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Area bounded by polar curves

Develop intuition for the area enclosed by polar graph formula.

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  • leaf green style avatar for user alvinthegreatsh
    Isn't it easier to just integrate with triangles?
    (14 votes)
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    • leaf blue style avatar for user Stefen
      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.
      (24 votes)
  • starky ultimate style avatar for user Marko Arezina
    I cannot find sal's lectures on polar cordinates and graphs. Where could I find these topics?
    (7 votes)
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  • blobby green style avatar for user Tim S
    What does the area inside a polar graph represent (kind of like how Cartesian graphs can represent distance, amounts, etc.)
    (6 votes)
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  • blobby green style avatar for user seanernestmurray
    At , Sal writes r(theta). Does he mean f(theta) as r = f(theta) or am I getting mixed up?
    (5 votes)
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  • aqualine tree style avatar for user CodeLoader
    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)
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    • starky ultimate style avatar for user Michele Franzoni
      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)
  • blobby green style avatar for user Stephen Mai
    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)
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    • piceratops ultimate style avatar for user JensOhlmann
      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)
  • ohnoes default style avatar for user Eugene Choi
    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)
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    • winston baby style avatar for user Home Instruction and JuanTutors.com
      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)
  • blobby green style avatar for user ameerthekhan
    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)
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    • female robot ada style avatar for user vbin
      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)
  • blobby green style avatar for user Pramod Eyyunni
    Why can't we apply the same reasoning and conclude that the arc length of a polar curve between theta = alpha and theta = beta equals the integral{a to b} r d{theta}? We usually use the parametric formula for arc length and end up with something different than the above integral. Thanks!
    (2 votes)
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  • blobby green style avatar for user dohafaris98
    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)
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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.