This deck covers double integrals in polar coordinates and the general change of variables. It builds the area element from a polar rectangle, explains why the extra factor of r must appear, and introduces the Jacobian with polar coordinates as the special case. It targets the classic errors: dropping the r in the area element, using the wrong angle range, leaving the integrand in x and y, and omitting the absolute value of the Jacobian.
Subject: Calculus III · 106 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. Convert between rectangular and polar coordinates fluently.
2. Explain why the area element in polar coordinates carries an extra factor, and state it correctly.
3. Set up and evaluate a double integral over a disk, annulus, sector, or region bounded by a polar curve.
4. Convert a rectangular double integral to polar form.
5. Compute a Jacobian and use the general change-of-variables formula, seeing polar as one special case.
Warm-up
Discussion prompt
Before we open Week 10 - Change of Variables: Polar Coordinates: without looking back, what was the main idea of Week 9 - Double Integrals & Volume, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck builds the double integral from Riemann sums up to volume under a surface. It covers the properties of the integral, Fubini's theorem in either order, the volume between two surfaces as top minus bottom, average value, and setting up Type I and Type II regions. It targets the classic traps: forgetting top minus bottom, blindly swapping the limits over a region that is not a rectangle, and confusing a signed integral with a true volume.
Concept
Rectangular coordinates locate a point by how far right and how far up it sits.
Polar coordinates locate the same point a different way: how far it is from the origin, and in which direction.
\[ (x, y) \quad \longleftrightarrow \quad (r, \theta) \]
Counterexample
Discussion prompt
Rectangular coordinates locate a point by how far right and how far up it sits.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Polar coordinates locate the same point a different way: how far it is from the origin, and in which direction.
Concept
radial coordinate r — The distance from the origin to the point. In integration we take it to be zero or positive.
angular coordinate theta — The angle, measured counterclockwise from the positive x-axis, to the ray through the point.
\[ r \ge 0, \qquad \theta \text{ measured from the positive } x\text{-axis} \]
Intuition
A radar screen does not report a blip as so-many-right and so-many-up. It reports a range and a bearing.
Range is the distance out; bearing is the direction. That is exactly the radial coordinate and the angle.
Whenever a problem is naturally about distance from a center, polar is the friendly language.
Analogy
Discussion prompt
Explain Think of a radar screen by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
A radar screen does not report a blip as so-many-right and so-many-up. It reports a range and a bearing.
Concept
Drop a right triangle from the point to the x-axis. The horizontal leg and vertical leg give the two conversion formulas.
\[ x = r\cos\theta, \qquad y = r\sin\theta \]
Explain it
Discussion prompt
Explain Rectangular from polar to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Drop a right triangle from the point to the x-axis. The horizontal leg and vertical leg give the two conversion formulas.
Concept
Going the other way, the Pythagorean relation gives the distance, and the ratio of the legs gives the angle.
\[ r^2 = x^2 + y^2, \qquad \tan\theta = \frac{y}{x} \]
The single relation that will save you again and again in integrals is the one for the sum of squares.
\[ x^2 + y^2 = r^2 \]
Concept
The tangent relation alone does not pin down the angle, because two opposite directions share the same tangent.
\[ \tan\theta = \frac{y}{x} \text{ leaves two candidates a half-turn apart} \]
Always look at which quadrant the point is in, and choose the angle that actually points there.
Ranking
Put in order
Put the moves of Convert a polar point to rectangular into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Horizontal leg is the distance times the cosine of the angle.
Worked example
Convert the point given in polar form to rectangular coordinates.
\[ (r, \theta) = \left(4, \tfrac{\pi}{6}\right) \]
Apply the x formula
Why: Horizontal leg is the distance times the cosine of the angle.
\[ x = 4\cos\tfrac{\pi}{6} = 4\cdot\tfrac{\sqrt{3}}{2} = 2\sqrt{3} \]
Apply the y formula
Why: Vertical leg is the distance times the sine of the angle.
\[ y = 4\sin\tfrac{\pi}{6} = 4\cdot\tfrac{1}{2} = 2 \]
Verify with the sum of squares
Why: The rectangular point must sit at distance 4 from the origin.
\[ x^2 + y^2 = (2\sqrt{3})^2 + 2^2 = 12 + 4 = 16 = 4^2 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Convert a polar point to rectangular", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The rectangular point must sit at distance 4 from the origin.
Step zero
Discussion prompt
Convert a rectangular point to polar — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find the distance
Answer:
Worked example
Convert the rectangular point to polar coordinates.
\[ (x, y) = (-1, 1) \]
Find the distance
Why: Use the Pythagorean relation for the radial coordinate.
\[ r = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \]
Find the angle, watching the quadrant
Why: The point is up and to the left, so it lives in the second quadrant; the reference angle is a quarter turn's eighth.
\[ \theta = \pi - \tfrac{\pi}{4} = \tfrac{3\pi}{4} \]
Verify by converting back
Why: The polar answer must reproduce the original rectangular point.
\[ \sqrt{2}\cos\tfrac{3\pi}{4} = -1, \quad \sqrt{2}\sin\tfrac{3\pi}{4} = 1 \;\checkmark \]
Concept
Holding the radial coordinate fixed traces a circle centered at the origin.
\[ r = a \;\Rightarrow\; \text{a circle of radius } a \]
Holding the angle fixed traces a ray out from the origin.
\[ \theta = \alpha \;\Rightarrow\; \text{a ray at angle } \alpha \]
Concept
Rectangular graph paper is a grid of horizontal and vertical lines. Polar graph paper is a grid of concentric circles and rays.
Figure (svg): Concentric circles crossed by rays from the origin, forming the polar grid.
Concept
Reach for polar when the region or the integrand has circular symmetry.
Regions that love polar: disks, circles, annuli (rings), sectors, and anything centered on the origin.
Integrands that love polar: any expression built from the sum of squares, because it collapses to a single squared radial term.
\[ x^2 + y^2 = r^2 \]
Intuition
A round region described with straight-line limits fights you: the boundary keeps changing formula as you slide across.
In polar the same round region becomes a plain box: the radial coordinate runs between two numbers and the angle runs between two numbers.
Choosing coordinates that match the shape turns curved limits into constants.
Concept
To integrate, we chop the region into small pieces. In polar the natural small piece is a polar rectangle.
polar rectangle — The little patch caught between two nearby circles and two nearby rays: the radial coordinate runs across a small width, and the angle sweeps a small amount.
Definition probe
Sort into buckets
Every line below is part of the definition of radial coordinate r or of polar rectangle — one or the other, never both. Put each where it belongs.
Picture it
Figure (svg): A curved patch bounded by two arcs at radius r and r plus delta-r and two rays a small angle apart.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
The inner edge is a short arc; the outer edge is a longer arc. That difference is the whole story of the area element.
Concept
Figure (svg): A curved patch bounded by two arcs at radius r and r plus delta-r and two rays a small angle apart.
The inner edge is a short arc; the outer edge is a longer arc. That difference is the whole story of the area element.
Intuition
Sweep a fixed small angle near the origin and you barely move. Sweep the same angle far out and you travel a long way.
So a polar patch is not a true rectangle of side-lengths equal to the two small changes. Its width in the angular direction grows with distance from the origin.
That growth is exactly proportional to the radial coordinate. Hold that thought; it becomes the extra factor.
Concept
The length of a circular arc is the radius times the angle it subtends.
\[ \text{arc length} = r \, \Delta\theta \]
This single fact is why distance from the origin shows up as a multiplier in the area of the patch.
Concept
Treat the little patch as almost-straight: one side is the radial width, the other side is the arc it sweeps.
\[ \Delta A \approx (\text{radial width})\times(\text{arc length}) = \Delta r \cdot (r\,\Delta\theta) \]
Rearranging, the small area is the radial coordinate times the two small changes.
\[ \Delta A \approx r \, \Delta r \, \Delta\theta \]
Concept
Passing to the limit of tiny pieces, the small changes become differentials and we get the area element.
\[ dA = r \, dr \, d\theta \]
the extra factor r — The multiplier that converts a coordinate patch in the r-theta plane into real area in the x-y plane. It is not optional decoration; it is the area of the patch.
Intuition
Rings far from the origin enclose more area per unit of angle than rings near the origin.
The extra factor is the bookkeeper that gives distant rings their fair, larger weight.
Leave it out and you are pretending every ring is as thin as the innermost one. The answer comes out wrong every time.
Estimation
Predict first
Find the area of a disk of radius R by integrating the area element. We already know the answer, so this is a test of the method.
Commit before you compute: what does Area of a disk, the honest way come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against the known formula
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The area of a disk of radius R is famously pi times the radius squared; the method reproduces it exactly.
Worked example
Find the area of a disk of radius R by integrating the area element. We already know the answer, so this is a test of the method.
Set up the integral of 1 over the disk
Why: Integrating the constant 1 over a region gives its area; the disk is a full sweep of the angle and a radial run from center to rim.
\[ A = \int_0^{2\pi}\!\!\int_0^{R} r \, dr \, d\theta \]
Do the inner radial integral
Why: Integrate the extra factor with respect to the radial coordinate, holding the angle fixed.
\[ \int_0^{R} r \, dr = \frac{R^2}{2} \]
Do the outer angular integral
Why: The inner result is constant in the angle, so sweeping a full turn just multiplies by that sweep.
\[ \int_0^{2\pi} \frac{R^2}{2}\, d\theta = 2\pi\cdot\frac{R^2}{2} = \pi R^2 \]
Verify against the known formula
Why: The area of a disk of radius R is famously pi times the radius squared; the method reproduces it exactly.
\[ \pi R^2 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Area of a disk, the honest way", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The area of a disk of radius R is famously pi times the radius squared; the method reproduces it exactly.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A student computes the area of a disk of radius R but writes the coordinate patch as if it were an ordinary rectangle, with no extra factor.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The result matches the circumference, not the area.
Include the extra factor. The coordinate patch is not a plain rectangle; its true area carries the radial coordinate.
Why: The result matches the circumference, not the area. Missing the factor collapses an area into a perimeter-like quantity.
Trap
A student computes the area of a disk of radius R but writes the coordinate patch as if it were an ordinary rectangle, with no extra factor.
\[ A \stackrel{?}{=} \int_0^{2\pi}\!\!\int_0^{R} \, dr \, d\theta = \int_0^{2\pi} R \, d\theta = 2\pi R \]
Notice the answer is a length, not an area
Why: The result matches the circumference, not the area. Missing the factor collapses an area into a perimeter-like quantity.
\[ 2\pi R \ne \pi R^2 \]
Include the extra factor. The coordinate patch is not a plain rectangle; its true area carries the radial coordinate.
\[ A = \int_0^{2\pi}\!\!\int_0^{R} r \, dr \, d\theta = 2\pi\cdot\frac{R^2}{2} = \pi R^2 \]
Check the units of the answer
Why: Now the result scales like a squared length, which is what an area must do. This is the number one error in the whole topic.
\[ dA = r \, dr \, d\theta \;\checkmark \]
Notation
Annotate
From Trap: forgetting the extra factor — read this one piece at a time. What is each part doing?
On: \( A \stackrel{?}{=} \int_0^{2\pi}\!\!\int_0^{R} \, dr \, d\theta = \int_0^{2\pi} R \, d\theta = 2\pi R \)
Concept
For a disk of radius R centered at the origin, the angle sweeps a full turn and the radial coordinate runs from the center out to the rim.
\[ 0 \le \theta \le 2\pi, \qquad 0 \le r \le R \]
Both limits are constants. The curved boundary of the disk has become a plain box in the two polar variables.
Step zero
Discussion prompt
Integrate the sum of squares over a disk — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Convert both the integrand and the area element
Answer:
Worked example
Evaluate the double integral over the disk of radius 3 centered at the origin.
\[ \iint_D (x^2 + y^2) \, dA, \qquad D:\; x^2 + y^2 \le 9 \]
Convert both the integrand and the area element
Why: The sum of squares becomes the squared radial coordinate, and the area element brings its extra factor.
\[ \iint_D (x^2+y^2)\,dA = \int_0^{2\pi}\!\!\int_0^{3} r^2 \cdot r \, dr \, d\theta \]
Combine the powers of the radial coordinate
Why: The squared radial term times the extra factor gives a single cubic term.
\[ = \int_0^{2\pi}\!\!\int_0^{3} r^3 \, dr \, d\theta \]
Do the inner integral
Why: Integrate the cubic in the radial coordinate from center to rim.
\[ \int_0^{3} r^3 \, dr = \frac{r^4}{4}\Big|_0^{3} = \frac{81}{4} \]
Do the outer integral
Why: The inner result is constant in the angle, so a full sweep multiplies by two pi.
\[ \int_0^{2\pi}\frac{81}{4}\,d\theta = 2\pi\cdot\frac{81}{4} = \frac{81\pi}{2} \]
Verify by a rough size check
Why: The integrand ranges from 0 at the center up to 9 at the rim over a disk of area nine pi; a value near forty pi is entirely reasonable, and the exact answer is forty and a half pi.
\[ \frac{81\pi}{2} = 40.5\pi \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Integrate the sum of squares over a disk", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The integrand ranges from 0 at the center up to 9 at the rim over a disk of area nine pi; a value near forty pi is entirely reasonable, and the exact answer is forty and a half pi.
Concept
Switching to polar is a full change of language. Every x and every y in the integrand must also become polar.
\[ x = r\cos\theta, \quad y = r\sin\theta, \quad x^2 + y^2 = r^2 \]
After the switch, nothing in the integral should still mention the old variables.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A student changes the area element to polar but forgets to rewrite the integrand, leaving old variables tangled with new limits.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The limits are in the radial and angular variables, but the integrand still names the old variables.
Translate the integrand at the same time as the element. Replace the sum of squares with the squared radial coordinate.
Why: The limits are in the radial and angular variables, but the integrand still names the old variables. There is nothing consistent to integrate.
Trap
A student changes the area element to polar but forgets to rewrite the integrand, leaving old variables tangled with new limits.
\[ \int_0^{2\pi}\!\!\int_0^{3} (x^2 + y^2)\, r \, dr \, d\theta \]
See why this is stuck
Why: The limits are in the radial and angular variables, but the integrand still names the old variables. There is nothing consistent to integrate.
\[ x, y \text{ have no meaning against } r, \theta \text{ limits} \]
Translate the integrand at the same time as the element. Replace the sum of squares with the squared radial coordinate.
\[ \int_0^{2\pi}\!\!\int_0^{3} r^2 \cdot r \, dr \, d\theta = \frac{81\pi}{2} \]
Check that only polar variables remain
Why: A clean polar integral mentions only the radial coordinate and the angle, never the old ones.
\[ \text{integrand and limits both in } r, \theta \;\checkmark \]
Translation
\( \int_0^{2\pi}\!\!\int_0^{3} r^2 \cdot r \, dr \, d\theta = \frac{81\pi}{2} \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
An annulus is a ring: everything between an inner circle and an outer circle.
The angle still sweeps a full turn, but the radial coordinate now starts at the inner radius, not at zero.
\[ 0 \le \theta \le 2\pi, \qquad a \le r \le b \]
Hypothesis
Predict first
Integrate over a ring is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Convert the integrand and the element
Why: The reciprocal of the sum of squares becomes the reciprocal of the squared radial coordinate, and the element brings its factor.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Evaluate the integral over the annulus between radius 1 and radius 2.
\[ \iint_D \frac{1}{x^2 + y^2} \, dA, \qquad D:\; 1 \le x^2 + y^2 \le 4 \]
Convert the integrand and the element
Why: The reciprocal of the sum of squares becomes the reciprocal of the squared radial coordinate, and the element brings its factor.
\[ \int_0^{2\pi}\!\!\int_1^{2} \frac{1}{r^2}\cdot r \, dr \, d\theta \]
Simplify the radial part
Why: The extra factor cancels one power, leaving the reciprocal of the radial coordinate.
\[ = \int_0^{2\pi}\!\!\int_1^{2} \frac{1}{r}\, dr \, d\theta \]
Do the inner integral
Why: The antiderivative of the reciprocal is the natural logarithm.
\[ \int_1^{2}\frac{1}{r}\,dr = \ln 2 - \ln 1 = \ln 2 \]
Do the outer integral
Why: A full sweep of the angle multiplies by two pi.
\[ \int_0^{2\pi}\ln 2 \, d\theta = 2\pi\ln 2 \]
Verify positivity and scale
Why: The integrand is positive over the ring, so the answer must be positive; two pi times the log of two is about four and a third, a sensible size.
\[ 2\pi\ln 2 \approx 4.36 > 0 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Integrate over a ring", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The integrand is positive over the ring, so the answer must be positive; two pi times the log of two is about four and a third, a sensible size.
Concept
Not every region is a full turn. A half-disk or a wedge only occupies part of the sweep.
The right half of a disk, where the horizontal coordinate is nonnegative, uses angles from a quarter turn below the axis to a quarter turn above.
\[ -\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2} \]
The upper half, where the vertical coordinate is nonnegative, uses angles from zero to a half turn.
\[ 0 \le \theta \le \pi \]
Anomaly
Predict first
A student writes this, and it looks reasonable:
The region is only the right half of a disk of radius 2, but a student sweeps a full turn anyway.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: A full sweep covers the whole disk, not the half that was asked for.
Match the angle range to the actual region. The right half uses only a half-turn of sweep, from a quarter turn below to a quarter turn above the axis.
Why: A full sweep covers the whole disk, not the half that was asked for. The area comes out twice too big.
Trap
The region is only the right half of a disk of radius 2, but a student sweeps a full turn anyway.
\[ \text{Area} \stackrel{?}{=} \int_0^{2\pi}\!\!\int_0^{2} r \, dr \, d\theta = 2\pi\cdot 2 = 4\pi \]
Spot the doubling
Why: A full sweep covers the whole disk, not the half that was asked for. The area comes out twice too big.
\[ 4\pi = 2\times(\text{true half-disk area}) \]
Match the angle range to the actual region. The right half uses only a half-turn of sweep, from a quarter turn below to a quarter turn above the axis.
\[ \text{Area} = \int_{-\pi/2}^{\pi/2}\!\!\int_0^{2} r \, dr \, d\theta = \pi\cdot 2 = 2\pi \]
Check against half of the full disk
Why: A disk of radius two has area four pi, so its right half is two pi. The corrected sweep matches.
\[ 2\pi = \tfrac{1}{2}(4\pi) \;\checkmark \]
Notation
Annotate
From Trap: the wrong angle range — read this one piece at a time. What is each part doing?
On: \( 2\pi = \tfrac{1}{2}(4\pi) \;\checkmark \)
Intuition
The single most reliable habit is to draw the region before writing a single limit.
Ask two questions of the picture: through what angles does the region live, and for each angle, from what inner radius to what outer radius does it run.
The answers to those two questions are the outer and inner limits, in that order.
Concept
Sometimes the outer boundary is itself a polar curve, where the reach depends on the angle.
\[ 0 \le r \le f(\theta), \qquad \alpha \le \theta \le \beta \]
A circle of radius one sitting to the right of the origin, tangent to the vertical axis, has a compact polar equation.
\[ r = 2\cos\theta \]
Ranking
Put in order
Put the moves of Area inside a polar circle into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The radial coordinate stays nonnegative only from a quarter turn below the axis to a quarter turn above; that sweep draws the whole circle.
Worked example
Find the area enclosed by the polar curve given below, a circle of radius one centered at the point one unit to the right of the origin.
\[ r = 2\cos\theta \]
Choose the angle range that traces the curve once
Why: The radial coordinate stays nonnegative only from a quarter turn below the axis to a quarter turn above; that sweep draws the whole circle.
\[ -\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2} \]
Set up the area integral
Why: Area is the integral of the element, with the radial coordinate running from the origin out to the curve.
\[ A = \int_{-\pi/2}^{\pi/2}\!\!\int_0^{2\cos\theta} r \, dr \, d\theta \]
Do the inner integral
Why: Integrating the extra factor gives half the square of the upper limit.
\[ \int_0^{2\cos\theta} r\,dr = \frac{(2\cos\theta)^2}{2} = 2\cos^2\theta \]
Use the power-reduction identity
Why: Rewrite the squared cosine so it can be integrated term by term.
\[ 2\cos^2\theta = 1 + \cos 2\theta \]
Do the outer integral
Why: Integrate over the sweep; the cosine term integrates to a sine that vanishes at both endpoints.
\[ \int_{-\pi/2}^{\pi/2}(1 + \cos 2\theta)\,d\theta = \left[\theta + \tfrac{1}{2}\sin 2\theta\right]_{-\pi/2}^{\pi/2} = \pi \]
Verify against the known circle
Why: The curve is a circle of radius one, whose area is pi times one squared; the integral agrees.
\[ \pi(1)^2 = \pi \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Area inside a polar circle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The curve is a circle of radius one, whose area is pi times one squared; the integral agrees.
Concept
Often an integral arrives already written in rectangular form, and polar is the escape hatch.
The procedure has three moves: recover the region from the rectangular limits, translate the integrand, and replace the area element.
\[ dx\,dy \;\longrightarrow\; r\,dr\,d\theta \]
Intuition
The rectangular limits are a description of the region in disguise. Decode them into a shape before switching.
Inner limits that involve a square root of a difference of squares are the tell-tale sign of a circular boundary.
\[ y = \sqrt{4 - x^2} \;\Longleftrightarrow\; x^2 + y^2 = 4,\; y \ge 0 \]
Once you see the circle, the polar limits almost write themselves.
Estimation
Predict first
Evaluate the rectangular integral over the quarter disk in the first quadrant. The integrand has no elementary antiderivative in the old variables.
Commit before you compute: what does A rectangular integral that only polar can finish come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the answer is a sensible positive number
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The integrand is positive and below one over a region of area pi; the result is a small positive fraction of pi, about seven tenths.
Worked example
Evaluate the rectangular integral over the quarter disk in the first quadrant. The integrand has no elementary antiderivative in the old variables.
\[ \int_0^{2}\!\!\int_0^{\sqrt{4-x^2}} e^{-(x^2+y^2)} \, dy \, dx \]
Identify the region
Why: The horizontal coordinate runs from zero to two and the vertical coordinate from zero up to the circle of radius two; that is the first-quadrant quarter disk.
\[ 0 \le \theta \le \tfrac{\pi}{2}, \qquad 0 \le r \le 2 \]
Translate the integrand and element
Why: The exponent is the sum of squares, which becomes the squared radial coordinate; the element brings its factor.
\[ \int_0^{\pi/2}\!\!\int_0^{2} e^{-r^2}\, r \, dr \, d\theta \]
Do the inner integral by substitution
Why: Let the exponent be the new variable; the extra factor is exactly what its differential needs.
\[ \int_0^{2} e^{-r^2} r\, dr = \left[-\tfrac{1}{2}e^{-r^2}\right]_0^{2} = \tfrac{1}{2}\left(1 - e^{-4}\right) \]
Do the outer integral
Why: The inner result is constant in the angle, so multiply by the quarter-turn sweep.
\[ \int_0^{\pi/2}\tfrac{1}{2}(1-e^{-4})\,d\theta = \frac{\pi}{4}\left(1 - e^{-4}\right) \]
Verify the answer is a sensible positive number
Why: The integrand is positive and below one over a region of area pi; the result is a small positive fraction of pi, about seven tenths.
\[ \frac{\pi}{4}(1 - e^{-4}) \approx 0.771 > 0 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "A rectangular integral that only polar can finish", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The integrand is positive and below one over a region of area pi; the result is a small positive fraction of pi, about seven tenths.
Concept
In the old variables the integrand had no elementary antiderivative, so the inner integral was a dead end.
The extra factor from the area element supplied precisely the piece a substitution needed, turning an impossible integral into a routine one.
\[ e^{-r^2}\, r \, dr \;\text{integrates cleanly, } e^{-x^2}\,dx\;\text{does not} \]
This is the deeper reason polar is so powerful: the factor is not a tax, it is often the key.
Concept
Polar coordinates are one instance of a bigger idea: swap the variables for new ones chosen to simplify the region or the integrand.
Every such swap needs a correction factor for how the swap stretches or shrinks area. For polar that factor was the radial coordinate; in general it is the Jacobian.
Concept
A change of variables is a pair of formulas giving the old variables in terms of the new ones.
\[ x = g(u, v), \qquad y = h(u, v) \]
It carries a region in the new plane over to the region you actually want to integrate in the old plane.
Intuition
Picture a fine square grid in the new plane. The transformation bends and stretches it into a mesh of little curved cells in the old plane.
A unit cell in the new plane does not map to a unit cell in the old plane. It maps to a cell of some other area.
The ratio of the new area to the old area, cell by cell, is the number we must track.
Intuition
In single-variable calculus, a substitution comes with a derivative factor that rescales the differential.
\[ \int f(x)\,dx = \int f(g(u))\, g'(u)\, du \]
In two variables the rescaling factor is not a single derivative but a determinant of four partial derivatives: the Jacobian.
Concept
Jacobian — The determinant of the matrix of partial derivatives of the old variables with respect to the new ones. It measures the local area-scaling of the transformation.
\[ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[6pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} \]
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of radial coordinate r, angular coordinate theta, polar rectangle, the extra factor r, Jacobian as Week 10 - Change of Variables: Polar Coordinates uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Concept
Written out, the determinant is the main-diagonal product minus the off-diagonal product.
\[ \frac{\partial(x,y)}{\partial(u,v)} = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u} \]
Intuition
Its value at a point tells you how much a tiny cell there is magnified or shrunk as it passes through the transformation.
A Jacobian of three means cells come out three times as large; a Jacobian of one-half means they come out half as large.
That is precisely the factor the area element needs so that new-plane area converts to correct old-plane area.
Concept
Area is never negative, but a determinant can be. A transformation that flips orientation produces a negative Jacobian.
So the area element uses the absolute value of the Jacobian, never the signed value.
\[ dA = \left| \frac{\partial(x,y)}{\partial(u,v)} \right| du \, dv \]
Step zero
Discussion prompt
The Jacobian of polar coordinates — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Compute the four partial derivatives
Answer:
Worked example
Confirm that the general machinery reproduces the extra factor we found geometrically. Here the new variables are the radial coordinate and the angle.
\[ x = r\cos\theta, \qquad y = r\sin\theta \]
Compute the four partial derivatives
Why: Differentiate each old variable with respect to the radial coordinate and with respect to the angle.
\[ \frac{\partial x}{\partial r} = \cos\theta,\; \frac{\partial x}{\partial \theta} = -r\sin\theta,\; \frac{\partial y}{\partial r} = \sin\theta,\; \frac{\partial y}{\partial \theta} = r\cos\theta \]
Form the determinant
Why: Main-diagonal product minus off-diagonal product.
\[ \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = \cos\theta\,(r\cos\theta) - (-r\sin\theta)(\sin\theta) \]
Simplify with the Pythagorean identity
Why: The two terms share the radial coordinate and combine into a single squared-trig sum equal to one.
\[ = r\cos^2\theta + r\sin^2\theta = r(\cos^2\theta + \sin^2\theta) = r \]
Verify against the geometric result
Why: The Jacobian is the nonnegative radial coordinate, so its absolute value is itself; this matches the area element built from the polar rectangle.
\[ \left|\tfrac{\partial(x,y)}{\partial(r,\theta)}\right| = r \;\Rightarrow\; dA = r\,dr\,d\theta \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "The Jacobian of polar coordinates", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The Jacobian is the nonnegative radial coordinate, so its absolute value is itself; this matches the area element built from the polar rectangle.
Concept
With the Jacobian in hand, the whole method is a single formula: rewrite the integrand in the new variables and multiply by the absolute value of the Jacobian.
\[ \iint_R f(x,y)\,dA = \iint_S f\big(g(u,v), h(u,v)\big)\,\left|\frac{\partial(x,y)}{\partial(u,v)}\right| du\,dv \]
The region S in the new plane is whatever maps onto the original region R.
Estimation
Predict first
Find the area of the ellipse whose semi-axes are the positive numbers a and b.
Commit before you compute: what does Area of an ellipse by substitution come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the circle case
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. When the two semi-axes are equal to a radius, the formula collapses to the area of a disk, a strong sanity check.
Worked example
Find the area of the ellipse whose semi-axes are the positive numbers a and b.
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1 \]
Choose a stretching substitution
Why: Rescaling each axis turns the ellipse into a plain unit disk in the new variables.
\[ x = a u, \qquad y = b v \;\Rightarrow\; u^2 + v^2 \le 1 \]
Compute the Jacobian
Why: The off-diagonal partials are zero, so the determinant is just the product of the two scale factors.
\[ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} a & 0 \\ 0 & b \end{vmatrix} = ab \]
Set up the transformed integral
Why: Area is the integral of one, times the absolute value of the Jacobian, over the unit disk.
\[ A = \iint_{u^2+v^2\le 1} ab \, du\, dv = ab \iint_{u^2+v^2\le1} du\,dv \]
Use the known area of the unit disk
Why: The remaining integral is the area of a disk of radius one, which is pi.
\[ A = ab\cdot \pi = \pi a b \]
Verify the circle case
Why: When the two semi-axes are equal to a radius, the formula collapses to the area of a disk, a strong sanity check.
\[ a = b = R \;\Rightarrow\; \pi R^2 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Area of an ellipse by substitution", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: When the two semi-axes are equal to a radius, the formula collapses to the area of a disk, a strong sanity check.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A student uses a substitution whose determinant comes out negative, and carries the signed value straight into the area integral.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: An area cannot be negative. The signed determinant only reports orientation, not size.
Take the absolute value of the Jacobian before integrating. The sign is discarded; only the magnitude scales area.
Why: An area cannot be negative. The signed determinant only reports orientation, not size.
Trap
A student uses a substitution whose determinant comes out negative, and carries the signed value straight into the area integral.
\[ \frac{\partial(x,y)}{\partial(u,v)} = -2 \;\Rightarrow\; \text{Area} \stackrel{?}{=} \iint_S (-2)\, du\, dv < 0 \]
See the impossible sign
Why: An area cannot be negative. The signed determinant only reports orientation, not size.
\[ \text{Area} < 0 \text{ is impossible} \]
Take the absolute value of the Jacobian before integrating. The sign is discarded; only the magnitude scales area.
\[ \left|\frac{\partial(x,y)}{\partial(u,v)}\right| = 2 \;\Rightarrow\; \text{Area} = \iint_S 2\, du\, dv > 0 \]
Check that the result is positive
Why: With the absolute value, area is positive as it must be, whatever orientation the transformation had.
\[ \text{Area} > 0 \;\checkmark \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
Everything about polar integration is one line of the general theory.
The transformation is the polar-to-rectangular map, and its Jacobian is the radial coordinate, which is already nonnegative.
\[ \left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| = r \]
So the famous extra factor is nothing but the absolute value of a Jacobian, seen up close.
Explain it
Discussion prompt
Explain Polar is just the special case to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Everything about polar integration is one line of the general theory.
Step zero
Discussion prompt
A linear substitution over a parallelogram — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Name the new variables after the boundaries
Answer:
Worked example
Find the area of the parallelogram bounded by the four lines listed, using a substitution that turns it into a square.
\[ x - y = 0,\; x - y = 2,\; x + y = 0,\; x + y = 2 \]
Name the new variables after the boundaries
Why: Choosing the combinations that appear in the boundary lines makes the region a plain box in the new plane.
\[ u = x - y, \qquad v = x + y \;\Rightarrow\; 0 \le u \le 2,\; 0 \le v \le 2 \]
Solve for the old variables
Why: Invert the linear system to express the old variables in terms of the new ones.
\[ x = \frac{u+v}{2}, \qquad y = \frac{v-u}{2} \]
Compute the Jacobian
Why: Take the four partial derivatives and form the determinant.
\[ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\ -\tfrac{1}{2} & \tfrac{1}{2} \end{vmatrix} = \tfrac{1}{4} + \tfrac{1}{4} = \tfrac{1}{2} \]
Integrate over the square
Why: Area is one integrated with the absolute value of the Jacobian over the two-by-two box.
\[ A = \int_0^{2}\!\!\int_0^{2} \tfrac{1}{2}\, du\, dv = \tfrac{1}{2}\cdot 4 = 2 \]
Verify with the shoelace area
Why: The parallelogram has vertices at the origin and three mapped corners; its direct area agrees with the substitution result.
\[ \text{vertices } (0,0),(1,1),(2,0),(1,-1)\;\Rightarrow\; A = 2 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "A linear substitution over a parallelogram", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The parallelogram has vertices at the origin and three mapped corners; its direct area agrees with the substitution result.
Concept
The average value of a function over a region is its integral divided by the area of the region.
\[ \bar{f} = \frac{1}{\text{Area}(D)} \iint_D f \, dA \]
Polar coordinates make both the integral and the area easy when the region is round.
Analogy
Discussion prompt
Explain Average value over a region by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
The average value of a function over a region is its integral divided by the area of the region.
Ranking
Put in order
Put the moves of Average of the sum of squares over a disk into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. We already found the integral of the sum of squares over this disk in polar coordinates.
Worked example
Find the average value of the sum of squares over the disk of radius 3 centered at the origin.
\[ f(x,y) = x^2 + y^2, \qquad D:\; x^2 + y^2 \le 9 \]
Reuse the integral computed earlier
Why: We already found the integral of the sum of squares over this disk in polar coordinates.
\[ \iint_D (x^2+y^2)\,dA = \frac{81\pi}{2} \]
Divide by the area of the disk
Why: The disk of radius three has area nine pi; the average is the integral over that area.
\[ \bar{f} = \frac{81\pi/2}{9\pi} = \frac{81}{18} = \frac{9}{2} \]
Verify the value is in range
Why: The function runs from zero at the center to nine at the rim, so an average of four and a half sits sensibly in between.
\[ 0 \le \tfrac{9}{2} \le 9 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Average of the sum of squares over a disk", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The function runs from zero at the center to nine at the rim, so an average of four and a half sits sensibly in between.
Pattern
1. Sketch the region and read its polar limits
Why: Find the angle range, then for each angle the inner and outer radial limits.
2. Replace every x and y in the integrand
Why: Use the conversion formulas; the sum of squares becomes the squared radial coordinate.
3. Replace the area element
Why: Swap the rectangular element for the radial coordinate times both differentials. Never drop the extra factor.
4. Integrate inner then outer, and sanity-check
Why: Evaluate the radial integral first, then the angular one, and confirm sign and size make sense.
Pattern
1. Choose a substitution that simplifies the region or integrand
Why: Often the boundary lines or the integrand suggest the new variables.
2. Solve for the old variables and find the new region
Why: You need the old variables in terms of the new ones to differentiate and to describe the transformed region.
3. Compute the Jacobian and take its absolute value
Why: The determinant of partial derivatives is the area-scaling factor; area needs its magnitude.
4. Rewrite the integrand and integrate over the new region
Why: Multiply the transformed integrand by the absolute value of the Jacobian and evaluate.
Real world
Discussion prompt
Outside this lesson: where does Week 10 - Change of Variables: Polar Coordinates actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Recipe: general change of variables is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
Double integrals in polar coordinates and the general change of variables. Builds the area element from a polar rectangle, explains why the extra factor r must appear, and introduces the Jacobian with polar as the special case.
Elimination
Eliminate the wrong options
Which expression is the correct area element in polar coordinates?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: A small polar patch has radial width dr and arc length equal to the radial coordinate times d-theta, so its area is the radial coordinate times both differentials. That single factor of the radial coordinate is the whole point.
Check
Pick the correct area element for a double integral in polar coordinates.
Check your understanding
Which expression is the correct area element in polar coordinates?
Answer: A
Why: A small polar patch has radial width dr and arc length equal to the radial coordinate times d-theta, so its area is the radial coordinate times both differentials. That single factor of the radial coordinate is the whole point.
Concept
Before trusting a polar answer, ask whether the sign and size are believable, and whether symmetry could have predicted anything.
A positive integrand over a real region must give a positive number. An area must scale like a squared length. These quick checks catch most slips.
Counterexample
Discussion prompt
Before trusting a polar answer, ask whether the sign and size are believable, and whether symmetry could have predicted anything.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
A positive integrand over a real region must give a positive number. An area must scale like a squared length. These quick checks catch most slips.
Prediction
Predict first
Which iterated integral correctly expresses the double integral of f over this disk in polar coordinates?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: the angle from 0 to 2 pi, the radial coordinate from 0 to 3, of f times the radial coordinate, dr d-theta
Why: The disk of radius three needs a full angular sweep from zero to two pi, a radial run from zero to three, the integrand converted to polar, and the extra radial factor in the element. Choice A has all four.
Check
A function is integrated over the disk described below. Choose the correct polar setup.
\[ D:\; x^2 + y^2 \le 9 \]
Check your understanding
Which iterated integral correctly expresses the double integral of f over this disk in polar coordinates?
Answer: A
Why: The disk of radius three needs a full angular sweep from zero to two pi, a radial run from zero to three, the integrand converted to polar, and the extra radial factor in the element. Choice A has all four.
Check
Evaluate the iterated polar integral.
\[ \int_0^{2\pi}\!\!\int_0^{1} r \, dr \, d\theta \]
Check your understanding
What is the value of this integral?
Answer: A
Why: The inner integral of the radial coordinate from zero to one is one half. Multiplying by the full angular sweep of two pi gives two pi times one half, which is pi. This is also the area of the unit disk.
Check
A transformation stretches each axis by a different factor. Compute its Jacobian.
\[ x = 3u, \qquad y = 2v \]
Check your understanding
What is the Jacobian of this transformation?
Answer: A
Why: The partial of x with respect to u is three and the partial of y with respect to v is two, while the off-diagonal partials are zero, so the determinant is three times two, which is six.
Elimination
Eliminate the wrong options
Which angle range describes the right half of the disk?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The right half of the disk is where the horizontal coordinate is nonnegative, which the rays cover from a quarter turn below the axis to a quarter turn above it, that is from negative pi over 2 to pi over 2.
Check
A region is the right half of a disk of radius 2, meaning every point has a nonnegative horizontal coordinate. Choose its angle range.
Check your understanding
Which angle range describes the right half of the disk?
Answer: A
Why: The right half of the disk is where the horizontal coordinate is nonnegative, which the rays cover from a quarter turn below the axis to a quarter turn above it, that is from negative pi over 2 to pi over 2.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Recipe: convert a double integral to polar · Recipe: general change of variables · Polar names a point by distance and direction · What r and the angle mean · Think of a radar screen. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You built the polar area element from the geometry of a small patch, saw why the extra factor must be there, and used it to integrate over disks, annuli, sectors, and polar curves.
You also met the general change of variables: pick a transformation, multiply by the absolute value of its Jacobian, and integrate. Polar coordinates are simply the case whose Jacobian is the radial coordinate.
The four errors to guard against: dropping the extra factor, using the wrong angle range, leaving the integrand in the old variables, and forgetting the absolute value on the Jacobian.
| Situation | What to use |
|---|---|
| Region is round / integrand has the sum of squares | Polar coordinates |
| Area element in polar | The radial coordinate times both differentials |
| General substitution | Multiply by the absolute value of the Jacobian |
| Polar as a special case | Its Jacobian equals the radial coordinate |
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