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‘ŸΒ² = 8 cos 2πœƒ but outside the polar curve π‘Ÿ = 2.

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Video Transcript

Find the area of the region that lies inside the polar curve π‘Ÿ squared is equal to eight cos of two πœƒ but outside the polar curve π‘Ÿ is equal to two.

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

Let’s now draw a sketch of π‘Ÿ 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œ‹ by three. The second is at five πœ‹ by three. The third is at seven πœ‹ by three. And the fourth intersection point is at 11πœ‹ by three. Now, we can attempt to sketch the curve of π‘Ÿ squared is equal to eight cos of two πœƒ. We can do this by first rewriting the curve as π‘Ÿ is equal to the square root of eight cos of two πœƒ. Next, we can find some points on the curve by inputting some values of πœƒ.

We have that at πœƒ is equal to zero, cos of zero is equal to one. Therefore, π‘Ÿ is equal to the square root of eight, which is also equal to two root two. At πœƒ is equal to πœ‹ by two, we have cos of two πœƒ is equal to cos of πœ‹, and cos of πœ‹ is equal to zero. Therefore, at πœƒ is equal to πœ‹ by two, π‘Ÿ is equal to zero. Continuing this on, at πœƒ is equal to πœ‹, we have that π‘Ÿ is equal to two root two. And at πœƒ is equal to three πœ‹ by two, π‘Ÿ 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 π‘Ÿ is equal to two but inside our curve of π‘Ÿ squared is equal to eight cos of two πœƒ. 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 πœƒ one and πœƒ two of one-half π‘Ÿ one squared minus π‘Ÿ two squared dπœƒ. For the regions which we’re concerned with, π‘Ÿ squared is equal to eight cos of two πœƒ is greater than or equal to π‘Ÿ is equal to two. Therefore, π‘Ÿ one squared is equal to eight cos of two πœƒ, and π‘Ÿ 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œ‹ by six to seven πœ‹ by six of one-half of eight cos of two πœƒ minus four dπœƒ. Next, let’s consider the area of the region on the right of our graph. This region is between the angles of 11πœ‹ by six and πœ‹ by six. However, if we were to integrate between 11πœ‹ by six and πœ‹ by six, we would be jumping from an angle of two πœ‹ 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œ‹ by six to negative πœ‹ by six. Since on our graph, negative πœ‹ by six is equal to 11πœ‹ by six.

We obtained that the area on the right is equal to the integral from negative πœ‹ by six to πœ‹ by six of one-half of eight cos of two πœƒ minus four dπœƒ. We can simplify both of these integrands. And now, we’re ready to integrate. We have that the integral of four cos of two πœƒ minus two is two sin of two πœƒ minus two πœƒ. Next, we substitute in our upper and lower bounds. Now, we can use that sin of seven πœ‹ by three and sin of πœ‹ by three are both root three over two. And that sin of five πœ‹ by three and sin of negative πœ‹ 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‘Ÿ squared is equal to eight cos of two πœƒ but outside π‘Ÿ is equal to two is four root three minus four πœ‹ by three.

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