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Worked example: Area between two polar graphs

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.D (LO)
,
CHA‑5.D.1 (EK)
,
CHA‑5.D.2 (EK)
More practice with the area encloised by polar functions, this time with two graphs.

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  • piceratops seed style avatar for user David Robinson
    At , why does he neglect the 1/2 shown in the previous equation. He jumps from 1/2 (3sintheta)^2 to simply 9 (sintheta)^2. Why would it not be 9/2....
    (17 votes)
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    • mr pants teal style avatar for user ishav.shukla
      So first he sets up two different equations for the two different regions but then he discusses that both the regions have the same area hence he only uses one equation and multiplies it by 2 which cancels out the 1/2 in front and hence he just takes out the 9 as is and as there is no 1/2 out front, it remains as 9 and not 9/2. Hope that helps.
      (30 votes)
  • aqualine seed style avatar for user Lowie Verschelden
    at the exam we have to make the graph by ourselves, then how can we calculate the integration intervals? Like without a graph how can you calculate it's from 0 to pi/4 and fro pi/4 to pi/2?
    great vids btw!
    (8 votes)
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    • duskpin tree style avatar for user Kat
      In my course we were given the following steps to graph a polar function: 1) recognize what kind of graph you are dealing with first. The general forms of polar graphs are good to know. For example, r = asin𝛉 and r = acos𝛉 are circles, r = cos(n𝛉) is a rose curve, r = a + bcos𝛉 where a=b is a cardioid, r = a + bcos𝛉 where a<b is a limaçon, and r^2 = a^2sin(2𝛉) and r^2 = a^2cos(2𝛉) are lemniscates. Knowing what the generic graph looks like will help you make sure that your graph is correct. 2) Set the polar expression equal to zero and solve for theta. This will give you the various "pitch lines." These pitch lines and the coordinate axes will also tell you your limits of integration, depending on what curve you are trying to integrate. For example, the rose curve cos(2𝛉) equals zero when theta is equal to π/4, 3π/4, 5π/4, and 7π/4. If you look at the graph of cos(2𝛉), you will see that each of the petals of the rose curve are nestled between these lines. If this is too much to graph at once for you, then try just graphing the first quadrant and going on from there. 3) set dr/d𝛉 equal to zero. This will give you candidates for where the curve is at a maximum. In the case of cos(2𝛉), it will give you the endpoints of each petal. In some cases you may also need to find where the tangent line is undefined. 4) sketch the pitch lines found in step two. Then plot the points found in step 3. Remember to keep in mind what the generic form of your graph looks like, and then draw your graph! I hope this helps some! :)
      (15 votes)
  • leaf green style avatar for user Gazi Kothiasin
    why is it that for rcos(theta) the radius is 0 at pie, when clearly cos(pie) is -1 and so the radius would have to be -3? where is the graph from when theta equals to pie/2 to 3pie/2?
    (6 votes)
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    • aqualine seed style avatar for user Sobhan.Bihan
      In the polar coordinate system, r denotes the distance of the point from the origin. Having -a for r means going a distance of a in the opposite direction. Suppose that at an angle of pi, r = -3. This means that you have to go 3 in the direction. This is just (in fact exactly) the same as going -3 along the number line in the negative direction ( -(-3) = 3 ). In general, mathematically:
      ( -r , theta ) = ( r , theta - 0.5pi )
      For more information follow the link below:
      https://en.wikibooks.org/wiki/Calculus/Polar_Introduction
      I apologize for the fact that much or all of the literature of the page is in degrees.
      (5 votes)
  • leaf green style avatar for user HernandezJocelyn97
    where does he get the limits pi/4 to pi/2 for the second integral?
    (9 votes)
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    • piceratops ultimate style avatar for user Andrei-Lucian Șerb
      If you look at the rays starting from the center, you can see that they represent various values of the angle theta. You can even split the whole polar plane into even more rays. All of those rays from 0 to pi/2 intersect the shaded region, so it means that theta goes from 0 to pi/2 (a total of pi/2 degrees) for our shaded region. Since we evaluate one integral from 0 to pi/4 (a total of pi/4 degrees), we evaluate the other integral from pi/4 to pi/2 (another total of pi/4 degrees). The total interval has a degree measure of pi/4 + pi/4, so we get our pi/2 total, for the whole shaded region.
      (0 votes)
  • aqualine ultimate style avatar for user Pipob Puthipiroj
    at , why did he choose pi/4?
    (3 votes)
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    • blobby green style avatar for user NJHarlen
      Because the angle pi/4 is when the area stops being bounded by the first circle (r = 3 sin theta) and starts being bounded by the second circle (r = 3 cos theta). With the first integral, he is trying to measure the red area, which is bounded by the first circle (r = 3 sin theta) from angle 0 to pi/4.
      (6 votes)
  • aqualine seedling style avatar for user Juanita Xiong
    At , why are the integrals multiplied by 1/2?
    (2 votes)
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  • leafers ultimate style avatar for user Edward Cerverizzo
    Sal, you forgot to double the answer! In case you forgot, it is because you only solved for one half of the area and it is symmetric.
    (1 vote)
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  • blobby green style avatar for user varun jha
    What would the limits of the integration be if we were to use double integration?
    (2 votes)
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    • leaf blue style avatar for user Stefen
      You can use the symmetry as in the video:
      The integral would look like ∫∫r dr dθ
      The bounds of the outer integral would be 0 to π/4
      The bounds of the inner integral would be 0 to 3cosθ.
      Evaluating the inner integral will give you a result that looks EXACTLY like the integral he writes at in the video where he has 1/2 ∫(3cosθ)^2 dθ, with 0 and π/4 being the limits.
      Hope that helped!
      (5 votes)
  • female robot amelia style avatar for user Anushya Iyer
    Can someone please explain to me how to get the limits for the integral cause this is the one thing that I need help with for finals.
    (2 votes)
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  • male robot hal style avatar for user matthias.estner
    So can you just use the formula, even if the center is not the origin? Made sense for me, in the previous examples, when the center was in the origin, but here the circles are (1.5, 0) and (0, 1.5).

    I mean the formula was derived for a circle section, measured from the origin...
    (1 vote)
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    • marcimus pink style avatar for user Alex Tran
      For integrating polar graphs, the integral should not depend on where the "center" of the graph is, especially the graph is a circle. This is because r doesn't depend on the center, it is just a distance from the origin. Moreover, r is only a function of theta, so the center is not involved.
      (2 votes)

Video transcript

- [Voiceover] We have two polar graphs here, r is equal to 3 sine theta and r is equal to 3 cosine theta and what we want to do is find this area shaded in blue. That's kind of the overlap of these two circles. So I encourage you to pause the video and give it a go. All right so I assume you've tried and what's interesting here is we're clearly bounded by two different polar graphs. And it looks like they intersect right over here. If we eyeball it, it looks like they're intersecting at when theta is equal to pi over four, and we can verify that. Cosine of pi over four is the same thing as sine of pi over four so it is indeed the case that these two things are going to be equal to each other. Their point of intersection happens at theta is equal to pi over four. And if that wasn't as obvious, you'd set these two equal to each other and figure out the thetas where this actually happened, but here it jumps out at you a little more. So this is theta is equal to pi over four. And so the key is to realize is that for theta being between zero and pi over four we're bounded by the red circle, we're bounded by r is equal to 3 sine theta and then as we go from pi over four to pi over two we're bounded by the black circle, we're bounded by r is equal to 3 cosine theta. So we can just break up our area into those two regions. So this first area right over here we already know is going to be one half times the definite integral from zero to pi over four of, what are we bounded by, 3 sine of theta and we're gonna square that thing d theta. That's the orange region and then this, I guess you could say this blue region right over here is going to be one half times the definite integral and now we're going to go from pi over four to pi over two of 3 cosine theta squared d theta. That's this region right over here. Now one thing that might jump out to you is that they're going to be the same area. These two circles they are symmetric around this line. Theta is equal to pi over four. So these are going to be the same area. So one thing that we could do is just solve for one of these and then double it and we will get the total region that we care about. So the total area, and you can verify that for yourself if you like, but I'm just going to say the total area, I'm just going to double this right over here. So the total area if I just double this, just the orange expression, I'm going to get the definite integral from zero to pi over four of... nine, three squared is nine sine squared theta d theta. And you could evaluate this by hand, you could evaluate this by calculator, let's evaluate this analytically. So sine square theta is the same thing as one half times one minus cosine of two theta. That's a trigonometric identity that we've seen a lot in trigonometry class, actually let me just write it up here. So sine square theta is equal to one half times one minus cosine of two theta. So if we replace this with this it's going to be equal to, let's take the one half out, so we're going to get nine halves times the definite integral from zero to pi over four of one minus cosine two theta d theta. And so this is going to be equal to nine halves antiderivative of one is theta and let's see cosine two theta, it's going to be negative sine of two theta, negative sine of two theta over two. Actually let me just, negative one half sine of two theta. And you could do u substitution and do it like this, but this you might be able to do in your head. And you can verify the derivative of sine two theta is going to be two cosine of two theta and then you multiply it times the negative one half, you just get a negative one right over there. And so then we are going to evaluate this at pi over four and at zero. So if you evaluate it at, well luckily if you evaluate this thing at zero this whole thing is going to be zero so we really just have to evaluate it at pi over four. So this is going to be equal to nine halves times pi over four minus one half sine of two times pi over four is pi over two, sine of pi over two. Sine of pi over two we already know is one, so it is really pi over four, so this right over here is just going to be equal to one. So this is going to be nine halves times, we could say pi over four minus one half or we could say pi over four minus two over four. So we could write it like that or we could multiply everything out or we could say this is going to be equal to nine pi minus 18 over eight and we are done.