Area enclosed by cardioid.
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- Is there a video on Cardioids?
I feel like Sal sort of introduced us to this without teaching us about it.(28 votes)
- It is surprisingly hard to find videos on these topics! Some of the best ones I've found on YouTube are by Krista King, PatrickJMT, and Blackpenredpen. Hope this helps!(3 votes)
- I don't get how Sal got alpha and beta.
Can someone explain to me how Sal got those numbers please?(8 votes)
- alpha and beta are the bounds of the integral. This integral has no visible bounds but the bounds can be set at 360 degrees = 2pi rad and 0 degrees = 0 rad since the shape covers the whole circle.(6 votes)
- When solving for Cardioid problems, are we always going to be presented with a graph, or will there be problems where we are only given the r equation, and we have to draw the graph ourselves or imagine the graph.(5 votes)
- How do you know if Beta is 2 pi, 4pi, 6pi, 8pi, 10pi or other multiples of 2pi radians?(3 votes)
- This depends on the specific function, here it makes a full loop at 2pi radians, s if you have beta be greater than 2pi you will be counting the area of a second loop. 4pi would essentially have you take the area of the shape twice, go on and try it.
So the takeaway is to always realize how many radians it takes for a curve to make a full cycle if there is one . If a curve doesn't go back to the start it's a little bit more tricky , but just be aware where it's taking the area of, and that it may double count parts of it.
Let me know if that didn't help(3 votes)
- How is the Beta for the cardioid 2pi?(2 votes)
- We're going from the start of the function(alpha) to the end of it(beta). In this case, the function starts at zero, and ends at 2pi. So, alpha is zero, beta is 2pi.
Hope it helps.(5 votes)
- ok, is trig identities important? do i have to memorize it(3 votes)
- Yes, they are important. Depending on the course they may allow a formula sheet with trig identities but in general its always good to memorize atleast a few important ones or know how to derive them on the spot. cos(x)^2 = (1/2)(1+cos(2x)) is pretty useful to memorize and for taking antiderivatives(1 vote)
- At5:06, isn't the cos^2 identity 1/2(theta+sin(theta)cos(theta))? Why is it different in this case?(2 votes)
- cos²u = ½(1 + cos(2u))
This is one of several Power Reduction Formulas.
- How do you know which trig identity useful? I don't want to memorize them, but I do want to know when and which to re-derive them from simpler identities.(2 votes)
- The full answer is, all the trig identities are useful, and they are not so many. Memorize them.(1 vote)
- is there any other way to integrate the r(0)^2 and solve for the area without using the trig identity besides just plugging the whole thing into the math 9 function into the calculator?(2 votes)
- I mistakenly chose pi/6 and 11pi/6 as my alpha and beta respectively. Did anyone else make the same mistake? And how would one know the graph encompassed an entire circle?(1 vote)
- This is definitely a mistake I've made on several problems. I've found that the best way to be sure that your bounds of integration are correct is to set the expression r(𝛉) equal to zero. This will give you candidates for the bounds of integration. Sometimes I like to do this without looking at the graph first because then I don't have any mental picture of what I think the bounds should be that might mess up my calculations. This is also much better than eyeballing the bounds of integration, which is very tempting for curves like these, but often can result in wrong answers due to very small regions where the function crosses a certain axes, or achieves a certain value that is very difficult to see just by eyeballing it. For this graph, if you were to zoom in very closely (try plotting this graph on demos.com if you would like) you would see that the graph just barely grazes the positive x-axis. Because of this, you must use 0 to 2π as your bounds of integration, otherwise you will cut off this tiny tiny piece on the positive x-axis, and your area will be wrong. I hope this helps some! It's definitely an easy mistake to make, so I think the best advice I can give is try to reason through it mathematically, not graphically, as the graph will sometimes trick you. :)(3 votes)
- So this darker curve in blue is the graph of r is equal to 1 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. Alright, 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 our starting theta to our ending theta, from alpha to beta, of r of theta squared d theta. And so we, essentially, 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? Well, 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 1 minus 1 we're right over there. 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? Maybe 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 zero to two pi. 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 anti-derivative of one, I know how to take the anti-derivitive of negative cosine of theta, but cosine squared theta, this is a little bit... It doesn't jump out at you that you can just do this, 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 learned this in trigonometry class. If you didn't, we'll, you've learned it just now. And that's why this is one of the more useful trigonometric identities if you're finding any type of anti-derivative or if you're integrating anything. And so let's do that. Let's rewrite this right over here as one half times one plus cosine of two theta. And let's see, and maybe we could... Yeah, let's just do it like that. I guess we could, if we wanted... Well, we'll just do it like that. So this is going to be equal to one half, and then we are going to... One half, now let's just start taking anti-derivatives. One half. Now the anti-derivative of one, with respect to theta, is just going to be theta. The anti-derivative of negative two cosine of theta, well that's just going to be negative two sine of theta. Negative two sine theta. You can take the derivative, the derivative of sine is cosine and the negative two, it'll just multiple it times the derivative of sine of theta so it's negative two cosine of theta. And then we're going to have, let's see... So let me distribute this. This is the same thing as one half plus one half cosine of two theta. So let's just assume it's this way. So the anti-derivative of one half, so the anti-derivative of one-half. So I'm really looking at that right over there. It's going to be one half theta. One half theta. And then the anti-derivative of one half cosine of two theta, let's see... The derivative of sine of two theta is two cosine of two theta. So the anti-derivative of this is... The anti-derivative of cosine of two theta, and you can do u-substitution if you like, but you might be able to do this in your head, the anti-derivative of cosine of two theta is going to be one half sine of two theta. And then you have this one half right over here. So this is going to be... Let me show you what I'm finding the anti-derivative of. Of that right over there, of this, and I guess this right over here. So this is going to be plus one fourth sine of two theta. And I encourage you to find the derivative here if that last part was a little bit confusing. The derivative of sine of two theta is two cosine of two theta, two over one fourth is one half, you get to one half cosine of two theta. And we're going to evaluate that at two pi, at two pi, and at zero. So when you evaluate it, one thing that might jump out at you is when you evaluate this at zero, this whole thing, everything, every term here is just going to be zero, so that simplifies things nicely. So we really just have to take one half of... It evaluated at two pi. So this is going to be one half times two pi, two pi, and then sine of two pi is zero, so that's just going to be zero. And then plus one half times two pi, so that's going to be plus pi. And then sign of two times two pi sine of four pi, that's still going to be zero. So this is going to be zero as well, and we are almost done. So this is going to be one half times three pi or three halves, three halves pi is the area, is the area of this region.