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Find the area of the crescent-shaped region (called a $ lune $) bounded by arcs of circles with radii $ r $ and $ R $. (See the figure.)

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 3

Trigonometric Substitution

Integration Techniques

Leighton E.

January 22, 2021

area of a sector of a circle

Nadia H.

September 22, 2020

I see how that could be confusing. In classical geometry, a radius of a circle or sphere is any of the line segments from its center to its perimeter, and in more modern usage, it is also their length.

Samantha T.

Anyone else confused by the Radius , can someone explain?

Lindsey P.

Hi Alex, An arc of a circle is a "portion" of the circumference of the circle. The length of an arc is simply the length of its "portion" of the circumference. The circumference itself can be considered a full circle arc length.

Alex G.

Can someone explain what the arcs of circles is?

Howie C.

Hey Cam, This problem involves finding the area of a crescent-shaped region also called a lune. This region can be imagined as the region bounded by a larger circle and a smaller circle, such that the larger circle touches the smaller circle internally.

Cam R.

What is crescent-shaped region?

Oregon State University

Harvey Mudd College

Baylor University

University of Nottingham

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

16:04

Area of a Lune The crescen…

06:35

A lune is a crescent-shape…

02:41

Find the area of the shade…

09:20

$\mathrm{A}$ lune is a cre…

Let's find the area of the crescent shaped region, which we call a loon bounded by the arcs of the circles with radius. Little are and big are and here's the corresponding figure for convenience. We could go ahead and just positioned the larger circle with Radius Capital R so that its center at the origin. The nice thing about doing that is that you have the equation X squared plus y squared equals R Square. This represents the large circle of radius are on black And then from there, actually, we can go ahead because in this problem where basically well, we'Ll find the area between these two circles that will just be an area between two curves. So we'LL actually want a function for this upper half of the black circle. So just take the square root here, positive square root because around the upper half circle now for the small circle, it's not quite centered at the origin The X value zero, but we've gone up by some value here. Let's just do notice, five b So notice that b distance between the sensors of the circles now here, looking at the picture, we could actually use the fact that if we look at this right part of the blue circle, the right most part that this is also on the bicycle let me go ahead and just draw right triangle here. If this is our little R Well, since this points on the black circle this new segment right here, the high pop news has links are the radius of the big circle. And then I could use that diagram to find me B Square equals, Let me actually take a step back B Square plus little Arthur Claire equals big r squared by pit, agree a CE go ahead and sell for B So that's the distance between the two centers. So now let's write an expression for the area that we want. So here we liked the area between the two circles That will be everything that underneath the other blue circle, but also at the same time above the blacks. Terrible. So really hear what we'LL need is where we have equation for the upper half as a black circle, which I'm doing right now. This was we got this earlier by solving for why and the Black Circle. So this is the lower function, the function that will be subtracting. But we need the upper function to subtract from. So in this case, the upper function is just the upper half of this blue circle. So I get that by solving the equation for the small circle which I should have written over here. Let me come back to the small circle we found B. So the equation for the small circle is X squared. Plus why minus b square equals little are square because we shifted up by a little bit in the wind direction and then we can go ahead and simplify this using our value. Be so First, let's just go ahead and write this integral. The integral will be from negative art are Then we have the upper circle, so the other and blue, the blue upper circle. So let's solve this small circle for why, and doing so, we should have B plus the square root of R squared minus x word. So this is the upper curve, and we know it B is equal to before I. Now let's just leave. It is B, but we're again way saw for be a few minutes ago. So This is the upper curve. And then I'm subtracting the curve in black and we saw that for why up here? So this is just this radical capital R squared minus little r squared no d x. So Edgar minus lower for the area between two curves. Let's make this a little easier by Instead of finding the theory of the entire loon, let's just find the area of the right half right hand side and then just multiplied by two. So instead of going from negative, our two are we'LL just go from zero are multiplied by two. Should be Sierra are there should be a two hour here. Silly, silly typo there and same expression over here. Same radicals. Let's go to the next page. I'm running out of room here. Oh, let's go ahead and distribute that integral to each of the three expressions So we end up with two. There are B dx two zero R little R squared minus X squared the X and then finally for the minus sign oh, capital R Square minus X square d x Now these last two hundred girls are very similar, so we could probably just go ahead and solve both of them at the same time, actually, But before we do that, let's do the simpler and a girl over here. So this and a girl of is just bx X is going from Sierra are, and so we just have to be our And then, if we want, we can go ahead and plug in the value of B. On the previous page, we saw that BEA was equal to the square root capital R squared minus R squared all inside their ankle. So that evaluates the first integral and the end points Now for the second and rule to avoid doing unnecessary repetition. Let's just really look at the integral a a squared minus x squared inside the radical. If we can evaluate this integral, then we can just plugin a equals little are and then plug in a equals big R and that also evaluate the other two hundred girls because they're so similar. So now let's ignore the ours for a moment and put in a day our tricks up for this in a rolling green, X equals a sign data. So D X is a co scientific data, and we know that this So this after the trigger. Natural trick substitution. We have a squared minus a square sine squared data. And then the ex, which was a co sign Taito. So I can pull a square outside the radical. I take the square room. I just get a Have another A over here. So give me a square and then I have one minus sign square on the inside and we know that one minus sign squared is co sign squared someone. We take the square room, which is have co sign left over another co sign over here. So we get co science where when you have ah, integral of a co signer assigned that has the even power. It will be best to use the half angle formula. So I pull out that one half, I get a squared over two and then I have, in a roll of data of one which is just data in the end, the integral of co sign tooth dater, which assigned to theta over too. And I guess here because I didn't put limits of integration for right now, let's just proceed. But way could ignore the sea when we actually evaluate these integral is at the end because these air definite intervals now coming back to this latest expression where we left off, we know this is equal to to scientific co signed data by the double angle formula so we could go ahead and cancel those tunes. We get a squared over two State A plus sign data, coastline data, plus he. And now finally, the last thing to do here. We're basically done, but we'd like to get things back in terms of X, so there will actually go ahead and use this trick. So draw the triangle, get everything back in terms of X, and then once we plug in little R or big are we could just go plug in our original limits. Zero's a little art. So let me go on to the next page, form our truth Substitution. We no sign of data is X ray, so let's draw that triangle. There's our theta so sine opposite over adjacent X over, eh? And we could use potash grand there up to find out opposite side nature that excuse me, that Jason Side a squared plus X squared is a square. So software age. So this is all due to the Pythagorean. Darryl. Now we have enough to say what? Co sign it. So we left off that one half a squared data plus Scientific co signed it on the previous speech. So now just go ahead and use the triangle. Here we have one half a square Seita. You could have get by just solving this equation up here for data. So take Sinan verse on the side. So here we have data equals sign Inverse X Over. Eh? Again, that's coming from this equation up here. Signed peoples ex operate. And then for the next term, we have signed data where we know that that's X operator. Let me go ahead and actually just distribute the one have a square. So that's the first term. And then we have for the next term, we have sign, which is X over, eh? And then we have co sign, which is h overlay Oh, each other and in our policy, and then just simplify. Here. All that's necessary is that cancel those a square and then to simplify this one more time. Yeah. Okay, so now is this is where, at the point, we're to evaluate those two in a girls that we were interested in. We want to go ahead. So this was from the previous page. Now we have to go ahead and take a to be a little R and eight to be big bar. And we could also go back to our usual limits of integration, which was zero little r So plugging this all in and going to the next page the first integral that we had was which was to be our And we also written that as two are radical, no r squared, minus little are square. So that was the first in a girl we already evaluated enough for the last two of the girls. These were the same in a girl. Just one had a little r one had big r so plugging in a equals r into our formula. We have our squared arc sine X over our plus ex radical and then a squared minus X squared becomes little R squared minus X squared. And this is our original and point zero r. So this is replacing the previous integral that represented this term here. This was coming from the two integral zero are radical R squared minus X squared. You might have noticed that there's the one half that we had in the previous formula. Those disappeared because we multiplied by this two year. So here we just plugged in a equals little R into our formula. And then now we go ahead and about and subtract the other in the girl. So then we have the minus. Then we have now its capital R squared. So a equals big are on this integral in this term sign in verse X over capital are this time plus X Basically the same term is last time in the second term that we just replaced. Little are with big are But this little are Here's the limit of integration is still our little R Now we're at the point where we could just go ahead and plug in R and zero into the second to terms. So this term out here we leave alone. This already been evaluated from zero dar. So we have to our route our square minus little are square. And then let's go ahead and plug in little r appear for X. So we have our squared off in the front and then we have sign in verse of our over our, which is just one. And actually we could even replace this. We can evaluate this. This is just pie over too Sign of fire or two is one, and pie or two is in the range of the signing Chris function. By definition, somebody race that, and then the next term will be plug in. Little are to the radical for X. We get squirt of Zero, which is zero. So there's nothing else to add. And then when I subtract, we plug in our equals. X equals zero. So then we have sign inverse of zero, which is era. And then we have a zero over here in front of the radical, so we just get zero again. So there's actually nothing else to write here. That's the second term, and then now we could do the same thing. But for the last term, so let's plug in the last one minus. So we have that capital R squared sign in verse, plugging that little are for X and then we also plug in. Little are for Exxon decide, and we'LL have a little R square room. Big R squared minus little r squared. And then when we plug in zero, we get signed inverse of zero, which is zero. And then we get a zero in front of the radical. So we get the Zero Kane. So this is our expression. And let's go ahead and simplify what we can. We see that here we have a two are radical. And then over here, we have a negative radical, so those could be combined. So we just have one left. So one R capital R squared minus R squared in the radical Plus we evaluated side inverse of one. That's a pie over to our square. And then we have minus capital are squares sign in, verse are over, little are over bigger. And this is our final answer. This is the area of the loon.

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