of x = uv, y = u.
3. Use x = 5u, y = 4v to find the area of the ellipse x²/25 + y²/16 ≤ 1.
4. Evaluate \iint_R (x - y)(x + y)\,dA, where R is the square bounded by x − y = 0, x − y = 2, x + y = 1 and x + y = 3.
5. Find the area of the region R in Problem 4 with a change of variables.
6. For u = x + y and v = x − 3y, find ∂(x, y)/∂(u, v) without solving for x and y.
7. Evaluate \iint_R x^2\,dA, where R is the ellipse x²/4 + y² ≤ 1.
8. Find the area of the region in the first quadrant between xy = 1 and xy = 3 and between the lines y = x and y = ex.
9. Use x = 2u, y = 3v, z = 4w to find the volume of the ellipsoid x²/4 + y²/9 + z²/16 ≤ 1.
10. An oval lawn has semi-axes 40 m and 25 m. One bag of fertilizer covers 400 m². Find the area of the lawn and the number of bags needed.

\dfrac{\partial(x, y)}{\partial(u, v)} = 3 \cdot 2 - (-1)(1)

= 7

\dfrac{\partial(x, y)}{\partial(u, v)} = v \cdot 0 - u \cdot 1

= -u

\dfrac{\partial(x, y)}{\partial(u, v)} = 5 \cdot 4 = 20

\text{Area} = 20 \cdot \pi = 20\pi

u = x - y, \quad v = x + y

x = \dfrac{u + v}{2}, \quad y = \dfrac{v - u}{2}

\dfrac{\partial(x, y)}{\partial(u, v)} = \dfrac{1}{2} \cdot \dfrac{1}{2} - \dfrac{1}{2} \cdot \Big(-\dfrac{1}{2}\Big) = \dfrac{1}{2}

\int_1^3 \int_0^2 uv \cdot \dfrac{1}{2}\,du\,dv

\dfrac{1}{2} \cdot 2 \cdot 4 = 4

\text{Area} = \int_1^3 \int_0^2 \dfrac{1}{2}\,du\,dv

= \dfrac{1}{2} \cdot 2 \cdot 2 = 2

\dfrac{\partial(u, v)}{\partial(x, y)} = 1 \cdot (-3) - 1 \cdot 1 = -4

\dfrac{\partial(x, y)}{\partial(u, v)} = -\dfrac{1}{4}

x = 2u, \quad y = v, \quad |J| = 2

\iint_R x^2\,dA = \iint_S 4u^2 \cdot 2\,du\,dv = 8\iint_S u^2\,du\,dv

\iint_S u^2\,du\,dv = \int_0^{2\pi} \cos^2\theta\,d\theta \int_0^1 r^3\,dr

= \pi \cdot \dfrac{1}{4}

8 \cdot \dfrac{\pi}{4} = 2\pi

u = xy, \quad v = \dfrac{y}{x}, \quad |J| = \dfrac{1}{2v}

\text{Area} = \int_1^e \int_1^3 \dfrac{1}{2v}\,du\,dv

= 2 \cdot \dfrac{1}{2}(\ln e - \ln 1) = 1

\dfrac{\partial(x, y, z)}{\partial(u, v, w)} = 2 \cdot 3 \cdot 4 = 24

V = 24 \cdot \dfrac{4\pi}{3} = 32\pi

\text{Area} = \pi(40)(25) = 1000\pi \approx 3141.6

\dfrac{3141.6}{400} \approx 7.85

Jacobian and change of variables FAQ

What is a Jacobian in calculus?

It is the determinant of the partial derivatives of a transformation. Its absolute value is the factor by which the transformation stretches small areas (or volumes in 3D).

Why do you take the absolute value of the Jacobian?

A negative Jacobian only means the transformation flips orientation, like a mirror. Area is always positive, so the integral uses |J|.

How do you choose u and v?

Look at the boundary curves. If R is bounded by x + y = 0 and x + y = 2, then u = x + y turns those into u = 0 and u = 2. Sometimes the integrand suggests u and v instead.

What is the Jacobian for polar coordinates?

It is r, so dA = r dr dθ. For cylindrical coordinates it is also r, and for spherical coordinates it is ρ² sin φ.

Summary

• The Jacobian is ∂(x, y)/∂(u, v) = x_u y_v − x_v y_u.
• Change of variables: dA = |∂(x, y)/∂(u, v)| du dv, over the new region S.
• Pick u and v that turn the boundaries of R into constants.
• If u and v are given in terms of x and y, use the reciprocal of ∂(u, v)/∂(x, y).
• Polar: J = r. Spherical: J = ρ² sin φ.

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