Video: Finding the Area of a Region Inside the Larger Loop and Outside the Smaller Loop of a Curve

Consider the polar curve π‘Ÿ = (1/2) + cos πœƒ. Find the area of the region inside its larger loop but outside its smaller loop.

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Video Transcript

Consider the polar curve π‘Ÿ is equal to one-half plus the cos of πœƒ. Find the area of the region inside its larger loop but outside its smaller loop.

In this question, we’re asked about the area of a region defined by a polar curve. So the first thing we’re going to want to do is sketch a graph of our polar curve. We could do this by taking sample values of πœƒ and substituting this into the equation for our curve and then plotting these on a graph. However, we could also do this by using a graphing calculator. The result will look something like this. The next thing we’ll want to do is get a rough idea of the region we’re asked to evaluate.

We see this region is defined by the area inside of the larger loop of our curve, but outside of the smaller loop. And we can sketch this in. This region will be bound by our curve, but we don’t want to include the area inside of the smaller loop. And at this point, there are lots of different methods we could use to find this area. We’ll only go through one of these. First, we need to recall how we find the area of a region bound by a polar curve. We recall we can calculate the area of the region bounded by the polar curve π‘Ÿ is equal to 𝑓 of πœƒ and the rays πœƒ one and πœƒ two by using the following integral. It’s equal to the integral from πœƒ one to πœƒ two of one-half times 𝑓 of πœƒ all squared with respect to πœƒ.

In our case, we already know our function 𝑓 of πœƒ. It’s equal to one-half plus the cos of πœƒ. We just need to find suitable values for πœƒ one and πœƒ two. To do this, let’s consider what happens when we substitute πœƒ is equal to zero into our polar curve. Doing this, we get one-half plus the cos of zero, which is equal to three over two. We can actually plot this point on our curve. Remember, πœƒ will be equal to zero and π‘Ÿ will be equal to three over two. This gives us the following starting point. In fact, this can give us some interesting information. We can see as we increase the value of πœƒ, our curve will be plotted in the following direction.

And we can keep going all the way until our value of π‘Ÿ is equal to zero. This will give us some value of πœƒ. We’ll call this πœƒ two and draw on the ray πœƒ two. And now, we can see something interesting. If we were to calculate the area bounded by our curve between πœƒ is equal to zero and πœƒ is equal to two, doing this, we would get the area bounded by our outer loop above the polar axis and the area inside the inner loop above the polar axis. And what’s more, we can see our curve is symmetric in the horizontal axis. This is because the cos of π‘₯ is an even function. So the cos of πœƒ will be equal to cos of negative πœƒ.

So we can use all of this information to start finding the area given to us in the question. First, the area we’ve shaded in blue will be given by the integral from zero to πœƒ two of one-half times one-half plus the cos of πœƒ all squared with respect to πœƒ. But remember, our curve is symmetric in the horizontal axis. When we multiply this by two, we’ll get the entire region bounded by our outer loop. But remember, this is only the area of the outer loop. We also need to subtract the area given by our inner loop. To do this, we’re going to, once again, need to use our integral rule.

First, we already know that our loop will start at the value of πœƒ two. We just need to find the value of πœƒ where this loop ends. And in fact, we already know where this will happen from our sketch. It will happen when π‘Ÿ is equal to zero. So we can find the value of πœƒ three and the value of πœƒ two by solving zero is equal to one-half plus the cos of πœƒ. And in fact, by using our definitions for πœƒ two and πœƒ three and by using our integral rule, we can find an expression for the area of our inner loop. So now, we have the area of everything inside of our outer loop and then we subtract the area of our inner loop. This gives us the area of the region given to us in the question.

But before we can start evaluating this, we need to find values for πœƒ two and πœƒ three. And these were where our curve touched the origin. So the value of π‘Ÿ would be equal to zero. So we need to solve the equation zero is equal to one-half plus the cos of πœƒ. And if we subtract one-half from both sides of the equation, we get the cos of πœƒ is equal to negative one-half. There’s a few different ways of finding the solutions to this equation. We’ll use our inverse trigonometric functions. We get one solution by taking the inverse cos of negative one-half which is equal to two πœ‹ by three. And in fact this is in our second quadrant. So we know that this is the solution πœƒ two.

We could find our other solution graphically or otherwise. However, we’re going to do this by remembering the cosine is an even function. So if two πœ‹ by three is a solution, negative two πœ‹ by three is also a solution. Finally, we remember the cos function is periodic about two πœ‹. So if negative two πœ‹ by three is a solution, then negative two πœ‹ by three plus two πœ‹ is also a solution. And if we simplify this expression, we get four πœ‹ by three, which we know is in the third quadrant, which means this is our solution πœƒ three.

We can now substitute these values of πœƒ two and πœƒ three into our integrals to give us an expression for the area of the region given to us in the question. Now, we could start simplifying and evaluating these integrals. However, the integrands of both of these integrals are the same. So we’ll just find an antiderivative for this integrand separately. We need to integrate one-half times one-half plus the cos of πœƒ all squared with respect to πœƒ. And we’ll start by evaluating the square of our parentheses. So by evaluating this either by using the FOIL method or binomial expansion, we now have the integral of one-half times one-quarter plus the cos of πœƒ plus the cos squared of πœƒ with respect to πœƒ.

Next, we’ll multiply through by one-half. This then gives us the integral of one-eighth plus the cos of πœƒ over two plus the cos squared of πœƒ over two with respect to πœƒ. However, we still can’t integrate this easily because this contains the cos squared of πœƒ. So we’ll rewrite this by using our double-angle formula. Recall the double-angle formula for cos tells us the cos of two πœƒ is equivalent to two times the cos squared of πœƒ minus one. We want to rearrange this to give us an equation for the cos squared of πœƒ. Adding one to both sides of our equivalents and then dividing through by two, we get the cos of two πœƒ plus one all over two is equivalent to the cos squared of πœƒ.

And this is a much easier expression to integrate. So we’ll use this in our integrand. Remember, we’re going to need to divide this by two. So by substituting this into our integrand and simplifying slightly, we get the integral of one-eighth plus the cos of πœƒ over two plus the cos of two πœƒ plus one over four with respect to πœƒ. And before we evaluate this integral, there’s one more piece of simplification we’ll do. We have one-quarter plus one-eighth which we know is equal to three over eight. And now, we can integrate each term separately.

First, the integral of three over eight with respect to πœƒ is three πœƒ over eight. Next, we know the integral of the cos of πœƒ with respect to πœƒ is the sin of πœƒ. So the integral of our second term is the sin of πœƒ over two. Similarly, we can evaluate the integral of our third term. We get the sin of two πœƒ over eight. And of course, we should add our constant of integration 𝐢. However, we won’t be using this because we’re using this in a definite integral.

We can now use this expression to evaluate both of our definite integrals. So by using the antiderivative we just found, we can evaluate both of our definite integrals. We get the following expression. And now, all that’s left to do is evaluate both of these at the limits of integration. However, there’s a small bit of simplification we can do before we do this. For example, consider what happens when we substitute πœƒ is equal to zero into our antiderivative. Each term evaluates to give us zero, so we don’t need to worry about this term. So let’s start by substituting πœƒ is equal to two πœ‹ by three into our antiderivative. This gives us the following expression.

And we can start simplifying. In our first term, inside of our parentheses, we have three over three which is one. We also have two divided by eight which is equal to four. So our first term simplifies to give us πœ‹ by four. Next, let’s simplify our second term. We know the sin of two πœ‹ by three is root three over two and we multiply this by one-half. So this gives us root three over four. Finally, we can simplify our third term. We know the sin of four πœ‹ by three is negative root three over two. And we multiply this by one-eighth to get negative root three over 16. This gives us two times πœ‹ by four plus root three over four minus root three over 16. And we’ll combine the second and third term inside of our parentheses to give us three root three over 16.

Now, we could do exactly the same to evaluate the limits of integration in our second expression. This would give us the following expression. And it’s important to note we evaluate each term separately in our calculator so we get an exact answer. Now, all that’s left to do is group like terms and simplify to give us our final answer. Doing this and then taking out the factor of one-quarter, we get one-quarter times πœ‹ plus three root three, which is our final answer.

Therefore, we were able to show the area inside the larger loop but outside of the smaller loop of the polar curve π‘Ÿ is equal to one-half plus the cos of πœƒ is equal to one-quarter multiplied by πœ‹ plus three root three.

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