Video: Finding the Area of a Region Bounded by a Given Curve

Find the area of the region enclosed by the inner loop of the polar curve π‘Ÿ = 1 + 2 sin πœƒ.

04:53

Video Transcript

Find the area of the region enclosed by the inner loop of the polar curve π‘Ÿ is equal to one plus two sin πœƒ.

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 πœƒ one to πœƒ two of one-half π‘Ÿ squared dπœƒ. We have been given π‘Ÿ in the question. And it’s equal to one plus two sin πœƒ. We just need to find the values of πœƒ one and πœƒ two.

Now, the values of πœƒ one and πœƒ 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 π‘Ÿ is equal to zero, since this point is on the origin. And so, we can find πœƒ one and πœƒ two by solving π‘Ÿ is equal to zero. When π‘Ÿ is equal to zero, one plus two sin πœƒ is equal to zero too. We can rearrange this to find that sin πœƒ is equal to negative one-half.

We need to find solutions here within the range, where πœƒ is between zero and two πœ‹. We can use a graph of sin πœƒ 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œ‹ by six. However, we’re looking for solutions between zero and two πœ‹. Since sin is periodic in two πœ‹, we can see from our sketch that one of the solutions will be πœ‹ by six less than two πœ‹, which is 11πœ‹ by six. And the other solution will be πœ‹ by six more than πœ‹, which is simply seven πœ‹ by six.

Therefore, we’ve now found our two values for πœƒ one and πœƒ two. We have that πœƒ one is equal to seven πœ‹ by six and πœƒ two is equal to 11πœ‹ 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œ‹ by six to 11πœ‹ by six of one-half multiplied by one plus two sin πœƒ squared dπœƒ. We can distribute the square and multiply through by one-half. Which gives the integral from seven πœ‹ by six to 11πœ‹ by six of one-half plus two sin πœƒ plus two sin squared πœƒ dπœƒ.

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 πœƒ is equal to one minus two sin squared πœƒ. This can be rearranged to make sin squared πœƒ the subject. Which gives that sin squared πœƒ is equal to one minus cos of two πœƒ 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œ‹ by six to 11πœ‹ by six of one-half plus two sin πœƒ plus one minus cos of two πœƒ dπœƒ. 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 πœƒ over two. When we integrate the second term of two sin πœƒ, we get negative two cos πœƒ. And for our final term of negative cos two πœƒ, we get negative one-half sin of two πœƒ. And we mustn’t forget that this is between the bounds of seven πœ‹ by six and 11πœ‹ by six. Now, we start by substituting in 11πœ‹ by six, giving us 11πœ‹ by four minus two cos of 11πœ‹ by six minus one-half sin of 11πœ‹ by three. Next, we substitute in seven πœ‹ by six. But you mustn’t forget to subtract this since seven πœ‹ by six is the lower bound. Giving us negative seven πœ‹ by four minus two cos of seven πœ‹ by six minus one-half sin of seven πœ‹ by three.

Next, we use the fact that cos of 11πœ‹ by six is equal to root three over two. Sin of 11πœ‹ by three is equal to negative root three over two. Cos of seven πœ‹ by six is equal to negative root three over two. And sin of seven πœ‹ by three is equal to root three over two. We can now multiply through all of the terms. And we end up with 11πœ‹ by four minus seven πœ‹ 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 π‘Ÿ is equal to one plus two sin πœƒ is equal to πœ‹ minus three root three over two.

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