College Calculus: Level II Polar Coordinates

OK, let us try another area problem.0000

We want to find the area inside the graph of r = 8sin(θ) and outside the graph of r = 4.0003

This one is a little tricky because we have not been given limits of integration,0011

And we have been given two different functions.0016

Somehow, we have to figure out what those functions look like.0018

Also, figure out the limits of integration ourselves.0022

r = 4 is probably the easier one, so I will start with that.0027

That is just all points at radius 4 from the origin.0031

That is just a circle.0036

r = 8sin(θ) is a little trickier.0039

If you graph r = 8sin(x), you get an exaggerated sin curve.0042

That goes up to 8 and down to -8.0053

Comes back to 0 at π, and again at 2π.0061

If you graph r = 8 sin(θ),0066

That starts out when θ = 0, it starts out at 0.0072

When θ = π/2, it goes up to 8.0084

That is 8.0092

When θ = π, it comes back to 0.0094

When θ is 3π/2, it is gone to -8.0101

It is just, let me show this in blue, retracing itself again.0110

When θ gets to 2π, it has come back to 0.0116

That is what the graph of r = 8sin(θ) is.0131

Its intercept on the π/2 axis is 8.0140

I want to show those 2 graphs together.0144

Remember r = 4 had a radius of 4.0146

Let me put that graph on there.0150

We are actually looking at two circles here.0156

We want to find the area inside the graph of r = 8sin(θ)0162

Outside the graph of 3=4.0167

Let me try to color that area in, in red.0170

That is the area that we are trying to find.0175

This means we have to figure out where those two graphs intersect.0177

Then integrate from 1 intersection point to the other.0185

To figure out where those two graphs intersect, I am going to set them equal to each other.0193

4 = 8sin(θ).0197

That means 1/2 = sin(θ).0203

There are two angles that have sin = 1/2.0207

That means θ = π/6, and 5π/6.0210

You can kind of see that in the picture.0216

This is π/6.0218

This is 5π/6.0221

What we are going to do is integrate using our area formula.0225

I think what I am going to do is just find 1/2 of this area.0235

I am just going to find that segment.0240

I will color that in black.0243

I will find that area and then multiply it by 2, then integrate it from π/6 to π/2.0245

Now I want to find the outside area - the inside area.0256

Well our formula for area is f(θ)²/2.0262

Here the outside area is 8sin(θ).0267

The outside curve is 8sin(θ),0269

So I am going to find 8sin(θ)²/2, that is the outside area,0272

- the inside area which is just 4²/2.0284

That is what I am going to integrate.0291

That is nice because the 2's cancel right away.0294

So we get the integral from π/6 to π/2 of 8 sin(θ)²,0298

So that is 64sin²(θ) - 4² is 16.0303

Remember the 2's cancel on the outside d(θ).0313

That is the integral of sin²(θ),0319

We are going to use a half-angle formula to deal with that.0323

Remember sin² is 1 - cos(2θ)/2.0324

So, 64sin²(θ) is 32 × 1 - cos(2θ).0330

We still have - 16 dθ.0340

Here we have 32 - 16.0347

I will put those together and get a 16 - 32 cos(2θ), just 16(θ).0352

Now the integral of cos(2θ) is sin(2θ)/2.0362

The integral of 32cos(2θ) is 16sin(2θ).0370

We said we wanted to integrate this from θ = π/6 to θ = π/2.0378

We will plug those bounds in, and we get 16 × π/2 minus,0388

Now, if we plug π/2 into sin(2θ),0397

That is giving us sin(π), and that is just 0, - 16/π/6.0401

Which is 16 × π/6 for now + 16 × sin(2) × π/6.0413

Which is the sin(π/3).0423

We can simplify this.0428

π/2 - π/6 = π/3.0430

If we combine those two terms, we get 16 π/3.0433

Now, the sin(π/3) is sqrt(3)/2.0441

So 16 × sin(π/3) = 8×sqrt(3).0447

We get our answer is 16 π/3 + 8×sqrt(3).0460

This problem was pretty tricky but it was really the geometry that made it tricky and not the calculus.0465

We were given two curves and we were not really told how to handle them.0470

The good thing to do is to start out by graphing the two curves, putting them together on the same graph,0475

Then trying to find these intersection points.0480

We actually found the intersection point algebraically by setting the two equation equal to each other.0486

Once we found the intersection points, those gave us the bounds on the integrals.0493

Then we used this area formula.0498

Remember the area was f(θ)²/2 dθ.0500

Since we had to find the area between two curves, we looked at the area of the outside one - the area of the inside one.0505

Then we worked out the integration and we got our answer.0514

So our last example is another length of the polar curve r = 3+3cos(θ).0000

Again, this is one where we are not given limits of integration.0008

We have to figure it out by ourselves.0011

The way we are going to figure it out is by graphing the thing.0014

So, the way we can graph it is we can plug in some easy values of θ.0028

When θ = 0, cos(θ) = 1 so this is 6.0033

When θ = π/2, cos(θ) has gone down to 0, so we are just talking about r = 3.0043

By the time θ = π, cos(θ) = -1, so that has gone down to 0.0054

At 3π/2, cos(θ) is back up to 0, so r has come back up to 3.0063

At 2π, cos(θ) is back up to 1.0073

So 3 + 3cos(θ) is back up to 6.0084

That is the curve we are looking at.0088

We are trying to find the length of that curve.0091

I think it will be easier if we just find the length of the top part of this curve, the part I am showing in red right now.0096

So, we are actually going to integrate from 0 to π, and then we will multiply our final answer by 2.0100

We are going to do 2 × the integral from 0 to π.0110

Now, remember we have to use the arc length formula, which is this Pythagorean Formula.0120

We take sqrt(f'(θ)² +f(θ)²).0126

f'(θ) is very easy.0139

The first 3 goes away because it is just a constant.0143

The derivative of 3cos(θ) is just -3sin(θ).0147

If we square that out we get 9sin²(θ).0156

f(θ)² is a little bit more messy.0162

We have to square out 3 + 3cos(θ).0164

That is 9 + 3 × 3 × 2 is 18 cos(θ) + 9cos²(θ) dθ.0168

That looks like it is going to be a pretty messy integral,0187

But good things happen here because we have 9sin²(θ) and 9cos²(θ)0190

Those combine just to be 9.0196

We get 2 × integral from 0 to π of square root of,0198

Well the cos² and the sin² give us 9,0206

And we have another 9 right there,0209

So that is just 18 + 18cos(θ) dθ.0212

Now, I can simplify this a little bit if I pull out the sqrt(18), so 2 sqrt(18) × integral from 0 to π.0223

Of 1 + cos(θ) dθ.0236

This is kind of a tricky integral.0242

There is a trick for doing integrals like this and it may not be completely obvious.0246

The trick is to remember the half angle formula.0252

Let me write that over here.0255

cos²(u), remember that that is 1 + cos(2u)/2.0258

If I solve that for cos(u), that says that cos(u) = sqrt(1+cos(2u)/2).0268

That is pretty nice because we have the sqrt of 1 + cos(θ).0281

We almost have this half angle formula set up for us.0286

All we have to do is let the θ be 2u.0290

dθ = 2du.0297

2 sqrt(18), what I would like to do is create this formula exactly.0310

I am going to put a sqrt(2) in here.0322

Then, multiply on a sqrt(2) on the outside.0325

So we get the sqrt(18) × sqrt(2).0330

Now the integral of 0 to π, of 1 + cos,0336

Now θ is 2u so this is 2u/2.0340

dθ is 2 du.0347

Let me simplify this a bit.0355

I am going to combine this 2 and this 2 and get a 4.0357

sqrt(18) × sqrt(2) is sqrt(36).0363

Now we have the integral from,0365

This 0 and π, those were values for θ0374

If θ = π, then u, since θ = 2u, u will be π/2.0379

If θ = 0, then 0 will also be 0.0388

In terms of u, this is the integral from u = 0 to u = π/2.0394

Now, sqrt(1+cos(2u)/2), the whole point of making this substitution was that that would convert into cos(u).0400

This is just cos(u).0409

This 2 I already moved to the outside du.0413

That took what looked like a very difficult integral, and converted it into a very easy one.0419

The sqrt(36) is just 6, so this is 24 × integral of cos(u) is sin(u).0423

We are evaluating this from u=0 to u=π/2.0434

That just gives 24 × sin(π/2) = 1 - sin(0) = 0.0440

Our final answer there is 24.0452

Again this qualifies as probably a pretty tricky polar coordinate problem.0457

What we are given there is just a polar curve.0460

And we were not given anything about any boundaries.0465

We had to first draw the graph of the curve and figure out what we were dealing with.0470

Once we draw the graph of the curve, it is not too hard to find out which part of the curve we want to integrate,0479

And multiply by an appropriate factor to set up our integral.0485

Then we used the formula for arc length, which is this Pythagorean Formula,0492

With sqrt(f'(θ)² + f(θ)²).0500

We plugged that into the integral, simplified it a little,0505

And then we discovered that what looked like a very difficult integral, sqrt(1+cos(θ),0511

But in order to handle that, we remembered this half angle formula, cos²(u) = 1 + cos(2u)/2.0516

We did a little manipulation to get the integral to match the half angle formula,0527

And once we did, it turned into a fairly easy integral in terms of u.0533

Then it was easy to finish and get the entire length of the curve.0540

Hi, this is educator.com and we are here to talk about polar coordinates.0000

The idea of polar coordinates is that we are not going to keep track of things in terms of x's and y's anymore.0006

Instead, we are going to keep track of points in terms of the radius r and the angle θ.0014

Every point now will have coordinates in terms of r and θ.0025

We will talk about functions r = f(θ).0030

There are sort of two places that calculus comes in in polar coordinates.0037

If you have a function, here I am talking about plugging in different values of θ and getting different values of r as your output.0042

There are two things you might be interested in calculating.0054

One is the area, before we calculate under, here it makes more sense to talk about calculating the area inside a curve.0060

The equation we have for that is the integral from, these are values of θ, θ = a to θ = b.0070

Of f(θ²/2).0085

If you are given f(θ), this f(θ) is just whatever the r is, you square that and divide by 2.0090

Then you take the integral with respect to θ.0098

The second thing we are interested in calculating is the arc length.0103

What is the length of that curve?0111

The way you figure that out is you find f(θ²), f'(θ²).0114

Then you use the Pythagorean theorem, you add them, take their square root,0122

And integrate from θ = a to θ = b.0127

Let us try those out using some examples.0135

The first example is the area inside the graph of r = θ, from θ going to 0 to pi/2.0138

If r = θ, then when θ = 0, r is just 0 but as θ increase to pi/2, r gradually increases.0145

We are trying to find that area there.0160

Let us work it out.0165

Our formula for the area is f(θ²/2) dθ0170

In this case, our f(θ) is just θ itself, so this is the integral of θ = 0 to pi/2.0180

Of θ/2 dθ.0192

The integral of θ² is θ³/30194

This whole thing is theta³/3/2, so over 6, evaluated from θ = 0 to θ = pi/2.0196

That is just pi/2³/6.0211

Now, 2³ is 8 and 8 × 6 is 48 so our answer is pi³/48.0220

That represents that area inside that curve.0226

To recap here, what we did was look at the function we were given,0235

R = f(θ), and then we just plugged it into this integral formula f(θ²)/2.0242

Then we worked out the integral.0249

Let us try that out with a slightly harder example.0252

We want to find the area inside one loop of the graph of r = cos(2θ).0255

Perhaps the first thing that makes this example difficult is we have not been told what the boundaries of θ are.0260

We really need to look at a graph to figure this out.0266

What does a graph of cos(2θ) look like.0273

Well if we graph cos(x) to warm up, it starts at 1, it goes down to -1, then it comes back to 1 at 2pi.0276

If we graph y = cos(2x), we get a graph with the same basic shape, but it oscillates twice as fast.0290

If that is pi, and that is 2pi, it does a complete period in the space of pi, and then another complete period by 2pi.0305

If we graph r = cos(2θ), we take that graph and we wrap it around a circle.0320

In the sense that when θ = 0, we start out at 1 at radius 1.0330

By the time θ gets to be pi/4, it has gone to radius 0.0341

This goes down to radius 0.0350

At pi/2, r = -1 so you know that if pi/2 is up here, the radius has gone down to -1.0353

It comes back to 0.0365

Now we are graphing this part of the graph.0368

By the time we get up to pi it has come up to 1 again, so here is pi in this direction.0370

Then between pi and 5pi/4 it goes back down to 0.0379

That is graphing that part of the graph and if we graph this part of the graph going from pi/4 to 3pi/2,0387

It is negative so the graph ends up here and then it spirals back down to 0.0400

And back to 1 again when it comes back to 2pi.0412

We get this interesting 4-leaf clover and we are trying to find the area inside one of those loops.0417

A good way to do it might be to find that area inside there, inside half of one of those loops.0425

That is between θ = 0 and θ = pi/4.0432

Then we will multiply our answer by pi/2.0441

Our area is 2 × the integral from 0 to pi/4.0445

Remember f(θ²)/2 so our f(θ) = cos(2θ) this is cos²(2θ)/2 dθ0454

Our 2's will cancel, that is convenient, so we get the integral from 0 to pi/4.0468

Remember how to integrate cos² of something.0475

You write that as 1+cos(2 × that thing), but the thing is 2θ, so this is actually 1 + cos(4θ).0478

All of that over 2.0484

So this 2 over here was not the 2 above, that cancelled, but this came from the half angle formula.0492

Now we integrate this with respect to θ.0497

I am going to pull the 1/2 outside now, we get 1/2.0500

Now we have to integrate 1 + cos(4θ).0505

The integral of 1 is just θ + the integral of cos(4θ) is just sin(4θ)/4.0509

Then we evaluate that from θ = 0 to θ = pi/4.0525

Then we get 1/2 of pi/4.0540

Plugging in θ = pi/4 + 1/4 sin(4θ).0545

Sin(4θ) is sin(pi) which is just 0.0553

- plug in θ = 0, we get 0 - sin(4×0) is just 0 again.0557

Our final answer is just pi/8.0566

The tricky part there is that we were not given the boundaries of integration.0570

We really had to look at the graph of r = cos(2θ)0580

Then interpret from the graph what useful boundaries of integration we could use to find the area.0584

Once we found the boundaries of integration, we just plugged it into the formula f(θ²)/2.0590

Then we worked out the integral and it was not too bad.0600

We will try some more examples later, this is educator.com.0602

Another example is the length of the polar curve, r = e^5θ as θ goes from 0 to 2pi.0610

Fortunately we have been given the limits of integration.0618

We are going to set up our arc length formula.0623

Which remember is f'(θ²) + f(θ)² dθ.0629

Our f'(θ) is well, r = e^5θ.0640

f'(θ) would be (5e^5θ)² + just f(θ²) so that is (e^5θ)².0650

We want to square root that and integrate it.0665

5 squared is 25, e^5θ)² is e^10θ.0670

Square root that and integrate it.0685

This is sqrt(26e^10θ).0690

I am going to pull the sqrt of 26 all the way out of the integral.0704

Then we have the sqrt(e^10θ) which is just e^5θ.0710

We want to integrate this from θ = 0 to 2pi.0718

Now the integral of e^5θ is just e^5θ/50728

We want to evaluate that from θ = 0 to 2pi.0742

I can write this as sqrt(26/5) ×0750

If you plug 2π into e^5θ you get e^10π.0755

If you plug 0 into e^5θ you get e⁰, which is just 1.0763

That is our answer for the arc length.0770

The key to that problem is recognizing that it is a length problem and then going to the arc length formula.0776

Which is this Pythagorean formula f'(θ)² + f(θ)² square of that and then integrate it.0784

We use the f(θ) that we are given, work it through, and then plug in the limits that we are given.0792