Chapter 11 · Change of variables and the Jacobian

The hook

The integral even sideways slicing can't save

Chapter 10 left you with a promise: there is a famous integral that the order-swap trick cannot rescue. Here it is — the Gaussian, the most important integral in all of probability:

I = \int_{-\infty}^{\infty} e^{-x^2}\,dx.

Every Chapter 10 move fails. There is no elementary antiderivative for $e^{-x^2}$ — so the Fundamental Theorem is useless. And there is no inner integral to swap: this is a single integral, so changing the order of integration has nothing to change. We are stuck — until we do something that looks insane.

The bell curve's area (left) is exactly

\sqrt{\pi}

— but you cannot get there along the

x

-axis. Square the integral, read it as a double integral over the whole plane, and switch to polar (right): a factor

r

drops out of nowhere and makes the impossible a one-liner.

The punchline, which we will earn in full after the break:

\left(\int_{-\infty}^{\infty} e^{-x^2}\,dx\right)^{\!2} = \iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dA = \pi \qquad\Longrightarrow\qquad I = \sqrt{\pi}.

The whole trick turns on that $r$ appearing when we go polar. Where did it come from? Chapter 10 told you to write $dA = r\,dr\,d\theta$ and never drop the $r$ — but it never told you why the $r$ is there. That "why" is this entire chapter, and its name is the Jacobian.

Why this matters

Polar was a special recipe you memorised. The Jacobian is the general tool behind it: any substitution re-grids the plane, and the factor that corrects for the re-gridding is $|J|$ . Master it once and you never again have to remember a coordinate system's area element — you can derive it. And the payoff is not academic: the $\sqrt{\pi}$ you are about to compute is the exact reason the normal distribution — the bell curve under nearly every dataset she will ever model — normalises the way it does.

Intuition before formulas

A substitution warps the plane

In single-variable calculus, $u$ -substitution carries a $\dfrac{du}{dx}$ factor — the substitution stretches the line, and $du = \dfrac{du}{dx}\,dx$ corrects for the stretch. In two dimensions the same thing happens, but the plane can stretch differently in different directions, so the correction is richer.

Picture a clean $uv$ -grid of little unit squares. A change of variables $x = x(u,v)$ , $y = y(u,v)$ sends that grid to a warped grid in the $xy$ -plane. A small cell of area $du\,dv$ lands as a small parallelogram, and its area is no longer $du\,dv$ — it is scaled by some local factor:

The map sends the square

uv

-grid (left) to a warped grid (right). One unit cell of area

du\,dv

becomes a parallelogram of area

|J|\,du\,dv

. The number

|J|

is just the local area-scaling factor — nothing more mystical than that.

The honest truth about "seeing" the warp

Nobody pictures the global warp of the plane — the way the whole grid bends is genuinely hard to hold in your head, and you never need to. You compute the stretch factor locally, at the cell, with a determinant, and trust it everywhere. That single local number is the entire idea; the rest of the chapter is learning to compute it and remembering to keep its absolute value.

Formal statement

The Jacobian, stated and used

That local area-scaling factor has a formula. It is the determinant of the matrix of partial derivatives of the old coordinates with respect to the new ones:

The Jacobian, and the change-of-variables theorem

J = \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} x_u & x_v \\ y_u & y_v \end{vmatrix}, \qquad \iint_R f(x,y)\,dA = \iint_S f\big(x(u,v),y(u,v)\big)\,|J|\,du\,dv.

The region $R$ in the $xy$ -plane is described, in the new coordinates, by a simpler region $S$ in the $uv$ -plane — usually a rectangle. Three things to hold onto:

The three load-bearing facts

The area element is $|J|\,du\,dv$ — the absolute value, always. A negative determinant means the map flips orientation; it never means negative area.
The integrand must be rewritten in $u$ and $v$ — every $x$ and $y$ inside $f$ is replaced via the substitution.
Reciprocal shortcut. If $\dfrac{\partial(x,y)}{\partial(u,v)}$ is awkward but $\dfrac{\partial(u,v)}{\partial(x,y)}$ is easy, compute the easy one and flip it: $\ \dfrac{\partial(x,y)}{\partial(u,v)} = \dfrac{1}{\partial(u,v)/\partial(x,y)}.$

And now the chapter's spine, paid off. For polar coordinates $x = r\cos\theta$ , $y = r\sin\theta$ , with $(u,v) = (r,\theta)$ :

J = \frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r.

So $|J| = r$ , and $dA = r\,dr\,d\theta$ — exactly the recipe you were handed in Chapter 10. The mysterious $r$ was the Jacobian of the polar map all along.

Worked example · WE-1 · compute a Jacobian, see the area scale

Take the linear map $u = x + y$ , $v = x - y$ . Invert it: $x = \tfrac{u+v}{2}$ , $y = \tfrac{u-v}{2}$ . Then

J = \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \tfrac12 & \tfrac12 \\[2pt] \tfrac12 & -\tfrac12 \end{vmatrix} = -\tfrac14 - \tfrac14 = -\tfrac12, \qquad |J| = \tfrac12.

Every area halves. Check it with the shortcut: $\dfrac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -2$ , and $\dfrac{1}{-2} = -\tfrac12$ — same answer. The sign of $J$ is bookkeeping (the map reflects); the area element takes $|J| = \tfrac12$ .

Worked example · WE-2 · a full evaluation, linear

Evaluate $\displaystyle\iint_R e^{(x-y)/(x+y)}\,dA$ over the triangle with vertices $(0,0)$ , $(1,0)$ , $(0,1)$ . The exponent begs for $u = x - y$ , $v = x + y$ ; from WE-1, $|J| = \tfrac12$ . The triangle $x \ge 0,\ y \ge 0,\ x + y \le 1$ becomes the clean region $0 \le v \le 1$ , $-v \le u \le v$ :

\int_0^1\!\int_{-v}^{v} e^{u/v}\,\tfrac12\,du\,dv = \tfrac12\int_0^1 v\,(e - e^{-1})\,dv = \frac{e - e^{-1}}{4} = \frac{e^2 - 1}{4e} \approx 0.5876.

The slanted boundaries $x - y = \text{const}$ and $x + y = \text{const}$ became horizontal and vertical lines in $uv$ — that straightening is the whole reason to change variables.

Worked example · WE-3 · non-linear, the chapter's teeth

Find the area of the region bounded by $xy = 1$ , $xy = 2$ , $y = x$ , and $y = 4x$ . The boundaries are products and ratios held constant, so substitute $u = xy$ , $v = y/x$ . Inverting, $x = \sqrt{u/v}$ , $y = \sqrt{uv}$ , and the Jacobian works out (via the shortcut, $\tfrac{\partial(u,v)}{\partial(x,y)} = 2y/x = 2v$ ) to

|J| = \frac{1}{2v}.

The four curves become four straight lines: $1 \le u \le 2$ , $1 \le v \le 4$ . So

\text{Area} = \int_1^4\!\int_1^2 \frac{1}{2v}\,du\,dv = \int_1^4 \frac{1}{2v}\,dv = \tfrac12\ln 4 = \ln 2 \approx 0.6931.

Notice $|J| = \tfrac{1}{2v}$ is not constant — it changes across the region. That is exactly what makes the non-linear case harder than WE-1, and exactly where marks are lost: you cannot pull a varying Jacobian out of the integral.

End of sitting 1 — you can now write down a Jacobian, use the reciprocal shortcut, and straighten a region with both a linear and a non-linear substitution. After the break: the Gaussian, paid in full, then the Jacobian Predictor to drill the one skill that matters, and the exam-shaped practice set.

Applications · the payoff

Cashing the hook: the Gaussian, in full

Now we earn the $\sqrt{\pi}$ . The trick is to compute $I^2$ instead of $I$ , because a product of two single integrals is a double integral over the plane:

Worked example · WE-4 · the polar payoff

Let $I = \displaystyle\int_{-\infty}^{\infty} e^{-x^2}\,dx$ . Writing the second copy in the variable $y$ ,

I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2}\,dx\right) \left(\int_{-\infty}^{\infty} e^{-y^2}\,dy\right) = \iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dA.

Over the whole plane, switch to polar: $x^2 + y^2 = r^2$ , and $dA = r\,dr\,d\theta$ — the Jacobian $|J| = r$ is what makes this work:

I^2 = \int_0^{2\pi}\!\int_0^{\infty} e^{-r^2}\,r\,dr\,d\theta = \int_0^{2\pi}\Big[-\tfrac12 e^{-r^2}\Big]_0^{\infty} d\theta = \int_0^{2\pi}\tfrac12\,d\theta = 2\pi\cdot\tfrac12 = \pi.

So $I^2 = \pi$ , and since $I > 0$ , $\ \mathbf{I = \sqrt{\pi}}$ . The $r$ that made the inner integral elementary — $e^{-r^2}\,r$ integrates instantly, while $e^{-x^2}$ never could — is the Jacobian. It is the same $r$ you were told never to drop in Chapter 10, now revealed as the hero of the whole calculation.

The anchor — why you actually care

That $\sqrt{\pi}$ is not a curiosity. The normal distribution is $\dfrac{1}{\sigma\sqrt{2\pi}}\,e^{-(x-\mu)^2/2\sigma^2}$ , and the $\sqrt{2\pi}$ downstairs is there for exactly one reason: to make the total probability integrate to $1$ . That constant is a rescaled version of the integral you just did. Every time a dataset is modelled as "normally distributed" — heights, measurement errors, exam scores, noise in a signal — the bell curve normalises because $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$ . You have just derived the cornerstone of statistics with a coordinate change.

Interactive playground

The Jacobian Predictor

One rule: commit your prediction of $|J|$ before the area-scale renders. Pick a substitution, watch the $uv$ -grid map to its warped $xy$ -image, then commit what you think the local area-scaling factor is. Only then does the answer light up. Committing — even to a wrong factor — is what installs the reflex; recognising $|J|$ once you are shown it is not the same skill.

The Jacobian Predictor

Substitution:

the u v grid

its warped image in x y

Commit your prediction · |J| =

For polar and the multiplicative map, $|J|$ is an expression, not a number — and equivalent spellings (e.g. $\tfrac12 v^{-1}$ and $\tfrac{1}{2v}$ ) are both accepted. The image cell's area only appears after you commit.

What to notice

Linear maps scale by a constant. The shear halves every cell ( $|J| = \tfrac12$ ); the stretch enlarges by $2 \times 3 = 6$ . The grid stays made of identical parallelograms.
Polar and multiplicative maps scale differently at different places. $|J| = r$ grows as you move out; $|J| = \tfrac{1}{2v}$ shrinks as $v$ grows. The image cells are visibly unequal.
The absolute value is not optional. Commit a negative value and the Predictor stops you: a negative determinant flips orientation, it does not make area negative.
The dropped factor is the classic loss. Predict $1$ for the polar map and you have made the exact Chapter 10 mistake — the $r$ is the Jacobian.

Retrieval gauntlet

Test yourself before moving on

Commit to an answer before reading the feedback. These interleave computation with decision-only items on purpose — knowing which Jacobian a problem wants is half the skill.

Quiz 1 · polar, through the Jacobian

$\displaystyle\iint_R (x^2 + y^2)\,dA$ over the unit disk $x^2 + y^2 \le 1$ equals:

Quiz 2 · the reciprocal trap

Find the area of the $xy$ -region that maps to the rectangle $0 \le u \le 4$ , $1 \le v \le 7$ under $u = x - 2y$ , $v = 3x - y$ .

Predict the next step. You can read

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

straight off

u = x - 2y

v = 3x - y

. Which determinant is that — and what must you do to it before it multiplies

du\,dv

Easy determinant first. $u$ and $v$ are given in terms of $x, y$ , so $\dfrac{\partial(u,v)}{\partial(x,y)}$ is immediate:

\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 1 & -2 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (-2)(3) = 5.

Reciprocate to get the factor that multiplies $du\,dv$ : $\ \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right| = \dfrac{1}{5}$ . The region is a rectangle of area $4 \times 6 = 24$ in $uv$ , so

\text{Area} = \iint_S \frac{1}{5}\,du\,dv = \frac15\cdot 24 = \frac{24}{5}.

Quiz 3 · name the substitution — do not integrate

For each, which substitution makes it cleanest? (i) the disk $x^2+y^2 \le 9$ with integrand $e^{-(x^2+y^2)}$ ; (ii) the region bounded by $xy=1,\ xy=4,\ y=x,\ y=3x$ ; (iii) the parallelogram with sides $x+2y=\text{const}$ , $3x-y=\text{const}$ .

Quiz 4 · the sign that isn't an error

A student uses $u = y$ , $v = x$ , correctly finds $\dfrac{\partial(x,y)}{\partial(u,v)} = -1$ , writes the area element as $-1\,du\,dv$ , and gets a negative area. What is the fix?

Exam translator — six cold problems

Exam shapes

In standard Indian engineering-math papers, and likely in MCC201A unless your class notes differ, change-of-variables wears four costumes — each one you have now seen worked:

"Evaluate using polar coordinates" → the Jacobian is $r$ ; $dA = r\,dr\,d\theta$ (WE-4, Q1). The dropped $r$ is the classic lost mark.
"Evaluate using the transformation $u = \ldots,\ v = \ldots$ " → compute $|J|$ , rewrite $f$ and the region (WE-2). Often a linear map straightening a slanted region.
"Find the area of the region bounded by $\ldots$ " → the curves name the substitution; $\iint |J|\,du\,dv$ (WE-3). Products/ratios ⇒ $u=xy,\,v=y/x$ .
"Evaluate $\int_{-\infty}^{\infty} e^{-x^2}dx$ " / a normal-distribution constant → square it, go polar, $\sqrt{\pi}$ (WE-4).

Worked solutions are complete; nothing points off this page. Do them as listed — interleaved on purpose — not blocked by method. Two of them, PR3 and PR4 (the non-constant and scaled Jacobians), open with a faded bridge: part of the work is set out, the key step left for you to predict.

PR1 · a radial integrand

Shape · disk + radial integrand ⇒ polar; |J| = r

PR1. Evaluate $\displaystyle\iint_R \sqrt{x^2+y^2}\,dA$ over the disk $x^2 + y^2 \le 4$ .

Polar. $\sqrt{x^2+y^2} = r$ , the disk is $0 \le r \le 2$ , $0 \le \theta \le 2\pi$ , and $dA = r\,dr\,d\theta$ :

\int_0^{2\pi}\!\int_0^{2} r\cdot r\,dr\,d\theta = \int_0^{2\pi}\frac{8}{3}\,d\theta = \frac{16\pi}{3}.

The integrand $r$ times the Jacobian $r$ gives $r^2$ — drop the Jacobian and you would integrate $r$ alone and get $\tfrac{8\pi}{...}$ wrong.

PR2 · a linear straightening

Shape · slanted parallel edges ⇒ linear map; |J| = ½

PR2. Evaluate $\displaystyle\iint_R (x+y)\,dA$ , where $R$ is bounded by $x+y=1$ , $x+y=2$ , $x-y=0$ , $x-y=1$ . Use $u = x+y$ , $v = x-y$ .

Set up. $|J| = \tfrac12$ (as in WE-1), the integrand $x+y = u$ , and the region is the rectangle $1 \le u \le 2$ , $0 \le v \le 1$ :

\int_1^{2}\!\int_0^{1} u\cdot\tfrac12\,dv\,du = \tfrac12\int_1^{2} u\,du = \tfrac12\cdot\frac{3}{2} = \frac{3}{4}.

Four slanted lines became four sides of a rectangle — the integral is then trivial.

PR3 · faded bridge · the multiplicative classic

Shape · products and ratios constant ⇒ u = xy, v = y/x; |J| varies

PR3. Find the area of the region bounded by $xy = 1$ , $xy = 3$ , $y = x$ , $y = 2x$ .

Bridge: the curves say $u = xy$ , $v = y/x$ , and the Jacobian is $|J| = \tfrac{1}{2v}$ (not constant). Before revealing, predict the next step: what are the $u$ - and $v$ -limits, and why can't you pull $|J|$ out of the integral?

Limits. $xy = 1,3 \Rightarrow 1 \le u \le 3$ ; $y/x = 1,2 \Rightarrow 1 \le v \le 2$ . Since $|J| = \tfrac{1}{2v}$ depends on $v$ , it stays inside:

\text{Area} = \int_1^{2}\!\int_1^{3} \frac{1}{2v}\,du\,dv = \int_1^{2}\frac{2}{2v}\,dv = \int_1^{2}\frac{1}{v}\,dv = \ln 2.

The inner $u$ -integral contributed the width $3 - 1 = 2$ ; the varying Jacobian turned the outer integral into a logarithm.

PR4 · faded bridge · scaled polar for an ellipse

Shape · an ellipse ⇒ scaled polar; the scale rides into |J|

PR4. Find the area of the ellipse $\dfrac{x^2}{4} + \dfrac{y^2}{9} \le 1$ using $x = 2r\cos\theta$ , $y = 3r\sin\theta$ .

Bridge: this is polar with the axes pre-scaled, so the boundary is just $r = 1$ , $0 \le \theta \le 2\pi$ . Before revealing, predict the next step: compute $|J|$ for $x = 2r\cos\theta$ , $y = 3r\sin\theta$ — how does the $2\times 3$ scaling show up?

Jacobian.

J = \begin{vmatrix} 2\cos\theta & -2r\sin\theta \\ 3\sin\theta & 3r\cos\theta \end{vmatrix} = 6r\cos^2\theta + 6r\sin^2\theta = 6r, \qquad |J| = 6r.

The $6 = 2\times 3$ is the ordinary polar $r$ multiplied by the axis scaling. With $0 \le r \le 1$ :

\text{Area} = \int_0^{2\pi}\!\int_0^{1} 6r\,dr\,d\theta = 6\cdot\tfrac12\cdot 2\pi = 6\pi.

Sanity check: the ellipse $\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}\le 1$ has area $\pi a b = \pi\cdot 2\cdot 3 = 6\pi$ . ✓

PR5 · the probability payoff

Shape · square it, go polar, read off the Gaussian

PR5. Evaluate $\displaystyle\iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dA$ , and hence state $\displaystyle\int_{-\infty}^{\infty} e^{-x^2}\,dx$ .

Polar over the whole plane. $x^2 + y^2 = r^2$ , $dA = r\,dr\,d\theta$ :

\int_0^{2\pi}\!\int_0^{\infty} e^{-r^2}\,r\,dr\,d\theta = \int_0^{2\pi}\tfrac12\,d\theta = \pi.

This double integral is $\left(\int_{-\infty}^{\infty} e^{-x^2}dx\right)^2$ , so $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$ .

PR6 · decision-only · name the move

Shape · no integration — choose the substitution and say why

PR6. For each, name the substitution and why — do not integrate. (a) $\iint (x^2+y^2)^{3/2}\,dA$ over $x^2+y^2 \le 1$ ; (b) the area bounded by $xy = 2,\ xy = 5,\ y = 3x,\ y = 7x$ ; (c) $\iint f\,dA$ over the parallelogram with sides $2x - y = \text{const}$ , $x + y = \text{const}$ .

(a) Polar. Circular region and a radial integrand $(x^2+y^2)^{3/2} = r^3$ — both signals point polar; $|J| = r$ .

(b) Multiplicative. Boundaries are products $xy = \text{const}$ and ratios $y/x = \text{const}$ , so $u = xy,\ v = y/x$ straightens all four; $|J| = \tfrac{1}{2v}$ .

(c) Linear. Slanted straight edges held constant ⇒ $u = 2x - y,\ v = x + y$ ; $|J|$ is a constant you read off the determinant. No circular symmetry, so polar would only make it worse.

Three things to carry out of this chapter

Change of variables re-grids the plane; the area element scales by $|J|$ . $dA = |J|\,du\,dv$ , absolute value always — rewrite the integrand and the region in $u, v$ too.
Polar's $r$ was the Jacobian all along. Same factor, now from one general rule: $x = r\cos\theta,\ y = r\sin\theta \Rightarrow |J| = r$ .
Compute the easy determinant and reciprocate. $\dfrac{\partial(x,y)}{\partial(u,v)} = \dfrac{1}{\partial(u,v)/\partial(x,y)}$ .

Pre-flight checklist — run it before you integrate

\textbf{1.}\ \text{Sketch drawn? Limits read off the sketch, not the statement?}

\textbf{2.}\ \text{Outer limits constant? Inner limits depend only on the outer variable?}

\textbf{3.}\ \text{Polar: is the } r \text{ in } dA = r\,dr\,d\theta \text{ present?}

\textbf{4.}\ \text{Change of variables: is } |J| \text{ present, and is it the } \textit{absolute} \text{ value?}

\textbf{5.}\ \text{Did you use } \tfrac{\partial(x,y)}{\partial(u,v)} \text{ — or reciprocate } \tfrac{\partial(u,v)}{\partial(x,y)} \text{ if that was the easy one?}

The anchor, one more time

Return to the bell curve. The $\sqrt{\pi}$ you derived is why the normal distribution normalises — the curve that sits under effectively every dataset she will model, from lab errors to exam marks. One coordinate change, and the cornerstone of statistics fell out. Re-do PR3 and PR4 from a blank page tomorrow — not re-read, re-do. Computing a Jacobian and keeping its absolute value is muscle memory, built with the pencil, not by recognising a finished solution.

What's next. Chapter 12 takes the same idea up a dimension — triple integrals, where the Jacobian becomes $r$ (cylindrical) and $\rho^2\sin\phi$ (spherical). The headline mark-loser there is the dropped $\sin\phi$ , the three-dimensional cousin of the dropped $r$ .

Same material, another voice

If a different explanation would help, this one maps one-to-one onto this chapter — free, from MIT OpenCourseWare:

Read / watch: MIT 18.02SC, Session 53 — Change of Variables — the general Jacobian with the same $u=x+y/v=x-y$ and $u=xy$ -type examples; Sessions 54–55 cover the polar Jacobian and a worked example if you want more.

✓ Chapter complete. Your progress, and every quiz answer, is saved on this computer — revisit any time.