Areas and Arc Length in Polar Coordinates | Free Calculus 2 Lesson (4 College Credits) | UPI Study

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Areas and Arc Length in Polar Coordinates

Calculus 2 · Chapter 9: Parametric Equations & Polar CoordinatesSection 9.6 · lesson 50 of 5410 min · free

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In this lesson

A sector, not a rectangle.

  • Build area from the sector element \frac{1}{2}r^{2}d\theta
  • Choose limits that sweep the region exactly once
  • Subtract to find the area between two polar curves
  • Use the polar arc length formula - and not confuse the two

Transcript

Everything said in the video, section by section (1,548 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 9.6

Areas and Arc Length in Polar Coordinates

In Cartesian coordinates, the area element is a thin rectangle. Height times width. In polar coordinates, that shape is wrong. Because polar coordinates sweep out angles from a central point, the natural building block is a thin circular sector. A wedge. Bounded by two rays separated by a small angle d theta, with radius r equals f of theta. And that single change — sector instead of rectangle — is the whole of this section.

The Element

The Building Block Is a Wedge

Let's derive the element. A full circle of radius r has area pi r squared. A sector spanning an angle d theta, out of the full two pi, takes the fraction d theta over two pi of that area. So d A equals pi r squared, times d theta over two pi. The pi's cancel. Leaving one half r squared d theta. That's the polar area element. And integrating it gives the area theorem. If r equals f of theta is continuous and nonnegative on alpha to beta, with beta minus alpha at most two pi, then the area bounded by the curve and the two rays is one half, the integral of r squared, d theta. Now — the limits must be chosen so the region is swept exactly once. If the curve retraces part of itself over the chosen interval, the integral counts that portion twice. That's the main trap in this section, and we'll see it immediately.

Worked Example 1

Area of One Petal

Find the area enclosed by one petal of the four-petaled rose, r equals two sine two theta. The first task is identifying the interval that traces exactly one petal. A petal begins and ends where the curve returns to the pole. Where r equals zero. Set two sine two theta equal to zero. So sine two theta is zero, meaning two theta is zero or pi. Hence theta is zero, or theta is pi over two. And between those values, sine two theta is positive — so the radius grows and returns to zero exactly once. One petal. Now apply the theorem with alpha zero and beta pi over two. Square the integrand. Squaring the coefficient two gives four. And squaring sine two theta gives sine squared two theta. The one half and the four combine to give two, times the integral of sine squared two theta. But a squared sine can't be integrated directly. So use the power-reduction identity. Sine squared u equals one minus cosine two u, over two. With u equal to two theta — which makes two u equal four theta. The integral becomes the integral of one minus cosine four theta. Which integrates to theta, minus one quarter sine four theta. Evaluate from zero to pi over two. Sine of two pi is zero, so the second term vanishes at both ends. The area is pi over two. And notice what would have happened with careless limits. Integrating from zero to two pi here would sweep all four petals. And the answer would be four times too large. Find where r equals zero first. Always.

Figure 9.6

One Petal, and a Region Between Two Curves

The two pictures show what we're computing. On the left, a single petal of the rose, swept as theta runs from zero to pi over two. The remaining three petals are dotted, to show what we're deliberately excluding. On the right, the setup for the next example. The cardioid r equals two plus two cosine theta, in teal. And the circle r equals three, in red. The region we want lies inside the cardioid and outside the circle. You can see it — that lens-shaped sliver on the right, where the cardioid bulges past the circle. And it's bounded by the two rays where the curves intersect.

Between Two Curves

Subtract the Squares

For a region between two polar curves, subtract the sector areas. Squaring each radius separately. A equals one half, the integral of r outer squared minus r inner squared. And note that's exactly the washer-method structure from section four point two. Subtract the squares. Never square the difference. Let's find the area inside the cardioid two plus two cosine theta, and outside the circle r equals three. Find the intersection rays first. That's always the first step. Set two plus two cosine theta equal to three. So two cosine theta is one, and cosine theta is one half. Which gives theta equals plus or minus pi over three. Between those angles, the cardioid lies outside the circle. So those are the limits. Set up the integral: one half, from minus pi over three to pi over three, of the cardioid squared minus nine. Expand the binomial square. Two plus two cosine theta, squared, is four plus eight cosine theta plus four cosine squared theta. Subtract nine and combine constants. Since four minus nine is minus five. Then apply power reduction to the cosine squared term. Four cosine squared theta becomes two plus two cosine two theta. Combining, the integrand is two cosine two theta, plus eight cosine theta, minus three. Integrate. The antiderivative is sine two theta, plus eight sine theta, minus three theta. Now here's a useful shortcut. Every term is odd in theta. So the bracket evaluates to twice its value at pi over three. At pi over three: sine two theta is root three over two. Eight sine theta is four root three. And minus three theta is minus pi. Doubling and multiplying by one half cancels out, leaving exactly the value at pi over three, doubled and halved. The area is nine root three over two, minus pi. About four point six five three square units.

Arc Length

A Polar Curve Is a Parametric Curve in Disguise

Now arc length in polar form. Section four point four established the arc length element. And section nine point three adapted it to parametric curves. And a polar curve is a parametric curve in disguise. Take theta as the parameter. The conversion equations give x equals f of theta cosine theta, and y equals f of theta sine theta. Differentiate each with respect to theta, using the product rule. Writing r prime for d r d theta. d x d theta is r prime cosine theta, minus r sine theta. d y d theta is r prime sine theta, plus r cosine theta. Now square each and add, expanding both binomials. And here's the elegant part. The cross terms — minus two r r prime sine cosine, and plus two r r prime sine cosine — are exact opposites. They cancel. Grouping what remains: r prime squared, times cosine squared plus sine squared. Plus r squared, times sine squared plus cosine squared. Both parentheses equal one, by the Pythagorean identity. Leaving the clean result: r squared plus r prime squared. So the polar arc length is the integral of root r squared plus d r d theta squared, d theta. Now — a warning worth taking seriously. Unlike the area formula, the arc length integrand carries no factor of one half. And it involves the derivative. Confusing the two formulas is the single most frequent source of error in polar problems. Area has the one half and no derivative. Arc length has the derivative and no one half.

Worked Example 3

Length of a Cardioid

Find the total length of the cardioid r equals two plus two cosine theta. The full curve is traced exactly once as theta runs from zero to two pi. Differentiate the radius. d r d theta is minus two sine theta. Square the radius and the derivative, then add. You get four, plus eight cosine theta, plus four cosine squared theta, plus four sine squared theta. Group the squared trig terms and apply the Pythagorean identity. Four cosine squared plus four sine squared is four. So the sum is four plus eight cosine theta, plus four. Which is eight plus eight cosine theta. Or eight times one plus cosine theta. Now the crucial step. The half-angle identity says one plus cosine theta equals two cosine squared of theta over two. So the expression becomes eight times two cosine squared theta over two. Sixteen cosine squared of theta over two. A perfect square. Which is exactly what allows the radical to be removed. The square root is four times the absolute value of cosine theta over two. Splitting at theta equals pi, where the cosine changes sign, and using symmetry, the total length is sixteen. And notice — that half-angle conversion is what made the whole thing work. Watch for it in every polar arc-length problem. As with parametric arc length, workable problems are engineered to collapse this way.

Key Takeaways

What to Carry Forward

Five takeaways. The polar area element is a sector — one half r squared d theta. Not a rectangle. So area is one half, the integral of r squared. Choosing the limits is the main difficulty. Find where r equals zero, or where two curves intersect, so the region is swept exactly once. Between two curves, subtract the squared radii. Outer squared minus inner squared. Never the squared difference. Arc length is the integral of root r squared plus d r d theta squared. Which comes out clean because the cross terms cancel when a polar curve is treated parametrically. And do not confuse the two formulas. Area has the one half and no derivative. Arc length has the derivative and no one half. That closes Chapter Nine. Next up is Chapter Ten — Vectors and the Geometry of Space.

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FAQ

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About Areas and Arc Length in Polar Coordinates

What is the formula for area in polar coordinates?
The area bounded by r=f(\theta) and the rays \theta=\alpha and \theta=\beta is A=\frac{1}{2}\int_\alpha^\beta r^2\,d\theta. The building block is a thin circular sector, not a rectangle. A sector of angle d\theta is the fraction \frac{d\theta}{2\pi} of a circle's area \pi r^2, so dA=\frac{1}{2}r^2\,d\theta. The formula assumes r is continuous and nonnegative and that \beta-\alpha\le 2\pi.
How do you find the area of one petal of a rose curve?
One petal's area is \frac{1}{2}\int_\alpha^\beta r^2\,d\theta, where \alpha and \beta are consecutive angles at which r=0, so the petal is swept exactly once. For r=2\sin 2\theta, r=0 at \theta=0 and \theta=\pi/2. The area is 2\int_0^{\pi/2}\sin^2 2\theta\,d\theta=\int_0^{\pi/2}(1-\cos 4\theta)\,d\theta=\pi/2. Integrating from 0 to 2\pi instead would count all four petals.
How do you find the area between two polar curves?
Use A=\frac{1}{2}\int_\alpha^\beta (r_{\text{outer}}^2-r_{\text{inner}}^2)\,d\theta, squaring each radius separately rather than squaring their difference. Take the limits from the intersection points. For the region inside r=2+2\cos\theta and outside r=3, setting 2+2\cos\theta=3 gives \theta=\pm\pi/3. The area is \frac{1}{2}\int_{-\pi/3}^{\pi/3}\left[(2+2\cos\theta)^2-9\right]d\theta=\frac{9\sqrt{3}}{2}-\pi\approx 4.653.
What is the difference between the polar arc length and polar area formulas?
Polar arc length is L=\int_\alpha^\beta\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta, which involves the derivative and has no factor of \frac{1}{2}, while area has the \frac{1}{2} and no derivative. Mixing them up is a very common error. For the cardioid r=2+2\cos\theta on [0,2\pi], r^2+(r')^2=16\cos^2(\theta/2). The integrand becomes 4|\cos(\theta/2)|, and the total length is 16.

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