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# Worked example: Area enclosed by cardioid

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.D (LO)
,
CHA‑5.D.1 (EK)
,
CHA‑5.D.2 (EK)

## Video transcript

so this darker curve in blue is the graph of R is equal to one minus cosine of theta of course we're dealing in polar coordinates here and what I'm interested in is to see if we can figure out the area enclosed by this curve and I encourage you to pause the video and try it on your own all right let's work through it together so we've already seen we've already given ourselves the intuition for the formula that the area enclosed by a polar graph is going to be equal to one half the definite integral from you should say are starting our starting theta to our ending data from alpha to beta of R of theta squared D theta and so we essentially just or we just have to apply this to this function right over here so in this case the area is going to be equal to one half the definite integral now what's our alpha and what's our beta what we're going from theta is equal to zero radians and we're essentially going all the way when theta is equal to zero radians is one minus one where right over here and then we go all the way around to theta is equal to two pi radians notice when we're back at two pi cosine of two pi is one one minus one is zero again so we get back to that point so we're going from theta is equal to zero radians to theta is equal to two pi radians now what's R of theta squared I mean I'll color code this a little bit R of theta squared well it's just going to be one minus cosine of theta one minus cosine theta squared and of course we have our D theta we have our D theta and now we just have to evaluate this integral so once again at any point you feel inspired try to evaluate this so let's do this alright so what I would do so this is going to be equal to one half times the definite integral from 0 to 2pi and let me expand this out this is going to be one minus two cosine theta plus cosine squared theta D theta D theta now I know how to take the antiderivative one I know how to take the antiderivative negative cosine of theta but cosine squared theta this is a little bit it's not it doesn't jump out at you that you can just do is use u substitution or something like that but lucky for us we have our trigonometric identities and so we know we know that cosine squared of theta is just the same thing as one half times one plus cosine of two theta you learn this in trigonometry class if you didn't what you learned it just now and so that's why that well this is one of the more useful trigonometric identities but if you're if you're finding any type of antiderivative if we're integrating anything and so let's do that let's rewrite this right over here as 1/2 times 1 plus cosine of two theta and let's see and maybe we could wait yeah let's just let's just do it like that I guess we could if we want what we'll do we'll just do it like that so this is going to be equal to this is going to be equal to 1/2 and then we are going to 1/2 now let's just start taking anti derivatives 1/2 now the antiderivative of 1 with respect to theta is just going to be theta the antiderivative of negative 2 cosine of theta well that's just going to be negative 2 sine of theta negative 2 sine theta you take the derivative the derivative of sine is cosine and the negative 2 just it'll just multiply it times the derivative of sine of theta so it's negative 2 cosine of theta and then we're going to have let's see actually let me distribute this this is the same thing as 1/2 plus 1/2 cosine of 2 theta so let's just assume well it's this way so the antiderivative of 1/2 so the antiderivative of 1/2 so I'm really looking at that right over there it's going to be 1/2 theta 1/2 theta and then the antiderivative of 1/2 cosine of 2 theta let's see sine the derivative of sine of 2 theta is 2 cosine of 2 theta so this is going to so the antiderivative is 1 the antiderivative of cosine of 2 theta and you can do u substitution if you like but you might be able to do this in your head the antiderivative of cosine of 2 theta is going to be 1/2 sine of 2 theta and then you have this 1/2 right over here so this is going to be let me show you what I'm finding the antiderivative of that right over there of this and I guess this right over here this is going to be plus 1/4 sine of 2 theta and I encourage you to define the derivative here too if you if that last part was a little bit confusing derivative of sine of 2 theta is 2 cosine of 2 theta 2 over 1/4 is 1/2 you get 2 1/2 cosine of 2 theta and we're going to evaluate that at 2 pi at 2 pi and at 0 so when you evaluate it well one thing that might jump out at is when you evaluate this at 0 the single whole thing is everything every term here is just going to be 0 so that simplifies things nicely so we really just have to take 1/2 of it evaluated at 2 pi so this is going to be 1/2 times 2 pi 2 pi and then sine of 2 pi is 0 so that's just going to be 0 and then plus 1/2 times 2 pi so that's going to be plus pi and then sine of 2 times 2 pi sine of 4 pi that's still going to be 0 so this is going to be 0 as well and we are almost done so this is going to be 1/2 times 3 pi or 3 halves 3 halves pi is the area is the area of this region