Lesson Video: Area Bounded by Polar Curves Mathematics

Video Transcript

Area Bounded by Polar Curves

In this video, we will learn how to calculate the area of a region enclosed by one or more polar curves. We’ll be looking at a variety of examples of how we can find integrals to find areas of this form. Let’s now consider the following polar curve. And we can call this curve 𝑟 or 𝑓 of 𝜃. Let’s suppose that we were asked to find the area bounded by this curve, 𝜃 one and the 𝜃 two. Now, typically, when we’re finding an area in Cartesian coordinates, the area we will be finding here would be the area between the curve 𝑟, the horizontal 𝑥-axis, and the 𝑥-values of 𝜃 one and 𝜃 two. So that’s this area here.

However, this is not a Cartesian graph. This is a polar graph. And when we say we’re finding the area between 𝜃 one and 𝜃 two, this, in fact, means the area enclosed by the lines 𝜃 is equal to 𝜃 one, 𝜃 is equal to 𝜃 two, and 𝑟 is equal to 𝑓 of 𝜃. Which gives us this shaded area in blue here. Which, as we can see, is not equal to the area we would have found using the Cartesian method, which is in yellow. Therefore, we have a new formula for finding this area. This formula tells us that the area is equal to the integral from 𝜃 one to 𝜃 two of one-half 𝑟 squared d𝜃. Let’s now look at an example of how this formula can be used.

Find the area of the region bounded by the polar curve 𝑟 is equal to two cos two 𝜃 between 𝜃 is equal to 𝜋 by two and 𝜃 is equal to three 𝜋 by two.

Now, we have a formula for finding area bounded by polar curves. And this formula tells us that the area is equal to the integral from 𝜃 one to 𝜃 two of one-half 𝑟 squared d𝜃. Now, we’ve been given 𝑟 in the question, and it’s equal to two cos of two 𝜃. And we’ve been asked to find the area between 𝜃 is equal to 𝜋 by two and 𝜃 is equal to three 𝜋 by two. Since three 𝜋 by two is greater than 𝜋 by two, we can say that 𝜋 by two must be equal to 𝜃 one. And three 𝜋 by two must be equal to 𝜃 two. We can substitute these values of 𝜃 one, 𝜃 two, and 𝑟 into our equation for the area.

We find that the area is equal to the integral from 𝜋 by two to three 𝜋 by two of one-half of two cos of two 𝜃 squared d𝜃. We can start by expanding the square term. We end up with the integral from 𝜋 by two to three 𝜋 by two of one-half times four cos squared of two 𝜃 d𝜃. And we can cancel the half with one of the twos from the four, which gives us this. Now, we cannot directly integrate a cos squared term like this on its own. However, there is a way to get rid of the square. We need to use the double angle formula for cos. This formula tells us that cos of two 𝜃 is equal to two cos squared 𝜃 minus one. We can rearrange this to make cos squared 𝜃 the subject, which gives us that cos squared 𝜃 is equal to cos of two 𝜃 plus one all over two.

Now, we’re nearly ready to substitute this formula in. However, in our integrand, we, in fact, have cos squared of two 𝜃. And in our formula, we only have cos squared of 𝜃. This can be solved by simply multiplying each 𝜃 in our formula by two. Which gives us that cos squared of two 𝜃 is equal to cos of four 𝜃 plus one all over too. And we can substitute this back into our integral to obtain the integral from 𝜋 by two to three 𝜋 by two of two multiplied by cos of four 𝜃 plus one all over two d𝜃. And now, we can cancel out the factor of two with the two in the denominator. And now, we’re ready to integrate our function.

Our first term, which is cos of four 𝜃, is a function within a function. The innermost function is four 𝜃. So we start by differentiating four 𝜃 with respect to 𝜃 to give us four. Since four is a constant, we’re able to do this integration. However, we mustn’t forget to divide by this four. And so, we start off with one over four. Then, we integrate the cos. And so, for our first term, we obtain one over four sin of four 𝜃. And our second term is simply one. Integrating one just gives us 𝜃. And we mustn’t forget that we’re integrating between 𝜋 by two and three 𝜋 by two. So these are the bounds which we need to substitute in.

Substituting in three 𝜋 by two, we obtain one-quarter sin of six 𝜋 plus three 𝜋 by two. And next, we substitute in 𝜋 by two, but we mustn’t forget to subtract this. Since 𝜋 by two is our lower bound. And so, we obtain minus one over four sin of two 𝜋 plus 𝜋 by two. Now, we know that sin of two 𝜋 is equal to zero. And we also know that sin is a periodic function with a periodicity of two 𝜋. Since six 𝜋 is a multiple of two 𝜋, this means that sin of six 𝜋 will also be equal to zero. Therefore, the two terms which we’re left with is three 𝜋 by two minus 𝜋 by two.

And it’s from here that we reach our solution. Which is that the area of the region bounded by the polar curve 𝑟 is equal to two cos of two 𝜃 between 𝜃 is equal to 𝜋 by two and 𝜃 is equal to three 𝜋 by two is equal to 𝜋.

In some questions, we may not be given the values of 𝜃 one and 𝜃 two, for which we need to find the area between. Sometimes, the region of which we need to find the area will simply be described. We will see this in the next example.

Find the area of the region enclosed by the inner loop of the polar curve 𝑟 is equal to one plus two sin 𝜃.

Let’s start by drawing a quick sketch of what this curve would look like. We could do this by using a graphing calculator, graphing software, or by plotting some points on the curve. Our curve would look something like this. We can see that the inner loop of our curve is this little loop at the bottom here. And so, this orange-shaded region is the area we’re trying to find. We do, in fact, have a formula for finding areas of regions enclosed by polar curves. This formula tells us that the area is equal to the integral from 𝜃 one to 𝜃 two of one-half 𝑟 squared d𝜃. We have been given 𝑟 in the question. And it’s equal to one plus two sin 𝜃. We just need to find the values of 𝜃 one and 𝜃 two.

Now, the values of 𝜃 one and 𝜃 two will occur at the end points of their closed inner loop. Since this is a loop, this will happen when the curve crosses over itself. Therefore, this is at this blue point here. We can see from our sketch of the curve that this occurs when 𝑟 is equal to zero, since this point is on the origin. And so, we can find 𝜃 one and 𝜃 two by solving 𝑟 is equal to zero. When 𝑟 is equal to zero, one plus two sin 𝜃 is equal to zero too. We can rearrange this to find that sin 𝜃 is equal to negative one-half.

We need to find solutions here within the range, where 𝜃 is between zero and two 𝜋. We can use a graph of sin 𝜃 to help us find our solutions. We draw in a line of negative one-half. Our calculator will give us a solution of negative 𝜋 by six. However, we’re looking for solutions between zero and two 𝜋. Since sin is periodic in two 𝜋, we can see from our sketch that one of the solutions will be 𝜋 by six less than two 𝜋, which is 11𝜋 by six. And the other solution will be 𝜋 by six more than 𝜋, which is simply seven 𝜋 by six.

Therefore, we’ve now found our two values for 𝜃 one and 𝜃 two. We have that 𝜃 one is equal to seven 𝜋 by six and 𝜃 two is equal to 11𝜋 by six. We’re now ready to use our formula for finding the area. We have that the area is equal to the integral from seven 𝜋 by six to 11𝜋 by six of one-half multiplied by one plus two sin 𝜃 squared d𝜃. We can distribute the square and multiply through by one-half. Which gives the integral from seven 𝜋 by six to 11𝜋 by six of one-half plus two sin 𝜃 plus two sin squared 𝜃 d𝜃.

Now, we know how to integrate each of these terms apart from the sin squared term. However, we can rewrite the sin squared term using the double-angle formula for cos. We have that cos of two 𝜃 is equal to one minus two sin squared 𝜃. This can be rearranged to make sin squared 𝜃 the subject. Which gives that sin squared 𝜃 is equal to one minus cos of two 𝜃 all over two. And this can be substituted back into our integral. And we can cancel the factor of two with the factor of two in the denominator. Leaving us with the integral from seven 𝜋 by six to 11𝜋 by six of one-half plus two sin 𝜃 plus one minus cos of two 𝜃 d𝜃. And we can group together the one and the one-half. So our integrand ends up looking like this. Which we’re now ready to integrate.

Integrating the first term, three over two, we simply get three 𝜃 over two. When we integrate the second term of two sin 𝜃, we get negative two cos 𝜃. And for our final term of negative cos two 𝜃, we get negative one-half sin of two 𝜃. And we mustn’t forget that this is between the bounds of seven 𝜋 by six and 11𝜋 by six. Now, we start by substituting in 11𝜋 by six, giving us 11𝜋 by four minus two cos of 11𝜋 by six minus one-half sin of 11𝜋 by three. Next, we substitute in seven 𝜋 by six. But you mustn’t forget to subtract this since seven 𝜋 by six is the lower bound. Giving us negative seven 𝜋 by four minus two cos of seven 𝜋 by six minus one-half sin of seven 𝜋 by three.

Next, we use the fact that cos of 11𝜋 by six is equal to root three over two. Sin of 11𝜋 by three is equal to negative root three over two. Cos of seven 𝜋 by six is equal to negative root three over two. And sin of seven 𝜋 by three is equal to root three over two. We can now multiply through all of the terms. And we end up with 11𝜋 by four minus seven 𝜋 by four minus root three plus root three over four minus root three plus root three over four. Combining these terms together, we reach our solution. Which is that the area of the region enclosed by the inner loop of 𝑟 is equal to one plus two sin 𝜃 is equal to 𝜋 minus three root three over two.

Next, we will learn how to find the area enclosed by two polar curves. Let’s consider the following scenario.

Let’s say we’ve been given two polar curves 𝑟 one and 𝑟 two and we’ve been asked to find the region enclosed by 𝑟 one and 𝑟 two between 𝜃 one and 𝜃 two. So that’s this shaded region here.

We can find a formula for this area by applying the formulas which we already know for finding areas of regions bound by polar curves. We know that we can find the area enclosed by 𝑟 two, 𝜃 one, and 𝜃 two, which is this area here. And it’s equal to the integral from 𝜃 one to 𝜃 two of one-half 𝑟 two squared d𝜃. We also know how to find the area enclosed by 𝑟 one, 𝜃 one, and 𝜃 two, which is this area here. We know that it’s equal to the integral from 𝜃 one to 𝜃 two of one-half 𝑟 one squared d𝜃. Now, we can see that this yellow area, which we’re trying to find, must be equal to the area encased by 𝑟 one between 𝜃 one and 𝜃 two minus the area encased by 𝑟 two between 𝜃 one and 𝜃 two. So that’s the integral between 𝜃 one and 𝜃 two of one-half 𝑟 one squared d𝜃 minus that integral between 𝜃 one and 𝜃 two of one-half 𝑟 two squared d𝜃.

Using integral properties, we’re able to combine these integrals since they have the same bounds, to give the integral between 𝜃 one and 𝜃 two of one-half 𝑟 one squared minus one-half 𝑟 two squared d𝜃. We can factor out the half to form our final formula. Which is that the area between two curves 𝑟 one and 𝑟 two between 𝜃 one and 𝜃 two is equal to the integral between 𝜃 one and 𝜃 two of one-half 𝑟 one squared minus 𝑟 two squared d𝜃. Let’s note that this only works when 𝑟 one is greater than or equal to 𝑟 two for 𝜃-values between 𝜃 one and 𝜃 two.

Now that we formed this new formula, let’s look at an example of how it works.

Find the area of the region that lies inside the polar curve 𝑟 squared is equal to eight cos of two 𝜃 but outside the polar curve 𝑟 is equal to two.

In order to find the regions that lie outside 𝑟 is equal to two but inside 𝑟 squared is equal to eight cos of two 𝜃, we’ll need to find the points of intersections of the two curves. Therefore, we can start by doing this. We need to solve 𝑟 squared is equal to eight cos of two 𝜃 and 𝑟 is equal to two, simultaneously. We can substitute 𝑟 equals two into the first equation. This will give us that four is equal to eight cos of two 𝜃. We end up with cos of two 𝜃 is equal to one-half. Now, we’re looking for the solutions for which 𝜃 is between zero and two 𝜋. Therefore, these will be the solutions for which two 𝜃 is between zero and four 𝜋. These solutions within this range are that two 𝜃 is equal to 𝜋 by three, five 𝜋 by three, seven 𝜋 by three, or 11𝜋 by three. And so, we find that our points of intersection are that 𝜃 is equal to 𝜋 by six, five 𝜋 by six, seven 𝜋 by six, and 11𝜋 by six.

Let’s now draw a sketch of 𝑟 is equal to two. This will simply be a circle with a radius of two. Next, we can mark on our points where the two curves will intersect. The first point is at 𝜋 by three. The second is at five 𝜋 by three. The third is at seven 𝜋 by three. And the fourth intersection point is at 11𝜋 by three. Now, we can attempt to sketch the curve of 𝑟 squared is equal to eight cos of two 𝜃. We can do this by first rewriting the curve as 𝑟 is equal to the square root of eight cos of two 𝜃. Next, we can find some points on the curve by inputting some values of 𝜃.

We have that at 𝜃 is equal to zero, cos of zero is equal to one. Therefore, 𝑟 is equal to the square root of eight, which is also equal to two root two. At 𝜃 is equal to 𝜋 by two, we have cos of two 𝜃 is equal to cos of 𝜋, and cos of 𝜋 is equal to zero. Therefore, at 𝜃 is equal to 𝜋 by two, 𝑟 is equal to zero. Continuing this on, at 𝜃 is equal to 𝜋, we have that 𝑟 is equal to two root two. And at 𝜃 is equal to three 𝜋 by two, 𝑟 is equal to zero. And we can add these points to our graph. We now have seven points of our curve drawn on our graph. And so, we can draw a rough sketch for what it should look like. As we can see, it sort of looks like a figure of eight.

Now, this sketch really helps us to find the regions of which we’re trying to find their areas. So it’s the regions which lie outside the curve 𝑟 is equal to two but inside our curve of 𝑟 squared is equal to eight cos of two 𝜃. There are, in fact, two regions which we’re interested in, which is these two regions here. We have a formula for finding the area of regions between two polar curves. And this formula tells us that the area is equal to the integral between 𝜃 one and 𝜃 two of one-half 𝑟 one squared minus 𝑟 two squared d𝜃. For the regions which we’re concerned with, 𝑟 squared is equal to eight cos of two 𝜃 is greater than or equal to 𝑟 is equal to two. Therefore, 𝑟 one squared is equal to eight cos of two 𝜃, and 𝑟 two is equal to two.

We can start by considering the area on the left of our diagram. This will be equal to the integral from five 𝜋 by six to seven 𝜋 by six of one-half of eight cos of two 𝜃 minus four d𝜃. Next, let’s consider the area of the region on the right of our graph. This region is between the angles of 11𝜋 by six and 𝜋 by six. However, if we were to integrate between 11𝜋 by six and 𝜋 by six, we would be jumping from an angle of two 𝜋 to an angle of zero. Which would occur when we cross the horizontal access. And, of course, we cannot do this. We can instead change the angle of 11𝜋 by six to negative 𝜋 by six. Since on our graph, negative 𝜋 by six is equal to 11𝜋 by six.

We obtained that the area on the right is equal to the integral from negative 𝜋 by six to 𝜋 by six of one-half of eight cos of two 𝜃 minus four d𝜃. We can simplify both of these integrands. And now, we’re ready to integrate. We have that the integral of four cos of two 𝜃 minus two is two sin of two 𝜃 minus two 𝜃. Next, we substitute in our upper and lower bounds. Now, we can use that sin of seven 𝜋 by three and sin of 𝜋 by three are both root three over two. And that sin of five 𝜋 by three and sin of negative 𝜋 by three are both negative root three over two. Next, we can expand everything here. And then, for our final step, we just simplify. Which gives us a solution that the area of the region that lies inside 𝑟 squared is equal to eight cos of two 𝜃 but outside 𝑟 is equal to two is four root three minus four 𝜋 by three.

We have now seen a variety of examples and techniques which we can use to find the areas bound by polar curves. Let’s cover some key points of the video.

Key Points

The area enclosed by the polar curve 𝑟 is equal to 𝑓 of 𝜃, 𝜃 one, and 𝜃 two is given by the integral from 𝜃 one to 𝜃 two of one-half 𝑟 squared d𝜃. The area enclosed by two polar curves 𝑟 one which is equal to 𝑓 of 𝜃 and 𝑟 two which is equal to 𝑔 of 𝜃, 𝜃 one, and 𝜃 two. Where 𝑟 one is greater than or equal to 𝑟 two for 𝜃 between 𝜃 one and 𝜃 two. Is given by the integral from 𝜃 one to 𝜃 two of one-half of 𝑟 one squared minus 𝑟 two squared d𝜃.