Video: Finding the Area of a Region in Polar Coordinates

Find the area of the region enclosed by one petal of π‘Ÿ = 3 cos (2πœƒ).

06:33

Video Transcript

Find the area of the region enclosed by one petal of π‘Ÿ is equal to three times the cos of two πœƒ.

We’re given a curve defined by a polar equation. And we need to determine the area of the region which is enclosed by one petal of this curve. The first thing we’ll want to do is get a picture of this curve. We’ll want to plot π‘Ÿ is equal to three times the cos of two πœƒ. And we could do this by substituting in values of πœƒ and then plotting individual points or we could use our graphing calculator. Both methods would work, and we should be comfortable with using either method. If we do this, we get a sketch which looks like this.

And the important thing to note here is all of the petals will have equal area, so we can find the area of any of these petals. Now to find the area enclosed by one of these petals, we’re going to first need to recall how we find the area enclosed by a polar curve. The area between the rays πœƒ one and πœƒ two of the polar curve π‘Ÿ, which sometimes you will see written as 𝑓 of πœƒ, is given by the definite integral from πœƒ one to πœƒ two of one-half π‘Ÿ squared with respect to πœƒ. In this case, we already know our value of π‘Ÿ is three times the cos of two πœƒ. We just need to find our values of πœƒ one and πœƒ two.

And to do this, we’re going to need to take a closer look at our sketch. We need to use this sketch to determine values of πœƒ one and πœƒ two, which will encompass one entire petal. There’s a few different ways of doing this. For example, if we had sketched our graph by taking values of πœƒ and substituting them into our expression and then plotting these onto our curve, we would be able to find the direction that our curve is plotted. We would be able to see that each petal starts at π‘Ÿ is equal to zero, changes some values of πœƒ, and then ends when π‘Ÿ is equal to zero.

To help us illustrate this, we’ll add the rays πœƒ one and πœƒ two. We want these to be where π‘Ÿ is equal to zero, since this is the start and end point of one of our petals. And it’s worth reiterating it doesn’t matter which petal we choose. We could’ve chosen any of the petals. We’ll choose the one in the first and second quadrant.

So we know πœƒ one is in the first quadrant and the solution to the equation three times the cos of two πœƒ is equal to zero. Similarly, we know that πœƒ two is the solution to three times the cos of two πœƒ is equal to zero and πœƒ two is in the second quadrant. We can use this information to solve for πœƒ. We just need to solve the equation three times the cos of two πœƒ is equal to zero. We’ll start by dividing through by three, giving us the cos of two πœƒ is equal to zero.

There’s a few different ways of solving this. We’ll do this using a graphical method. And remember, we’re only interested in the solutions in the first and second quadrant. And there’s a couple of choices of how we could sketch this. We could just straight up sketch the curve 𝑦 is equal to the cos of two πœƒ and solve it this way. However, we’ll do this by sketching the curve 𝑦 is equal to the cos of πœƒ. And in doing this, we know the cos of πœ‹ by two is equal to zero and the cos of three πœ‹ by two is equal to zero.

It’s worth pointing out we’re not including any of the other solutions, for example, at two πœ‹ or at negative πœ‹ by two, because these will correlate to solutions in the third and fourth quadrant in our polar diagram. Because when we divide these through by two, we will get solutions in the third and fourth quadrant of our polar diagram. And we’re not interested in these solutions.

And now we can solve for πœƒ one and πœƒ two, either by sketching the cos of two πœƒ from this sketch or by just solving two πœƒ is equal to πœ‹ by two and two πœƒ is equal to three πœ‹ by two. We get πœƒ one is πœ‹ by four and πœƒ two is three πœ‹ by four. And we’re now ready to use our integral method for finding the area enclosed by this region. We use our value of πœƒ one, πœƒ two, and our equation for our polar curve. We get the integral from πœ‹ by four to three πœ‹ by four of one-half times three cos of two πœƒ all squared with respect to πœƒ.

And we can start simplifying. First, we’ll distribute our square over our parentheses. And of course, we can simplify three squared to give us nine. And we’ll write this inside of our first coefficient. And since nine over two is a constant factor, we can take this outside of our integral. This gives us nine over two times the integral from πœ‹ by four to three πœ‹ by four of the cos squared of two πœƒ with respect to πœƒ.

And in fact, there are several different methods we could use to evaluate this integral. The easiest method is to use our double-angle formula. First, we’ll recall the double-angle formula for cosine. It tells us the cos of two π‘₯ is equivalent to two times the cos squared of π‘₯ minus one. Remember, we want to use this to rewrite the cos squared of two πœƒ in a way which we can integrate easily. And to do this, we want the cos squared of π‘₯ in our double-angle formula to be the cos squared of two πœƒ.

So to do this, we’re going to rewrite our double-angle formula with π‘₯ equal to two πœƒ. Doing this and simplifying, we get the cos of four πœƒ is equivalent to two cos squared of two πœƒ minus one. Now, all we need to do is rearrange this to give us an equivalence for the cos squared of two πœƒ. We just add one to both sides of our equivalence and divide through by two. This gives us the cos squared of two πœƒ is equivalent to the cos of four πœƒ plus one all divided by two. And now this expression is much easier to integrate than the cos squared of two πœƒ.

So we’ll use this to rewrite our integrand. We get nine over two times the integral from πœ‹ by four to three πœ‹ by four of the cos of four πœƒ plus one all over two with respect to πœƒ. Let’s now clear some space before we start evaluating this integral. In fact, we’ll do one more piece of simplification. We’ll take the constant factor of one-half outside of our integral, rewriting the coefficient in front of our integral as nine over four.

And after doing all of this, we can see our integrand is in a form which we can integrate easily. We know the integral of the cos of four πœƒ is the sin of four πœƒ over four and the integral of the constant one is equal to πœƒ. So we get the following expression. All we need to do now is evaluate this at the limits of integration. Doing this, we get nine over four times the sin of four times three πœ‹ by four all divided by four plus three πœ‹ by four minus the sin of four times πœ‹ by four over four minus πœ‹ by four.

And in this case, we don’t actually need a calculator to evaluate this expression. For example, both arguments inside of our sine function are integer multiples of πœ‹. And the sin of an integer multiple of πœ‹ is equal to zero. So both of these terms are equal to zero. So inside of our parentheses, we just have three πœ‹ by four minus πœ‹ by four, which we know is equal to πœ‹ by two. Finally, all we need to do is multiply this by nine over four, and we get nine-eighths πœ‹. Therefore, we were able to show the area of the region enclosed by one petal of π‘Ÿ is equal to three times the cos of two πœƒ is equal to nine-eighths times πœ‹.

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