33–40. Areas of regions Make a sketch of the region and its bo...

Question

33–40. Areas of regions Make a sketch of the region and its bounding curves. Find the area of the region.

The region inside the circle r = 8 sin θ

Accepted Answer

Welcome back, everyone. I'm the area of the region enclosed by the polar curve R equals 6, cosine theta. For this problem, let's recall that the area formula is 12 multiplied by the integral from alpha to beta of R squared the theta. All that we have to do in this problem is clearly define alpha and beta. Those are the limits of integration. So let's focus on the curve R equals 6 cosine theta, and let's recognize that this is a circle. Right? So, essentially what we have to understand is that we form a circle in the form of R equals a multiplied by cosine theta. When theta Ranges for a total of pi radiance. So it doesn't really matter at what initial alpha we begin. We must end with beta equals alpha plus pi. So let's assume that alpha is 0. Let's suppose that we start at 0 radiance, right? Then beta is going to be pi plus 0, which is pi. Meaning we can integrate from 0 to pi, or for example, we can also choose from, let's say negative pi divided by 2 up to pi divided by 2. In this problem, we can just choose alpha equals 0 and beta equals pi. So, Now we can just integrate. We get one half multiplied by the integral from 0 to pi. R squared is going to be 6 cosine theta squared, and then we have the theta. So the integral becomes 1/2, integral from 0 to i of 36, cosine squared the the theta. Now, 36 can be factored out because it's a constant, right? And what we can do is simply write 18 instead of 1/2 multiplied by 36. So we get 18 integral from 0 to pi cosine squared of theta theta. To integrate, we can use the reduction formula. We can reduce that square, and we get 18 integral from 0 to pi. What is cosine squared of theta? Well, it's 1 plus cosine of the double angle, that'd be cosine of 2 theta divided by 2. Now, let's divide 18 divided by 2 gives us 9, multiplied by the integral from 0 toi of 1 plus cosine of 2 theta the theta. Well then from here we can integrate. We got 9 in Cs. The integral of the theta is theta, and we're adding the integral of cosine of 2 theta, which is sin of 2 theta, and we're dividing by 2 according to the reverse chain rule. We want to find the result from 0 to pi. We got 9 respi plus. Sign of 2 multiplied by i gives us 2 pi divided by 2. Minus 90 plus sign of 2 multiplied by 0 gives us 0 divided by 2. We can now notice that we have sin of 2 pi, which is 0. We can also cross out all of the remaining zero terms. Sign of zero is also a 0. So we simply get 9 multiplied by pi, meaning the final answer is 9 pi square units. Right, this is the area. Thank you for watching.

33–40. Areas of regions Make a sketch of the region and its bounding curves. Find the area of the region.

The region inside the circle r = 8 sin θ

Key Concepts

Polar Coordinates and Graphing

Area Calculation in Polar Coordinates

Determining the Limits of Integration

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