7.4 Area and Arc Length in Polar Coordinates

[STARTS AT 0:38]

Instructor: Now we want to find the area of one petal of the rose defined by the equation r equals sine-- or equals 3 sine 2 theta. Now we actually looked at this same function in the last section. And so I already know something about its symmetry. But if we didn't know that, we could observe that sine is cyclic, certainly. And it returns to its original position every pi over 2, based on that 2 theta. So the area we're actually trying to find, OK, if I had that function, which that function is a leaf. It has four leaves, it's a rose. And I want to find the area of this one leaf. 

Well, this leaf begins at theta equals zero, and it ends at theta equals pi over 2. So what I want to find, according to that theorem, is the area which is equal to the integral, or one half the integral from alpha to beta. So this is from 0 to pi over 2, of f of theta squared, or r squared. In this case, it's defined in terms of r. So let's go and just say r squared, which is going to be 9 sine squared 2 theta d theta. That will give me the area of that one petal of that rose. 

Now it turns out with polar coordinates, all of those trigonometric intervals we did previously? Those come in very handy, those techniques we had. So those will likely come up again. All right, so we know that sine squared, we can rewrite that in a way. So, I'm going to go ahead and factor the nine out. 9/2 integral from 0 to pi over 2 of sine squared, which we said previously was 1/2, minus cosine of twice our angle. Well in this case, it's actually going to be 4 theta, because we had 2 theta. And we have a 1/2 half there. All right. 

Let's go ahead and actually factor that 1/2 out as well. So this will be 9/4 integral from 0 to pi r 2 of 1 minus cosine 4 theta d theta. All right. Well 9/4, this is going to be theta minus sine of 4 theta divided by 4, evaluated from 0 to pi over 2. Well, that will be sine of 2 pi. So this will become 9/2-- or 9/4 times pi over 2. 

And then evaluating that at 0. It is 0. So we get 9/8 pi, or 9 pi over 8. And that is the area of that one petal of that rose. 

[ENDS AT 4:18]