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The slice is a sector, so area is $\tfrac12\int r^2\,d\theta$ — and the angle range that traces the curve once decides the answer.
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 derive the polar area element from the area of a circular sector and say why it carries both a half and a square, find the angle range that traces a curve exactly once by sketching rather than by algebra, compute the area between two polar curves by subtracting the squares of the radii inside a single integral, use the polar arc length formula and see it as Pythagoras on a step with a radial and a circumferential component, tell $dr/d\theta$ apart from $dy/dx$, and check any polar answer against a region whose area you already know.
Unit 2 built every application out of one move: write what a slice contributes, then integrate. The slices were rectangles, plates, tubes and bands, and the elements were $(f-g)\,dx$, $\pi R^2 dx$, $2\pi rh\,dr$, $2\pi y\,ds$.
In polar coordinates a thin slice is a sector: a wedge with its point at the pole. Its area is not a height times a width, and that single fact changes every formula in the unit. Every application of the integral is the same move: write what one thin slice contributes, then add the slices up. Get the slice right and the integral writes itself; reach for a remembered formula instead and the first unfamiliar region defeats you.
A sector is a wedge of a disc: radius $r$, opening angle $\alpha$, area $\tfrac12 r^{2}\alpha$ because it is the fraction $\alpha/2\pi$ of the whole disc.
To sweep a region is to let the angle run from $\alpha$ to $\beta$, the radius reaching out to the curve at each angle. The region swept is bounded by the curve and by the two rays at the ends.
The range that traces once is the interval of $\theta$ over which no point of the curve is visited twice. It is found by sketching, never by algebra, and it decides the answer — a range that traces twice gives exactly twice the area.
Area swept. For $r = f(\theta) \ge 0$ on $[\alpha, \beta]$, $$A = \frac12\int_\alpha^\beta r^2\,d\theta.$$ The slice is a sector of radius $r$ and angle $d\theta$, whose area is $\tfrac12 r^2 d\theta$.
Between two curves. If $r_{\text{out}} \ge r_{\text{in}} \ge 0$ over the range, $$A = \frac12\int_\alpha^\beta\left(r_{\text{out}}^2 - r_{\text{in}}^2\right)d\theta.$$ The difference of the squares, inside one integral — the same structure as a washer, for the same reason.
Arc length. A polar curve is the parametric curve $x = r\cos\theta$, $y = r\sin\theta$ with parameter $\theta$. Differentiating and simplifying gives $$L = \int_\alpha^\beta\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta.$$ The two terms are the two ways a step can move: $r\,d\theta$ round and $dr$ outwards, at right angles, combined by Pythagoras.
Slope. Also from the parametric formulas: $$\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}.$$ Note that this is not $dr/d\theta$, which measures how fast the curve moves away from the pole and is a different quantity altogether.
The range is the problem. Sketch $r$ against $\theta$ first. One petal of a rose runs between consecutive zeros of $r$; a cardioid needs a full turn; a circle $r = 2a\cos\theta$ is traced once over $[-\pi/2, \pi/2]$ and twice over a full turn.
Another way: picture
A fan of thin wedges opening from the pole, each reaching out to the curve. Their tips are all at the origin and their outer edges are little arcs of different radii. Adding wedge areas is the polar integral; adding rectangle areas would be the Cartesian one, and the two pictures answer the same question with different slices.
Another way: steps
The polar formulas look unfamiliar and are the same ideas.
Area. In Cartesian coordinates the slice is a rectangle of height $f$ and width $dx$. In polar coordinates it is a sector of radius $r$ and angle $d\theta$. Both are "the measure of one slice, integrated"; only the shape of the slice changed, and with it the formula.
Between two curves. A Cartesian region between curves subtracts heights, $\int(f - g)dx$, because the slice is a rectangle whose height is a difference. A polar region between curves subtracts squares, because the slice is an annular sector whose area is a difference of two sector areas. The same distinction separated the area formula from the washer formula in unit 2, and it appears here for the same reason: areas of sectors and of discs go as the square of the radius, and lengths of segments do not.
Length. $\sqrt{r^2 + (r')^2}\,d\theta$ is Pythagoras on a step with two components, exactly as $\sqrt{1 + (y')^2}\,dx$ was. Here the components are $r\,d\theta$ (round) and $dr$ (out); there they were $dx$ (across) and $dy$ (up).
Surfaces. $\int 2\pi\,(\text{distance to the axis})\,ds$, with $ds$ the polar arc element. Lesson 12 unchanged.
So nothing in this lesson is a new principle. What is new is that the slice has a different shape, and that the range of integration has to be found by looking rather than by reading it off the question.
Forgetting the half or the square. $\int r\,d\theta$ is a length times an angle and is not an area; the units say so.
Squaring after integrating. The square belongs to the slice, so it goes inside.
A range that traces the curve twice. $r = 2\cos\theta$ over $[0, 2\pi]$ gives twice the circle's area. Sketch first.
Subtracting two separately computed areas. For a region between curves, subtract the squares inside one integral over the range where the ordering holds; two integrals over different ranges do not combine.
Confusing $dr/d\theta$ with $dy/dx$. The first measures moving away from the pole, the second is the slope in the plane; they are unrelated, and $dr/d\theta = 0$ does not mean a horizontal tangent.
Integrating over a range where $r$ changes sign. The formula assumes a definite outer boundary; split at the zeros of $r$.
Not checking against a known case. A polar area formula that does not give $\pi r^2$ for a full circle is wrong, and that takes one line to test.
There are four polar formulas and they look like four things to memorise, which is how they are usually held and why they get confused with each other. They are all the same two ideas from unit 2: the measure of one slice, integrated, and Pythagoras applied to a short step. Only the shape of the slice has changed, from a rectangle to a sector, and every difference between the Cartesian and polar formulas traces back to that.
What is genuinely new, and what actually costs marks, is the range. In every Cartesian problem the limits were either given or found by solving for crossings. In a polar problem they must be found by looking — at the curve, or at the graph of $r$ against $\theta$. A cardioid needs a full turn; a circle $r = 2a\cos\theta$ needs half of one; one petal of a four-petal rose needs a quarter of one. Get it wrong and the integral is computed perfectly and answers a different question, with nothing in the arithmetic to signal it.
So the habit that matters most in this last lesson is the one it opens with: sketch first. Graph $r$ against $\theta$ on ordinary axes, mark the zeros, note where $r$ goes negative, and read off the range. Then the calculus — which by now is every technique of unit 1, applied to a trigonometric integrand — is the routine part.
And when the answer arrives, check it against a region you already know. A polar area that does not give $\pi r^2$ for a full circle is wrong, and that check costs one line.
$r = \cos 2\theta$. Graph $r$ against $\theta$: it is zero at $\theta = \pm\pi/4$ and peaks at $\theta = 0$. Between those zeros, one petal is traced.
The range comes from the auxiliary graph.
$A = \dfrac12\displaystyle\int_{-\pi/4}^{\pi/4}\cos^{2}2\theta\,d\theta$, and the half-angle identity gives $\cos^2 2\theta = \dfrac{1 + \cos4\theta}{2}$.
Unit 1's identities finish a unit 6 problem.
$= \dfrac14\left[\theta + \dfrac{\sin4\theta}{4}\right]_{-\pi/4}^{\pi/4} = \dfrac{\pi}{8}$. Integrating over $[0, 2\pi]$ instead would give $\pi/2$ — the total of all four petals, a correct answer to a different question.
$r = 1 + \cos\theta$, traced once over $[0, 2\pi]$. Then $\dfrac{dr}{d\theta} = -\sin\theta$.
Both $r$ and its derivative are needed.
$r^2 + (r')^2 = (1+\cos\theta)^2 + \sin^2\theta = 2 + 2\cos\theta$, and the half-angle identity gives $2 + 2\cos\theta = 4\cos^2\tfrac\theta2$.
The square root comes out, which is why this example exists.
$L = \displaystyle\int_0^{2\pi}2\left|\cos\tfrac\theta2\right|d\theta = 8$, splitting at $\theta = \pi$ where the cosine changes sign. The absolute value is not decoration: dropping it gives $0$.
Graph $r$ against $\theta$: it is zero at $\pm\pi/2$ and negative between $\pi/2$ and $3\pi/2$, retracing the same circle. So the range that traces once is $[-\pi/2, \pi/2]$.
The range is the whole difficulty.
$A = \dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}4\cos^{2}\theta\,d\theta = 2\int_{-\pi/2}^{\pi/2}\cos^{2}\theta\,d\theta$.
Square before integrating.
$\displaystyle\int_{-\pi/2}^{\pi/2}\cos^2 = \dfrac{\pi}{2}$, so $A = \pi$ — which is $\pi \cdot 1^2$ for a circle of radius $1$, as the last lesson said this curve was. Using $[0, 2\pi]$ would have given $2\pi$.
The circle $r = 6$ is swept through an angle of $\dfrac{2\pi}{2}$. The area is $k\pi$. Find $k$.
Answer:
Why is polar area $\tfrac12\displaystyle\int r^{2}\,d\theta$ rather than $\displaystyle\int r\,d\theta$?
Put the steps of finding the area enclosed by $r = 3(1 + \cos\theta)$ into the order you do them.
Number the steps in order (write the number in the box):
For the circle $r = 2$ swept through $\dfrac{2\pi}{2}$, fill in the stages.
| Value | |
|---|---|
| The radius | 2 |
| Its square | |
| The area, divided by $\pi$ |
Match each quantity to its polar integral.
| $\tfrac12\displaystyle\int r^{2}\,d\theta$ | $\displaystyle\int\sqrt{r^{2} + \left(\frac{dr}{d\theta}\right)^{2}}\,d\theta$ | $\tfrac12\displaystyle\int\left(r_{\text{out}}^{2} - r_{\text{in}}^{2}\right)d\theta$ | $\displaystyle\int 2\pi r\sin\theta\sqrt{r^{2} + \left(\frac{dr}{d\theta}\right)^{2}}\,d\theta$ | |
|---|---|---|---|---|
| The area swept by $r = f(\theta)$ | ||||
| The length of $r = f(\theta)$ | ||||
| The area between two polar curves | ||||
| The surface swept by $r = f(\theta)$ about the polar axis |
Find the area between the circles $r = 3$ and $r = 7$, over a full turn. It is $k\pi$; find $k$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The circle $r = 7$ is swept through an angle of $\dfrac{2\pi}{5}$. The area is $k\pi$. Find $k$.
Answer:
You can compute areas and lengths for polar curves and choose the range they need. Say in your own words why the polar area between two curves subtracts squares where the Cartesian area between two curves subtracts heights. That is the end of Calculus II: you can integrate what can be integrated, decide what converges, represent a function by a series with an error bound, and do calculus on curves that are not graphs.
10. Your turn: the area inside $r = 2\cos\theta$, step 3
And the check is available.