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## Calculus 2

### Unit 5: Lesson 7

Area: polar regions (single curve)

# Area bounded by polar curves

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.D (LO)
,
CHA‑5.D.1 (EK)
,
CHA‑5.D.2 (EK)
Develop intuition for the area enclosed by polar graph formula.

## Want to join the conversation?

• Isn't it easier to just integrate with triangles?
• Well, the pie pieces used are triangle shaped, though they become infinitely thin as the angle of the pie slice approaches 0, which means that the straight opposite side, closer and closer matches the bounding curve.

I cannot stress enough how important a good foundation is in polar coordinates, which can make so many types of problems easier to solve. Once you are grounded in polar coordinates, it will be easier to learn the spherical and cylindrical coordinate systems. Some problems can only be solved using these coordinate systems.

There is nothing you learn in math that is just for the sake of learning. Everything is a foundation for what is to come, including trigonometry and polar coordinates.
• I cannot find sal's lectures on polar cordinates and graphs. Where could I find these topics?
• What does the area inside a polar graph represent (kind of like how Cartesian graphs can represent distance, amounts, etc.)
• At , Sal writes r(theta). Does he mean f(theta) as r = f(theta) or am I getting mixed up?
• Yup he just used both r (theta) and f (theta) as representations of the polar function.
• Do I get it right? We approximate the area with an infinite amount of triangles. Let's consider one of the triangles. The smallest one of the angles is dθ. Call one of the long sides r, then if dθ is getting close to 0, we could call the other long side r as well. The area of the triangle is therefore (1/2)r^2*sin(θ). Since θ is infinitely small, sin(θ) is equivalent to just θ. Then we could integrate (1/2)r^2*θ from θ=a to θ=b.
• You are correct, I reasoned the same way.

It's worth noticing that Sal's explanation exploits the same basic principle, although it's not clearly stated: as θ gets closer and closer to zero the curvature of a generic polar function enclosed by two radii that are θ apart approaches the curvature of a circle, that's why we are allowed to use the same formula.

It's a very verbose explanation and still not that rigorous, i think this is why Sal didn't talk about it, but i find it kinda cool so there you have it :)
• Why isn't it just ∫rdθ. I get the correct derivation but I don't understand why this derivation is wrong. This is my logic: as the angle becomes 0, R becomes a line. Using the integral, R acts like a windshield wiper and "covers" the area underneath the polar figure. Keep in mind that R is not a constant, since R describes the equation of the radius in terms of θ.
• Good question Stephen Mai. Start thinking of integrals in this way. If you see an integral like this ∫ f(x) dx. Interpret it like f(x) is a variable hight and dx is a small distance. Hight times distance is a rectangle . And if you sum (∫ is like a sum of infinity small steps) all rectangles up you get the complete area.

Now to your Integral ∫ r dθ. You calculate the radians times the angle. But the radius r times the angle θ is not an area. It is the arc length s.
r * θ = s this comes from the definition of radiance (rad) the angle unit. What you found was the arc length or circumference of the circle.

To become the area take the integral ∫ ds dr. Because for a small arc length ds times a small distance dr you become a rectangle. Some all rectangles up and you get the area of it. Remember that ds was your first Integral ∫ r dθ. If you understand double integrals you can write it like that ∫ (∫ r dθ) dr.
• Sal, I so far have liked the way you teach things and the way you try to keep it as realistic as possible, but the problem is, I CAN'T find the area of a circle. How am I supposed to 'know' that the area of a circle is [pi*r^2]? They didn't teach me that in school, but maybe you taught here, I don't know. My method for calculating the are is to divide the area to infinite number of triangles, the only problem I have is to calculate the sides that touch the f(theta) curve.
(1 vote)
• From basic geometry going forward, memorizing the formula for 1. the area of the circle, 2. circumference of a circle, 3. area of a rectangle, 4. perimeter of a rectangle, and lastly area of a triangle ,will make going to more complex math easier. Of course one can derive these all but that is like reinventing the wheel every time you want to go on a journey! :)
• At . why is the proportionality fraction theta over two pi? Wasn't the fraction always theta over 360 degrees? It doesn't make sense to me that we would compare degrees of part of the circle with the radians.
(1 vote)
• That fraction actually depends on your units of theta. In most cases in calculus, theta is measured in radians, so that a full circle measures 2 pi, making the correct fraction theta/(2pi). If theta were measured in degrees, then the fraction would be theta/360. Not for nothing, but in pie charts, circle angles are measured in percents, so then the fraction would be theta/100. If you dig down, you've actually learned quite a bit of ways of measuring angles ... percents of circles, percents of right angles, percents of straight angles, whole circles, degrees, radians, etc. In all these cases, the ratio would be the measure of the angle in the particular units divided by the measure of the whole circle.
• How do I know exactly which function to integrate first when asked about the area enclosed between two curves ? In order to get a positive result ? Please help ^_^