7.4 Area and Arc Length in Polar Coordinates

[STARTS AT 4:20]

Instructor: Next, find the area outside the cardioid R equals 2 plus 2 sine theta, and inside the circle r equals 6 sine theta. 

Well, we've actually been specifically told which one is the outer function and which one of the inner functions. So we don't have to worry about that. But we do need to know where these two intersect. So we'll set them equal to each other. All right. 

Well that would be 4 sine theta equals 2 or sine theta equals 1/2 and there are two places where that is true. Sine of pi over 6 and 5 pi over 6. All right. So those were intervals. That is our interval. Those are our end points for our integral, or integrand, here. So from pi over 6 to 5 pi over 6, and we want to find the area of each. 

Now, let's go ahead and just do these two separately. So 1/2 times the integral from pi over 6 to 5 pi over 6 of our outer function, versus our inner function. Well our outer function is going to be 6 sine theta squared, or 36 sine squared theta d theta. no equals sign there We're going to say minus 1/2 integral from pi over six to 5 pi over 6 of our inner function. We want to be outside of that function. So squared 2 plus 2 sine theta squared d theta. 

Let's clean some things up, slightly. It's going to be a 36 for the first, like I said. And we'll multiply those together. So that is 18 integral pi over 6 five pi over 6 sine squared theta d theta. Now this if I-- going to factor in my 1/2 half as I go. That's going to be 4 2 plus 2 sine theta, which we're distributing that. That'd be 4, so this would be minus 2. So, minus 2. Then that would be an 8 sine theta. So then, we will divide that by 2, so minus 4 sine theta. OK. And then we'll have a 4 sine squared, so this would be plus 2 sine squared theta. And then that's going to be the interval of the integral from pi over 6 to 5 pi over 6. OK. 

So we'll work these out separately. Although, I think we might actually be able to combine some things, there. If we took that as-- the second one as three separate integrals. We'd have 18 sine theta minus 2 sine squared theta. So we actually might be able to combine those. I'm not going to worry with that, but I think we could do that. All right. Now for the sine squared, we know that's going to be a 1/2 minus 1/2 cosine 2 theta. So, 1/2 minus 1/2 cosine 2 theta. Integral 18 pi over 6, 5 pi over 6. OK. There's our first minus. We'll go ahead and factor a 2 out of that. And I'm going to replace, what we just had there. So 1 minus 2 sine theta plus 1/2 minus cosine 2 theta minus 1/2 of that. And then we have our d theta. OK. 

Now we have 18 times theta minus. And that will be a 1/2 sine 2 theta. That'd be 1/4 sine 2 theta minus 2. And that'll be evaluated from pi over 6 to 5 pi over 6. And this last one, minus 2 times theta minus 2-- or plus 2. Hold on. Cosine to sine, that'd change the sign. So yes, that'd be plus cosine theta plus 1/2 theta, which we could combine those two, minus 1/4 sine 2 theta, evaluated pi over 6 to 5 pi over 6. And that should be 4 pi if I've done all of my math right there. 

[ENDS AT 10:50]