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

Video Transcript

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.