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Discs, rings and sectors as polar rectangles, the area element $r\,dr\,d\theta$ and where its extra factor comes from, the integrals this change makes possible, and the regions it makes worse.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to convert a double integral to polar coordinates, write the area element correctly, read the radial and angular limits off a sketch of a disc, ring or sector, and evaluate the result. You will also be able to say where the extra factor of $r$ comes from, and decide when converting to polar coordinates is the wrong move.
Polar coordinates from Calculus II, where a curve was described by $r$ as a function of $\theta$, and double integrals over general regions, where the limits came off a sketch. This lesson puts the two together, and the only genuinely new ingredient is the factor of $r$ that the area element carries.
Polar coordinates locate a point by its distance $r$ from the origin and the angle $\theta$ measured anticlockwise from the positive $x$ axis. A polar rectangle is a region $a \le r \le b$, $\alpha \le \theta \le \beta$ — a disc, a sector, a ring or a piece of one. The area element in polar coordinates is $dA = r\,dr\,d\theta$, and the extra $r$ in it is called the Jacobian factor, a name the change-of-variables lesson will explain.
The substitution is
$$x = r\cos\theta, \qquad y = r\sin\theta, \qquad x^2 + y^2 = r^2,$$
and the integral becomes
$$\int\!\!\int_D f(x,y)\,dA = \int_{\alpha}^{\beta}\!\!\int_{a}^{b} f(r\cos\theta, r\sin\theta)\;r\,dr\,d\theta.$$
Where the extra $r$ comes from. Take a small piece of the region between $r$ and $r + dr$ and between $\theta$ and $\theta + d\theta$. It is very nearly a rectangle, with one side $dr$ in the radial direction and the other an arc. The arc has length $r\,d\theta$, not $d\theta$, because an angle sweeps a longer arc the further out you are. So the area of the small piece is $r\,dr\,d\theta$.
This is not bookkeeping. Forgetting the $r$ makes the area of the unit disc come out as $2\pi$ rather than $\pi$ — the circumference in place of the area, which at least has the decency to be the wrong kind of quantity.
The regions this was made for. A boundary of constant $r$ (a circle about the origin) or constant $\theta$ (a ray from the origin) becomes a constant limit, so discs, rings, sectors and their combinations turn into polar rectangles with constant limits in both variables. In rectangular coordinates every one of them has square roots in its inner limits.
The integrands this was made for. Anything built from $x^2 + y^2$ collapses to a power of $r$. So does $\sqrt{x^2+y^2}$, and so does any function of distance from the origin.
When not to. A boundary that is a straight line not through the origin becomes $r = c/\cos\theta$ or worse, and the region usually has to be split. The test is always the region first and the integrand second.
Another way: picture
Draw the grid polar coordinates impose: circles about the origin and rays out from it. The small cells it makes are not all the same size — near the origin they are slivers and far out they are broad — and the factor $r$ in $dA$ is precisely the record of how the cell at distance $r$ has grown. Rectangular coordinates need no such factor because all their cells are the same.
Another way: steps
To convert a double integral to polar coordinates:
The most famous use of this change is $\int\int_D e^{-(x^2+y^2)}\,dA$ over the disc of radius $R$. In rectangular coordinates the inner integral would be $\int e^{-x^2}\,dx$, which has no elementary antiderivative, so the problem is closed before it begins.
In polar coordinates,
$$\int_0^{2\pi}\!\!\int_0^{R} e^{-r^2}\,r\,dr\,d\theta = 2\pi \cdot \left[-\tfrac12 e^{-r^2}\right]_0^{R} = \pi\left(1 - e^{-R^2}\right).$$
The factor of $r$ that the area element supplied is exactly the factor the substitution $u = r^2$ needs. That is worth pausing on: the $r$ people treat as an annoyance is often the thing that makes the integral possible.
Letting $R \to \infty$ gives $\pi$ for the whole plane, and from that one number the value $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$ follows — the normalising constant of the normal distribution, obtained by going up a dimension and coming back down. A one-dimensional integral was solved by turning it into a two-dimensional one, which is not a move that a first course prepares anybody to expect.
Losing the $r$. By a distance the commonest error in the unit. If an area comes out with the wrong units — a length where an area belongs — this is why.
Doubling the $r$. Writing $\int\int r^2\,dr\,d\theta$ for the area because the integrand was already $r$. The area element contributes exactly one factor of $r$, whatever the integrand happens to be.
Sweeping $\theta$ too far. A half disc is $0 \le \theta \le \pi$; taking $\theta$ to $2\pi$ with $r$ from $0$ to $R$ covers the whole disc and silently doubles the answer.
Allowing negative $r$. In an integral, $r$ runs from $0$ upwards. The curve-sketching convention that lets $r$ be negative has no place here; it traverses regions twice.
Converting because the integrand looks radial. If the region is a square, the new limits will cost more than the integrand saves.
It is the area of the region you are integrating over, and leaving it out does not give a slightly wrong answer to the right question — it gives the exact answer to a different one. $\int\int f\,dr\,d\theta$ is a perfectly well-defined integral; it is simply not an integral over the region in the plane, because it weights a sliver near the origin the same as a broad cell far out.
The habit that prevents the error is to write $dA = r\,dr\,d\theta$ as its own line every single time, before the integrand is touched, rather than trying to remember it at the end. Every change of variables in the rest of this course carries a factor of exactly this kind, and the next two lessons are mostly about what that factor is in general.
Find $\int\int_D \sqrt{9 - x^2 - y^2}\,dA$ over the disc of radius $3$. In polar form the integrand is $\sqrt{9 - r^2}$ and the region is $0 \le r \le 3$, $0 \le \theta \le 2\pi$.
Both the region and the integrand simplify at once.
The inner integral is $\int_0^3 \sqrt{9 - r^2}\;r\,dr$, and the substitution $u = 9 - r^2$, $du = -2r\,dr$ turns it into $\tfrac12\int_0^9 \sqrt{u}\,du = 9$.
The area element's $r$ is what the substitution needed.
Then $\int_0^{2\pi} 9\,d\theta = 18\pi$ — which is half of $\tfrac43\pi \cdot 27$, the volume of a ball of radius $3$, exactly as a hemisphere should be.
A known answer confirms the method.
Find the area between the circles of radius $1$ and $2$ about the origin. In polar coordinates it is $1 \le r \le 2$, $0 \le \theta \le 2\pi$: a polar rectangle.
A hole at the centre costs nothing here.
The area is $\int_0^{2\pi}\!\int_1^{2} r\,dr\,d\theta = 2\pi\left[\tfrac{r^2}{2}\right]_1^{2} = 2\pi \cdot \tfrac32 = 3\pi$.
Two constant limits and one line of work.
And $3\pi = \pi(2^2 - 1^2)$, the difference of the two disc areas. In rectangular coordinates the same region is not type one or type two and has to be cut into four pieces with square roots in every limit.
The coordinate system chose the amount of work.
In polar form $x^2 = r^2\cos^2\theta$, and with the area element the integrand is $r^3\cos^2\theta$, over $0 \le r \le 2$ and $0 \le \theta \le \pi/2$.
Integrand and area element together.
The integral separates: $\left(\int_0^2 r^3\,dr\right)\left(\int_0^{\pi/2}\cos^2\theta\,d\theta\right) = 4 \times \tfrac{\pi}{4}$.
So the answer is $\pi$. The separation worked because the limits were constants in both variables — a polar rectangle behaves exactly like an ordinary rectangle in that respect, which is another reason these regions are the easy ones.
Match each region to its polar description, with $5 < 9$.
| $0 \le r \le 9$, $0 \le \theta \le 2\pi$ | $5 \le r \le 9$, $0 \le \theta \le 2\pi$ | $0 \le r \le 9$, $0 \le \theta \le \pi/2$ | $0 \le r \le 9$, $0 \le \theta \le \pi$ | |
|---|---|---|---|---|
| The disc of radius $9$ about the origin | ||||
| The ring between radii $5$ and $9$ | ||||
| The first-quadrant quarter of that disc | ||||
| The half of that disc above the $x$ axis |
Put the stages of converting $\int\int_D f\,dA$ over the disc of radius $3$ into polar form.
Number the steps in order (write the number in the box):
The area of the disc of radius $7$ is a multiple of $\pi$. What is the multiplier?
Answer:
Over the disc of radius $6$, $\int\int (x^{2} + y^{2})\,dA$ is a multiple of $\pi$. What is the multiplier?
Answer:
Match each region to its polar description, with $5 < 8$.
| $0 \le r \le 8$, $0 \le \theta \le 2\pi$ | $5 \le r \le 8$, $0 \le \theta \le 2\pi$ | $0 \le r \le 8$, $0 \le \theta \le \pi/2$ | $0 \le r \le 8$, $0 \le \theta \le \pi$ | |
|---|---|---|---|---|
| The disc of radius $8$ about the origin | ||||
| The ring between radii $5$ and $8$ | ||||
| The first-quadrant quarter of that disc | ||||
| The half of that disc above the $x$ axis |
For which region would converting to polar coordinates make the work harder rather than easier?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the stages of converting $\int\int_D f\,dA$ over the disc of radius $7$ into polar form.
Number the steps in order (write the number in the box):
You can recognise a region that polar coordinates suit, convert the integrand and the area element, and evaluate the polar integral. Next: the same ideas one dimension higher, integrating over a solid instead of a region.
10. Your turn: $\int\int_D x^2\,dA$ over the quarter disc of radius $2$ in the first quadrant, step 3