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

Find the area of the region that lies inside the polar curve πΒ² = 8 cos 2π but outside the polar curve π = 2.

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Find the area of the region that lies inside the polar curve π squared is equal to eight cos of two π but outside the polar curve π is equal to two.

In order to find the regions that lie outside π is equal to two but inside π squared is equal to eight cos of two π, weβll need to find the points of intersections of the two curves. Therefore, we can start by doing this. We need to solve π squared is equal to eight cos of two π and π is equal to two, simultaneously. We can substitute π equals two into the first equation. This will give us that four is equal to eight cos of two π. We end up with cos of two π is equal to one-half. Now, weβre looking for the solutions for which π is between zero and two π. Therefore, these will be the solutions for which two π is between zero and four π. These solutions within this range are that two π 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 π is equal to π by six, five π by six, seven π by six, and 11π by six.

Letβs now draw a sketch of π 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 π squared is equal to eight cos of two π. We can do this by first rewriting the curve as π is equal to the square root of eight cos of two π. Next, we can find some points on the curve by inputting some values of π.

We have that at π is equal to zero, cos of zero is equal to one. Therefore, π is equal to the square root of eight, which is also equal to two root two. At π is equal to π by two, we have cos of two π is equal to cos of π, and cos of π is equal to zero. Therefore, at π is equal to π by two, π is equal to zero. Continuing this on, at π is equal to π, we have that π is equal to two root two. And at π is equal to three π by two, π 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 weβre trying to find their areas. So itβs the regions which lie outside the curve π is equal to two but inside our curve of π squared is equal to eight cos of two π. There are, in fact, two regions which weβ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 π one and π two of one-half π one squared minus π two squared dπ. For the regions which weβre concerned with, π squared is equal to eight cos of two π is greater than or equal to π is equal to two. Therefore, π one squared is equal to eight cos of two π, and π 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 π minus four dπ. Next, letβ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 π minus four dπ. We can simplify both of these integrands. And now, weβre ready to integrate. We have that the integral of four cos of two π minus two is two sin of two π minus two π. 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 π squared is equal to eight cos of two π but outside π is equal to two is four root three minus four π by three.