Phase 3 - Lesson 3.3

Multiple Integrals and Change of Variables

Integrating over regions of the plane and space, iterating with Fubini, and rescaling with the Jacobian determinant.

⏱ 55 min● Intermediate🔗 Prereqs: 3.2

Builds on: Partial derivatives and the Jacobian from 3.2.
Leads to: Change of variables underlies joint densities and the Gaussian normalization in probability (Phase 7).

Learning Objectives

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Set up and evaluate a double integral as an iterated integral over a region.
State Fubini’s theorem and swap the order of integration correctly.
Apply the change-of-variables formula with the Jacobian determinant.
Compute the Gaussian integral using a polar change of variables.

Key Vocabulary

Double integral: The limit \(\iint_R f\,dA\) of Riemann sums of \(f\) over a plane region \(R\).

Iterated integral: Evaluation \(\int\!\!\int f\,dy\,dx\) one variable at a time, inner limits possibly depending on the outer.

Fubini’s theorem: For integrable \(f\), the double integral equals either iterated integral, so the order may be swapped.

Jacobian determinant: The factor \(|\det J|\) by which a change of variables rescales area/volume elements.

Region of integration: The domain \(R\); its description (as x- or y-simple) sets the inner and outer limits.

Polar coordinates: The map \(x=r\cos\theta,\ y=r\sin\theta\) with area element \(dA=r\,dr\,d\theta\).

Intuition & Motivation

Intuition

A double integral adds up a quantity spread over a two-dimensional region - a volume under a surface, a total mass, or a probability. You compute it by slicing: integrate along one direction to collapse the region to a single variable, then integrate that. The change-of-variables formula is the multivariable substitution rule; the single-variable factor \(du/dx\) is replaced by the Jacobian determinant, which measures how much a coordinate map stretches area.

Double integrals and Fubini

Over a rectangle \(R=[a,b]\times[c,d]\), Fubini’s theorem lets us iterate in either order:

\[\iint_R f\,dA=\int_a^b\!\!\int_c^d f(x,y)\,dy\,dx=\int_c^d\!\!\int_a^b f(x,y)\,dx\,dy.\] (3.4)

Over a non-rectangular region we describe it as y-simple (\(g_1(x)\le y\le g_2(x)\)) or x-simple and let the inner limits depend on the outer variable. Swapping order then requires re-reading the region, not just flipping symbols.

Worked Example - Swapping the order of integration

Consider \(\int_0^1\!\!\int_x^1 e^{y^2}\,dy\,dx\). The inner antiderivative of \(e^{y^2}\) has no elementary form.

Re-describe the triangular region \(\{0\le x\le 1,\ x\le y\le 1\}\) as \(\{0\le y\le 1,\ 0\le x\le y\}\).

Swap: \(\int_0^1\!\!\int_0^y e^{y^2}\,dx\,dy=\int_0^1 y\,e^{y^2}\,dy\) since the inner integral is just \(e^{y^2}\cdot y\).

Substitute \(u=y^2\): \(\tfrac12\int_0^1 e^u\,du=\tfrac12(e-1)\). Choosing the right order made an impossible integral trivial.

Change of variables and the Jacobian

Theorem - Change of variables

If \(T:(u,v)\mapsto(x,y)\) is a smooth injective map with Jacobian determinant \(\det J_T=\partial(x,y)/\partial(u,v)\), then \(\iint_R f\,dx\,dy=\iint_{T^{-1}(R)} f(T(u,v))\,|\det J_T|\,du\,dv\).

The absolute value of the determinant is the local area-scaling factor: a coordinate map that doubles areas contributes a factor of 2. For polar coordinates \(\det J=r\), giving the familiar \(dA=r\,dr\,d\theta\).

The Gaussian integral

The most important integral in quantitative finance normalizes the normal distribution. It has no elementary antiderivative in one variable, yet a polar trick evaluates it exactly:

Worked Example - Evaluating \(I=\int_{-\infty}^{\infty} e^{-x^2}\,dx\)

Square it and write as a double integral: \(I^2=\int_{-\infty}^{\infty}\!\!\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\,dx\,dy\).

Switch to polar: \(x^2+y^2=r^2\) and \(dA=r\,dr\,d\theta\), so \(I^2=\int_0^{2\pi}\!\!\int_0^{\infty} e^{-r^2}\,r\,dr\,d\theta\).

Inner: \(\int_0^\infty e^{-r^2}r\,dr=\tfrac12\) (substitute \(s=r^2\)). Outer: \(\int_0^{2\pi}d\theta=2\pi\).

Hence \(I^2=\pi\) and \(I=\sqrt{\pi}\). Rescaling gives \(\int e^{-x^2/2}dx=\sqrt{2\pi}\), the normal constant.

This estimator throws random points in a square and counts the fraction inside a quarter-disc - a stochastic double integral. It converges at rate \(1/\sqrt{N}\), the same law that governs Monte Carlo option pricing (Phase 13).

Common Mistakes to Avoid

Swapping the order of integration without redrawing the region - the limits change, not just the differentials.
Forgetting the Jacobian factor \(r\) in polar coordinates; \(dA\neq dr\,d\theta\).
Using the signed determinant instead of \(|\det J|\); area scaling is always nonnegative.
Assuming every double integral factors into a product; that only happens when \(f(x,y)=g(x)h(y)\) on a rectangle.

Quant Practitioner Tips

When an inner integral is intractable, try swapping the order before anything fancier.
Radial symmetry (functions of \(x^2+y^2\)) almost always signals polar coordinates.
Memorize \(\int_{-\infty}^\infty e^{-x^2/2}dx=\sqrt{2\pi}\); it is the normalizing constant behind every Gaussian in the course.

Knowledge Check

Q1 Easy

Fubini’s theorem guarantees that for an integrable \(f\) on a rectangle:

The two orders of iterated integration give the same value

The integral factors as a product

f must be continuous

The region must be a disc

Q2 Easy

Under the polar change of variables, the area element \(dx\,dy\) becomes:

\(dr\,d\theta\)

\(r\,dr\,d\theta\)

\(r^2\,dr\,d\theta\)

\(\tfrac1r\,dr\,d\theta\)

Q3 Medium

The value of \(\int_{-\infty}^\infty e^{-x^2}\,dx\) is:

\(\pi\)

\(\sqrt{\pi}\)

\(2\pi\)

\(1\)

Practical Exercise

Evaluate \(\iint_R (x+y)\,dA\) where \(R\) is the triangle with vertices \((0,0),\ (1,0),\ (0,1)\). Set up the iterated integral, then compute.

▶ Show full solution

Describe \(R\) as \(0\le x\le 1,\ 0\le y\le 1-x\):

\[\int_0^1\!\!\int_0^{1-x}(x+y)\,dy\,dx=\int_0^1\Big[xy+\tfrac{y^2}{2}\Big]_0^{1-x}dx=\int_0^1\Big(x(1-x)+\tfrac{(1-x)^2}{2}\Big)dx.\]

Expand the integrand: \(x-x^2+\tfrac12(1-2x+x^2)=\tfrac12-\tfrac12x^2\). Integrate:

\[\int_0^1\Big(\tfrac12-\tfrac12 x^2\Big)dx=\tfrac12-\tfrac16=\tfrac13.\]

So the integral equals \(\tfrac13\). By symmetry in \(x\) and \(y\) you could also split into two equal pieces as a check.

After the reveal, answer for yourself: Re-do it with the order dx dy; the symmetry of the region should give the identical answer 1/3.

Lesson Summary

Double integrals are evaluated by iteration; Fubini lets us choose or swap the order (after re-reading the region). The change-of-variables formula rescales by the Jacobian determinant \(|\det J|\), and the polar trick delivers the Gaussian integral \(\int e^{-x^2}dx=\sqrt\pi\) that normalizes every normal distribution.

Formula Sheet Additions

Change of variables

\[\iint_R f\,dx\,dy=\iint_{T^{-1}(R)} (f\circ T)\,|\det J_T|\,du\,dv\]

Jacobian determinant = local area-scaling factor.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers

Q: What replaces du/dx from single-variable substitution in a 2-D change of variables?
A: The absolute Jacobian determinant |det J| of the coordinate map, the local area-scaling factor.

Q: How is the Gaussian integral evaluated?
A: Square it into a double integral, switch to polar (dA=r dr dθ), integrate to get π, so the integral is sqrt(π).

Completion Checklist

I can explain the core ideas in my own words
I worked the derivations/examples by hand
I completed the interactive workbench(es)
I passed the knowledge check

Confidence / mastery rating

Personal notes

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