Video: Finding the Area of a Region Lying Inside Two Polar Curves

Find the area of the region that lies inside the polar curve π‘Ÿ = 3 cos πœƒ but outside the polar curve π‘Ÿ = 1 + cos πœƒ.

07:39

Video Transcript

Find the area of the region that lies inside the polar curve π‘Ÿ is equal to three times the cos of πœƒ but outside the polar curve π‘Ÿ is equal to one plus the cos of πœƒ.

We’re given a pair of polar curves, and we’re asked to determine the area of the region which lies inside one curve but outside of the other curve. And the first thing we’re going to want to do whenever we’re asked a question like this is to sketch the region we’re asked to find the area of. And we’re going to want to do this by sketching both of our polar curves. There’s a few options for doing this. We can substitute values of πœƒ into our curves and then plot these onto a graph. Or we can just use a graphing calculator. Either method works. We should be comfortable with using both methods.

We should get a sketch which looks like the following. We want to find the area which lies inside our curve π‘Ÿ is equal to three cos of πœƒ but outside of the curve π‘Ÿ is equal to one plus the cos of πœƒ. And we can see this area on our sketch; it’s the following region. To find this region, we’re going to need to recall the following formula to help us find polar areas. We know the integral from πœƒ one to πœƒ two of one-half times π‘Ÿ squared with respect to πœƒ will give us the area of the polar curve defined by π‘Ÿ between the rays πœƒ one and πœƒ two. And at this point, it can be tricky to determine how we’re going to find the area of this specific region.

The first thing we need to notice is our area is entirely contained within π‘Ÿ is equal to three cos of πœƒ. However, we can’t use this alone. We’re going to need to subtract some area from π‘Ÿ plus one cos of πœƒ. And to help us, the easiest way is usually to find the intersections between our two curves. In this case, we can see that should be two. To find the places where our curves intersect, we just equate our curves. We get three cos of πœƒ should be equal to one plus the cos of πœƒ.

And we can solve this. We’ll start by subtracting the cos of πœƒ from both sides. This gives us two cos of πœƒ is equal to one. Next, we’ll divide both sides of the equation through by two. This gives us the cos of πœƒ should be equal to one-half. There’re several different ways of solving this. We could do this graphically, or we could use the inverse cosine function and the fact that the cosine function is even. However, the cos of πœƒ being equal to one-half is a standard angle that we should know. For πœƒ between zero and two πœ‹, this will only happen when πœƒ is πœ‹ by three or when πœƒ is equal to five πœ‹ by three. This means we can mark these values of πœƒ onto our diagram. And this gives us the following two rays which pass through the intersects between our two curves.

Before we find an expression for the area of our two curves, there’s one more thing we should do. We want to find the directions our two curves are plotted. We can do this by substituting values of πœƒ into our two curves. For example, if we plot πœƒ is equal to zero and then a few more values of πœƒ into the curve π‘Ÿ is equal to three cos of πœƒ, we can see it’s plotted counterclockwise. This can be useful to see. In fact, if we did the same in our other curve, we would see the same is true. We now need to set up our integral to find the area of the region given to us in the question.

And to find this region, we first need to find the area bounded by the curve π‘Ÿ is equal to three cos of πœƒ and our two rays πœƒ is equal to πœ‹ by three and πœƒ is equal to five πœ‹ by three. Then all we need to do is subtract the area bounded by our two rays and the other curve. Let’s start with the outermost area. That’s the area marked in green. We’re doing this by using our integral formula for area of polar curves. First, this is bounded by the curve π‘Ÿ is equal to three cos of πœƒ. It’s also important to realize that our values of πœƒ will start at five πœ‹ by three and end at πœ‹ by three.

We can see this from our sketch. If we had chosen our values of πœƒ to be the other way round, we would get the areas on the other side of our two rays. It’s also worth pointing out here some people prefer to write five πœ‹ by three as negative πœ‹ by three. It doesn’t matter which you will prefer. It would give us the same answer. In our case, we’re going to change this to negative πœ‹ by three. Remember, we then need to subtract the area bounded by these two rays and our inner curve. So we’re subtracting the integral from negative πœ‹ by three to πœ‹ by three of one-half times one plus the cos of πœƒ all squared with respect to πœƒ.

And we can simplify this. First, since both of these integrals have the same limits of integration, we can combine these into one integral. So we’ll combine these into one integral. Next, we’ll take the constant factor of one-half outside of our integral. Then we need to distribute the squares over both of our parentheses. Distributing the squares over our parentheses, we get one-half times the integral from negative πœ‹ by three to πœ‹ by three of nine cos squared of πœƒ minus one plus two cos of πœƒ plus the cos squared of πœƒ with respect to πœƒ.

And we can simplify this. We can distribute the negative over our parentheses. And then we see we have negative cos squared of πœƒ plus nine squared of πœƒ. And this is equal to eight cos squared of πœƒ. So we need to evaluate the following integral. And this means we need to integrate the cos squared of πœƒ. And this is difficult to integrate directly. The easiest way is to use the double angle formula for cosine.

Recall the following version of the double angle formula for cosine. The cos of two πœƒ is equivalent to two cos squared of πœƒ minus one. We want to use this to rewrite eight cos squared of πœƒ. So we’ll start by adding one to both sides of our equivalence, which gives us cos two πœƒ plus one is equivalent to two cos squared of πœƒ. Then we’ll multiply both sides of the equivalence through by four. This gives us eight cos squared of πœƒ is equivalent to four cos of two πœƒ plus four. And this is much easier to integrate, so we’ll write this into our integral. Doing this, we get one-half times the integral from negative πœ‹ by three to πœ‹ by three of four cos of two πœƒ plus four minus one minus two cos of πœƒ.

And of course, we can simplify this. Four minus one is equal to three. And now we’re ready to evaluate this integral term by term. First, to integrate cos of two πœƒ, we need to change cosine into sine and then divide through by a coefficient of πœƒ. So this is four sin of two πœƒ divided by two. And of course, we can simplify this. Four divided by two is equal to two. So the integral of our second term is two sin of πœƒ.

Next, we need to integrate the constant three with respect to πœƒ. We know this is just equal to three πœƒ. And to integrate our third term, we know the integral of negative cos of πœƒ with respect to πœƒ is negative sin of πœƒ. So the integral of our third term is negative two sin of πœƒ. And of course, we need to evaluate this at the limits of integration. To evaluate this expression, we’re going to first need to clear some space. So all we need to do is clear some space and then evaluate this at the limits of integration. We’ll start by substituting the upper limit of integration into our antiderivative. This gives us the following expression.

And we can evaluate this. First, two multiplied by the sin of two πœ‹ by three is equal to the square root of three. Next, three times πœ‹ by three is just equal to πœ‹. Finally, negative two multiplied by the sin of πœ‹ by three is equal to negative the square root of three. And of course, we can then cancel root three minus root three to give us zero. So our antiderivative evaluated at πœ‹ by three simplifies to just give us πœ‹. We then need to subtract our antiderivative evaluated at the lower limits of integration when πœƒ is equal to negative πœ‹ by three. This gives us the following expression.

And we can evaluate this Just as we did before. First, two sin of negative two πœ‹ by three is equal to negative the square root of three. Next, three times negative πœ‹ by three is equal to negative πœ‹. Finally, negative two multiplied by the sin of negative πœ‹ by three is equal to positive the square root of three. And we can simplify this in exactly the same way we did before. Negative root three plus root three is equal to zero. So our antiderivative evaluated at negative πœ‹ by three gave us negative πœ‹. Remember, we need to subtract this value. So we have one-half times πœ‹ minus negative πœ‹. And of course, this just evaluates to give us πœ‹. And this is our final answer.

Therefore, we were able to show the area of the region which lies inside of the polar curve π‘Ÿ is equal to three cos of πœƒ but outside of the polar curve π‘Ÿ is equal to one plus the cos of πœƒ is equal to πœ‹.

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