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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?(14 votes)
- 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.(21 votes)

- I cannot find sal's lectures on polar cordinates and graphs. Where could I find these topics?(7 votes)
- They are in the PreCalculus course. Here is a link to the first one. Also, there is a search box at the top, if you didn't notice it. :D

https://www.khanacademy.org/math/precalculus/parametric-equations/polar-coor/v/polar-coordinates-1(7 votes)

- What does the area inside a polar graph represent (kind of like how Cartesian graphs can represent distance, amounts, etc.)(6 votes)
- This gives a really good answer in my opinion:

https://answers.yahoo.com/question/index?qid=20110222165203AAcLScg(3 votes)

- At6:22, Sal writes r(theta). Does he mean f(theta) as r = f(theta) or am I getting mixed up?(5 votes)
- Yup he just used both r (theta) and f (theta) as representations of the polar function.(3 votes)

- 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.(5 votes)
- 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 :)(2 votes)

- 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.(3 votes)
- 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**.(3 votes)

- 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! :)(0 votes)

- At3:35. 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.(4 votes)

- 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 ^_^(2 votes)
- If you want to get a positive result, take the integral of the upper function first.(1 vote)

- What exactly is a polar graph, and how is it different from a ordinary graph?(1 vote)
- In any 2-dimensional graph, we indicate a point with two numbers. For the ordinary (Cartesian) graphs, the first number is how far left and right to go, and the other is how far up and down to go.

In a polar graph, ordered pairs look like (r, θ) instead of (x, y). r tells us the distance from the origin, and θ gives the angle from the positive x-axis. So the polar point (1, π/2) tells us to go right 1, then rotate π/2 radians (a quarter circle) around the origin. This lands us on the Cartesian point (0, 1).(1 vote)

## Video transcript

- [Voiceover] We now
have a lot of experience finding the areas under
curves when we're dealing with things in rectangular coordinates. So we saw we took the Riemann sums, a bunch of rectangles,
we took the limit as we had an infinite number of
infinitely thin rectangles and we were able to find the area. But now let's move on
to polar coordinates. And in polar coordinates
I won't say we're finding the area under a curve,
but really in this example right over here we have
a part of the graph of r is equal to f of theta and we've graphed it between theta is equal to alpha and theta is equal to beta. And what I wanna do in
this video is come up with a general expression
for this area in blue. This area that is bounded,
I guess you could say by those angles and the graph
of r is equal to f of theta. And I want you to come
up, or at least attempt to come up with an expression on your own, but I'll give you a
little bit of a hint here. When we did it in rectangular coordinates we divided things into rectangles. Over here rectangles don't
seem as obvious because they're all kind of coming to this point, but what if we could divide things into sectors or I guess we could
say little pie pieces? Someone is doing some
serious drilling downstairs. I don't if it's picking
up on the microphone. But anyway, I will continue. So what would happen if
we could divide this into a whole series of kind of pie pieces and then take the limit as if we had an infinite number of pie pieces? So we want to find the
area of each of these pie pieces and then take the
limit as the pie pieces I guess you could say
become infinitely thin and we have an infinite number of them. And I'll give you one more
hint, for thinking about the area of these pie, I guess you could say the area of these pie wedges. I'll give you another
hint, so if I have a circle I'll do my best attempt at a circle. Luckily the plumbing or
whatever is going on downstairs has stopped for now
allowing me to focus more on the calculus, which is
obviously more important. All right so if I have
a circle, that's my best attempt at a circle, and it's of radius r and let me draw a sector of this circle. It's a sector of a circle, so
that's obviously r as well. And if this angle right
here is theta, what is going to be the area of
this sector right over here? So that's my hint for you,
think about what this area is going to be and we're
assuming theta is in radians. Think about what this area
is going to be and then see if you can extend
that to what we're trying to do here to figure out, somehow I'm giving you a hint again. Using integration, finding
an expression for this area. So I'm assuming you've had a go at it. So first let's think about
this, what's the area of the entire circle,
well we already know that. That's going to be pi r squared, formula for the area of a circle. And then what's going
to be the area of this? Well it's going to be a
fraction of the circle. If this is pi, sorry if this
is theta, if we went two pi radians that would be the
whole circle so this is going to be theta over
two pi of the circle. So times theta over two pi would be the area of this sector right over here. Area of the whole circle
times the proprotion of the circle that we've kind of defined or that the sector is made up of. And so this would give
us, the pis cancel out, it would give us one half
r squared times theta. Now what happens if instead of theta, so let's look at each of these over here. So each of these things that I've drawn, let's focus on just one of these wedges. I will highlight it in orange. So instead of the angle
being theta let's just assume it's a really,
really, really small angle. We'll use a differential
although this is a bit of loosey-goosey mathematics
but the important here is to give you the
conceptual understanding. I could call it a delta
theta and then eventually take the limit as our delta
theta approaches zero. But just for conceptual
purposes when we have a infinitely small or super
small change in theta, so let's call that d theta,
and the radius here or I guess we could say this length right over here. You could view it as the radius of at least the arc right at that point. It's going to be r as a
function of the thetas that we're around right over
here, but we're just going to call that our r right over there. And so what is going to be the
area of this little sector? Well the area of this
little sector is instead of my angle being theta I'm calling my angle d theta, this
little differential. So instead of one half
r squared it's going to be, let me do that in a color you can see. This area is going to be
one half r squared d theta. Notice here the angle
was theta, here the angle was d theta, super, super small angle. Now if I wanted to take
the sum of all of these from theta is equal to alpha
to theta is equal to beta and literally there is an
infinite number of these. This is an infinitely small angle. Well then for the entire
area right over here I could just integrate all of these. So that's going to be the
integral from alpha to beta of one half r
squared d theta where r, of course, is a function of theta. So you could even write it this way, you could write it as
the integral from alpha to beta of one half r of
theta squared d theta. Just to remind ourselves or assuming r is a function of theta in this case.