45–60. Areas of regions Find the area of the following regions...

Question

45–60. Areas of regions Find the area of the following regions.

The region inside the lemniscate r² = 6 sin 2θ

Accepted Answer

Hello. In this video, we are going to be computing the total area enclosed by the Limiscot R 2 is equal to 12 multiplied by sin of 2 theta. Now, here, the, in order to compute the total area, the area of the Limiscot is defined as 1/2 multiplied by the definite integral from A to B of R2 D theta. Now here we are given that R 2 is equal to 12 multiplied by sine of 2 theta. But what we need to do is we need to go ahead and determine the bounds of integration. Now, in order to determine the bounds of integration, we want to first determine when 12 sin of 2 theta is greater than or equal to 0. By solving for sin of 2 theta, we will divide both sides of the equation by 12, leaving us with sin of 2 theta is greater than or equal to 0. This is only true when theta is equal to k pi divided by 2. And so what this means is that there's going to be 2 loops for the given Limiscon. The first loop is going to be defined from 0 to pi divided by 2, and the second loop is going to be defined as pi. To 3 pi divided by 2. Now in this case here. We just have to find the area of one loop and double that area afterward since both of the loops are going to have the exact same area. So let's just go ahead and integrate from 0 to pi divided by 2. That is going to give us the equation. A is equal to 1/2 multiplied by the integral from 0 to pi divided by 2 of 12 sin of 2 theta d theta. We can move the constant of 12 to the outside of the integral, leaving us with 6 multiplied by the integral from 0 to pi divided by 2 of sin of 2 theta d theta. Now what we can go ahead and do is integrate normally. The antiderivative of sine of 2 theta is going to be -1/2 cosine of 2 theta, so we will have 6 multiplied by -1/2 cosine of 2 theta, and this will be evaluated from 0 to pi divided by 2. Furthermore, multiplying the constants together leaves us with -3 multiplied by cosine of 2 theta evaluated from 0 to pi divided by 2. Now by plugging in the bounds, we get 3 cosine of 2 theta multiplied by pi divided by 2, which is just pi. -3, which is 3 cosine of 0. Now here, cosine at 0 is going to be 1. And cosine of pi is going to be -1. So we will end up with 3 plus 3, which is equal to 6. So this is going to be the area of one loop of the limiscot, but keep in mind that there are 2 loops in total. So we will take the value of 6 and multiply it by 2 because of two loops, and that is going to give us a total area of 12, which is going to be the solution to the problem. So the total area of the Limiscot is going to be 12, and the overall solution to the problem is going to be option B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

45–60. Areas of regions Find the area of the following regions.The region inside the lemniscate r² = 6 sin 2θ

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45–60. Areas of regions Find the area of the following regions.

The region inside the lemniscate r² = 6 sin 2θ