Polar curve area with calculator

What we're going to try to do is use our powers of calculus to find this blue area right over here. What this blue area is, is the area in between successive loops of the graph. The polar graph ( r(\theta) = 3\theta \sin(\theta) ) I'm graphing it in polar coordinates here.

Just to make sure we understand what's going on, this first loop gets traced out when ( \theta ) goes between ( 0 ) and ( \pi ). So this is ( 0 \leq \theta \leq \pi ). You can see that makes sense; when ( \theta = 0 ), this whole thing is ( 0 ). But then as ( \theta ) increases, our ( r ) increases as well. However, as ( \theta ) goes back to approaching ( \pi ), remember ( \sin(\pi) = 0 ), so then this thing will approach zero again.

That makes sense, but then as we go from ( \pi ) to ( 2\pi ), you might expect to see some type of a graph down here. But what you need to realize is when we go from ( \pi ) to ( 2\pi ), the sine of those ( \theta ) values are going to be negative. So what happens is it flips over. Instead of having an ( r ) right over here in the third quadrant, when ( \theta ) is just a little bit more than ( \pi ), since ( r ) would be negative, it flips over and then it ends up right over there.

So between ( \pi ) and ( 2\pi ), we actually trace out this second circle right over here, this outer one. It makes sense that the radius is getting larger because we have this ( 3\theta ) right over here. As ( \theta ) is getting larger and larger in magnitude, we should say the magnitude of our ( r ) is getting larger. It turns out that the ( r ) would have been up here if it was positive, but then it goes negative; it goes out here, but it has a larger and larger magnitude because of this ( 3\theta ).

So that out of the way, let's think about how we find this area. In general, if you're looking at a polar curve and you want to find an area, the general notion is, well, you would go from ( \theta = \alpha ) to ( \theta = \beta ). So these would be your bounds on your angle. What you do—and we've shown this, we've given you the intuition for it in previous videos—is it would be ( \frac{1}{2} \int r(\theta)^2 d\theta ), which is the general formula for finding the area inside of a polar curve.

What we're trying to do here is find the area between two successive loops for this polar curve, where ( \theta ) goes from ( 0 ) to ( 2\pi ). So how do we do that? One way to do that is to find the total area in this larger loop and then subtract from that the area in the smaller loop right over here.

So how do we do that? At any point, if you get inspired, pause the video; it's always good for you to try to work it through yourself and just passively watch me work through it. So let's do it together.

So what's the area of this outer circle? If we can include this right over here, well, that's going to be the integral—remember, the outer circle gets traced out as we go from ( \pi ) to ( 2\pi ). It would be ( \frac{1}{2} \int_{(\pi)}^{(2\pi)} (3\theta \sin(\theta))^2 d\theta ). That's the outer circle. From that, we want to subtract the area of this inner circle. So minus—well, what are the bounds there? Well, we said we trace out that inner circle as we go from ( 0 ) to ( \pi ): ( 0 ) to ( \pi ) gives us ( \frac{1}{2} \int_{(0)}^{(\pi)} (3\theta \sin(\theta))^2 d\theta ).

Now it turns out that these are not trivial integrals to solve on your own, but we have tools. In fact, if you're taking a lot of classes, they expect you to be able to use tools. If you're taking AP Calculus BC, they would expect you to be able to use a graphing calculator. So that's what we're going to do right now in order to calculate this quantity, in order to calculate this area.

Let me get out my TI-84 emulator. So there we have it. And let's see, I want to be able to, I want you to see my keystrokes, so let me see if I can remember that. So the first was the same thing over again, it's just different bounds. The first time—so what you want to do is go to the math functions, math, and then you see right over here this is your definite integral function. So I'm just going to scroll down there; it's a definite integral. This TI-84 Plus CE, I guess, writes it in nice definite integral notation.

Sometimes you'll just see that ( \text{fn int} ( ) open parentheses, and then you write the expression. You say what you're integrating with respect to, and then you give the bounds. But here it's pretty clear. So remember that blue integral right over there? I want to go from ( \pi ). So I'll do second—that's the ( \pi ) button right over here—( \pi ) to ( 2\pi ).

Then I'm going to have ( \frac{1}{2} \times ) and I'm just going to do it with respect to ( x ); it doesn't matter if I do all of the thetas instead of thetas I use ( x )'s. So I'm going to have—I'm just doing that to make it easy to type into the calculator. So this is going to be ( 3x \sin(x) ) and then I'm going to want to square this entire thing. So ( \sin(x) )—whoops, I don't want to go there—( \sin(x) ), let’s see, that doesn't close that parenthesis yet. Okay, so that closes that parenthesis and so I want to square that—so squared.

There you have it. And then I want to do ( dx ), so that's that first blue integral. And then from that, I want to subtract—so I'll get my math function again. I could just click ( 9 ); it does the integral again, and now I'm going to go from ( 0 ) to ( \pi ) of this same thing.

So I'm gonna go from ( 0 ) to ( 2\pi ) and I'm going to try to type in that same thing. I am doing ( \frac{1}{2} \times ) open parentheses ( 3x \sin(x) ) close that, close that—so that's that—and I'm going to square it, square it. And then I have my ( dx ), so then I do ( dx ), and if I typed in everything correctly, I should just be able to press enter and get the value.

I got ( 139 ) point—well, let's just go around to the nearest hundred point ( 5 ) three, so this is approximately ( 139.53 ) square units.