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A thin polar slice is a sector, so the area element is $\tfrac12 r^2\,d\theta$ and the area is $\tfrac12\int r^2\,d\theta$ — with the range over which the curve is traced exactly once doing more of the work than the formula, and $R^2 - r^2$ for the region between two curves.
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 can find the area a polar curve encloses with $\tfrac12\int r^2\,d\theta$, saying why the element carries both a half and a square, and you can check the formula against a circle. You can decide the range over which a curve is traced exactly once — a full turn, half a turn, or a single petal between two zeros of $r$ — and find the area between two polar curves by subtracting the squares of their radii.
You can read a polar curve and find where it reaches the pole, and in calculus AB you built every area integral the same way: write what one thin slice contributes, then add the slices. That method is all this lesson needs. The only thing that changes is the shape of the slice.
Sector: the slice of a disc between two radii — the polar area element.
Area element: $\tfrac12 r^2\,d\theta$, the area of one thin sector.
Polar area: $\tfrac12\displaystyle\int_\alpha^\beta r^2\,d\theta$.
Traced once: the range of $\theta$ over which the curve is drawn exactly one time — the limits the integral needs.
In Cartesian coordinates a thin slice is a rectangle of height $y$ and width $dx$, so the element is $y\,dx$. In polar coordinates a thin slice is a wedge from the pole — a sector — and a sector is a different shape with a different area.
A sector of radius $r$ opening through an angle $d\theta$ is the fraction $\dfrac{d\theta}{2\pi}$ of a whole disc of area $\pi r^2$, so its area is $\tfrac12 r^2\,d\theta$. Adding the slices:
$$A = \frac12\int_\alpha^\beta r^2\,d\theta.$$
Both the half and the square are needed. Check against a circle of radius $a$: $\tfrac12 a^2 \cdot 2\pi = \pi a^2$, which is right. $\int r\,d\theta$ would give $2\pi a$, a circumference, and $\int r^2 d\theta$ would give twice the area. Ten seconds of checking rules out both.
The limits are the difficulty. $\alpha$ and $\beta$ must be the range over which the curve is traced exactly once. A cardioid needs a full turn; $r = 2a\cos\theta$ is finished in half a turn, so integrating to $2\pi$ doubles the answer; a single petal of a rose runs between two consecutive zeros of $r$, which is why the last lesson spent time on where $r = 0$.
Between two curves, take the outer area minus the inner:
$$A = \frac12\int_\alpha^\beta \left(R^2 - r^2\right)d\theta$$
with the squares subtracting, not the radii, and $\alpha$ and $\beta$ the angles at which the two curves meet.
Another way: picture
Picture a fan of paper opening from a single point. Each leaf is a thin triangle with its apex at the pole, and a triangle of base $r\,d\theta$ and height $r$ has area $\tfrac12 r^2 d\theta$ — the half is the half of base times height you have always had. Cartesian slices are rectangles and have no half; polar slices are wedges and do. That single difference is the whole of the formula.
Another way: steps
| Curve | Traced once over | Area |
|---|---|---|
| $r = a$ | $0$ to $2\pi$ | $\pi a^2$ |
| $r = 2a\cos\theta$ | $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$ | $\pi a^2$ |
| $r = a(1 + \cos\theta)$ | $0$ to $2\pi$ | $\tfrac32\pi a^2$ |
| one petal of $r = a\cos(2\theta)$ | $-\tfrac{\pi}{4}$ to $\tfrac{\pi}{4}$ | $\tfrac{\pi a^2}{8}$ |
The second row is the one to stare at. Integrating $r = 2a\cos\theta$ from $0$ to $2\pi$ gives $2\pi a^2$ — twice the truth — and every line of the working is correct. Nothing in the algebra can tell you; only the geometry can. Sketching the curve before integrating is not a nicety here.
Dropping the half, or the square. $\int r\,d\theta$ is not an area at all and $\int r^2 d\theta$ is twice one. Test any formula against a circle of radius $a$ and the wrong ones fail immediately.
Using $0$ to $2\pi$ out of habit. For a curve finished in half a turn that doubles the answer, and for a single petal it multiplies it by the number of petals. The range is part of the problem, not a default.
Subtracting the radii instead of their squares. The region between two curves is $\tfrac12\int (R^2 - r^2)\,d\theta$. $(R - r)^2$ is a different expression and measures nothing.
$r = 5$ over $0 \le \theta \le 2\pi$: the integrand is the constant $25$.
Square the radius.
$\tfrac12\displaystyle\int_0^{2\pi} 25\,d\theta = \tfrac12 \cdot 25 \cdot 2\pi$.
Constant integrand.
$= 25\pi$, which is $\pi r^2$. The formula reproduces something you already knew, so it has been written down correctly.
$r = 4\cos\theta$ is a circle of radius $2$, drawn completely as $\theta$ goes from $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$.
Half a turn.
$\tfrac12\displaystyle\int_{-\pi/2}^{\pi/2} 16\cos^2\theta\,d\theta = 8 \cdot \tfrac{\pi}{2} \cdot \ldots = 4\pi$, using that the average of $\cos^2$ is $\tfrac12$.
The identity does the work.
$4\pi = \pi \cdot 2^2$, correct. Integrating to $2\pi$ instead would have given $8\pi$ — the circle counted twice.
The curve is traced once as $\theta$ runs from $0$ to $2\pi$, and the squared radius is the constant $9$.
Range first.
$\tfrac12\displaystyle\int_0^{2\pi} 9\,d\theta = \tfrac12 \cdot 9 \cdot 2\pi$.
Half, and squared.
$= 9\pi$, which is $\pi r^2$ with $r = 3$ — the check that tells you the half and the square are both in the right place.
Put the steps of finding the area enclosed by $r = 5(1 + \cos\theta)$ into the order you do them.
Number the steps in order (write the number in the box):
Find the area enclosed by the polar curve $r = 6\sin\theta$, traced once as $\theta$ runs from $0$ to $\pi$. Give the answer as a multiple of $\pi$.
Answer:
Which integral gives the area a polar curve $r = f(\theta)$ encloses between two angles?
Give the area each curve encloses, as a multiple of $\pi$.
| area, as a multiple of pi | |
|---|---|
| $r = 4$ | |
| $r = 8\cos\theta$ | |
| $r = 8(1 + \cos\theta)$ |
Match each curve to the range of $\theta$ over which it is traced exactly once.
| $0$ to $2\pi$ | $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$ | $-\tfrac{\pi}{4}$ to $\tfrac{\pi}{4}$ | $-\tfrac{\pi}{6}$ to $\tfrac{\pi}{6}$ | |
|---|---|---|---|---|
| $r = 3$ | ||||
| $r = 6\cos\theta$ | ||||
| One petal of $r = 4\cos(2\theta)$ | ||||
| One petal of $r = 4\cos(3\theta)$ |
Two circles about the pole have $r = 4$ and $r = 7$. Give the area inside the larger one, then the area of the ring between them — both as multiples of $\pi$.
outer disc = p times pi, ring = q times pi
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of finding the area enclosed by $r = 6(1 + \cos\theta)$ into the order you do them.
Number the steps in order (write the number in the box):
You can find an area swept in polar coordinates. Without looking: why does the polar area formula carry a half and a square, and what goes wrong if you integrate a circle through the pole from zero to a full turn?
9. Your turn: the area enclosed by $r = 3$, step 3