💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

Get the answer to your homework problem.

Try Numerade Free for 7 Days

Like

Report

Prove the formula $ A = \frac{1}{2} r^2 \theta $ for the area of a sector of a circle with radius $ r $ and central angle $ \theta $. [Hint: Assume $ 0 < 0 < \frac{\pi}{2} $ and place the center of the circle at theorigin so it has the equation $ x^2 + y^2 = r^2 $.Then $ A $ is the sum of the area of the triangle $ POQ $ and the area of the region $ PQR $ in the figure.]

$$\frac{1}{2} r^{2} \theta$$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 3

Trigonometric Substitution

Integration Techniques

Oregon State University

University of Nottingham

Idaho State University

Boston College

Lectures

01:11

In mathematics, integratio…

06:55

In grammar, determiners ar…

02:06

Prove that the area of a s…

01:26

PROOF Prove t…

01:18

Geometry Show that the are…

01:46

Area of a circular sector …

02:25

(a) Show that $r=2 \sin \t…

10:00

(a) Given a circular secto…

04:03

Find the area of the indic…

01:48

Group Activity Area of a S…

03:59

Areas of circles Use integ…

04:40

here we like to prove the formula A equals one half r squared data is the area of the sector of a circle with a radius R and central angle data. And so here we have a picture to represent that we can see that this sector consist of a triangle here and then also said this other region on the right, which we can just call P Q R. So to find the area of the sector, which is go ahead and find area of the triangle area of P Q. R. Add them together. Hold it. So here's a suggestion that we can just take the circle X squared plus y squared equals R Square. That just means that we're centering the circle at the origin. So this blue line out here is coming from a circle of radius little R senator at the origin. So we'LL get to that in a second. First, let's deal with the easier shape triangle. We know a formula for this one half base times height, so also because we're centering the circle at the origin, we see that this red line right here this is the radius of the blue circle and as a little r so radius equals little R. Let's call this base of the triangle from ODA Q. Let's just call that length B and then from Cuba. P. That's the height of the triangle is called that each So the area of the triangle. So Ariel, let's say ace of Triangle Triangle area, one half based time site. Now we could actually find being H in terms of our and data using the right triangle. And that will be fine tohave things in terms of our data, because that's the formula we want. We want ours and data. So let's do that. Using this triangle coastline data is be over are adjacent over. I promise. Good and saw that for B. Arcosanti data. Similarly sign H O R R. Opposite over Heaven knows and then solve that for H unemployed Everything in and we get one half b h. So we get our from each of these terms. So multiply those ours together our square and then sign times goes on. So that's the area of the triangle. There's no more to do here. Well, we could come back. We can rewrite this, but there's no need to simplify it. Now let's go ahead and find the area of the other region, so we're halfway there. We have this region p Q R to deal with. So this is just the area under the blue circle. But it's not the area under the whole blue circles, just this region that's in the right of the triangle, So this will affect our limits of integration. So also, we see that this is the upper half of the circle. So let's go ahead and solve the equation of the circle for why. So we have y equals square root R squared, minus X squared in the radical. So this is the upper half of the circle, which you can write as a function. So it's a circle out here. And then when we evaluate the integral so area of the region, let's call it P Q. R. So it's a region over here on the right. This will just be the area underneath this blue circle curve. So we have. It will be the integral for area area under this function, our circle. But now we have to deal with the limits so the lower limit should be over here a tte this point Q and so from the origin we already found this link be here. We know that bee is our coastline data. So that is our lower limit of integration depends on data. Different day knows they're going to give you different values of Q. So here's B or weaken right as our co signed data and then the upper limit is all the way out here to the edge. But this length out here, all the way to the circle, that's just little are because that's just the radius of the circle. So that is our upper limit of integration. Now we have an integral to evaluate here. Looks like we could do it. Tricks up here you can do X equals R signed Ada. You can also do X equals R co sign and actually let me not use the letter data here because that is already being used. It is shows up in our limit. That's the same data that was originally in the triangle. We can also try co sign you. This will work similarly and so we can choose either one. We see that there's a co sign in the limit of integration. Perhaps that might suggest that this thing can be easier to simplify if we use the second tricks up. But either one is fine. Let's go ahead and choose the second one. So let's go on to the next page running out of room here X equals our cause. I knew Take the derivative to get that negative. Our sign You deal. Look. So the mineral that we had in the previous page Oh, every under the blue curve, the circle. Okay. And then now we go ahead into our substitution. Let's not worry about the limits right now of integration. Let's just call them you want. And you too. They will change. Usually when you do a substitution. Our square minus r squared co sign squared you and then we have DX from our tricks up. And this gave us negative are sign you do. Let's just go ahead and simplify this. So the radical deal with this on the side Weaken pulling our square right. This is one minus co signed square squared of our square is just the radius are And then we get the square root of science Where using that that that Gwen identity for signing co sign. And then this is just our sign. So we have a negative out here in the front, and in the second term, so's pull that out. Then we have our and then we have another are here, so are square. You won you too. And then we have signed times time science. Where? Let's use a half angle identity Anytime. You haven't even power of sign Chenery best to use this identity. Havinga So it's quite and pull that too negative. R squared over two. You won you two one minus co sign to you to you. Morning. Gotta let me come over here to the right. So excuse me to the left. Somebody bring this from the right back over. Negative r squared over two. We can evaluate this integral, integral of one is just you. Then we have signed to you over too, you know? And this is where we have an issue. Because we left this. We didn't find the new limits of integration, so it's fine. You could try to find them. Sometimes it's easier, but in this case, we could just use the triangle, go back, write everything in terms of X and then use our original limits of integration which was our coastline data and our So here let's use the triangle first simplify this term here, let's use that double angle formula for sign. You going to sign you coz I knew on top two on the bottom and then cancel those twos So negative r squared over two You minus sign you coz I knew Now he's a trip substitution or really firm Our trick substitution will have a triangle We'LL get your other triangle Then we could find sign you on course I knew So this is what we'LL be using up here Exes are co sign you but Kosi eggs over our That's enough for a triangle, right triangle There's our you angle and then we have X over our fur co sign adjacent over hypotenuse Okay, then we have some other side here Let's just call it a tch by both agree anteroom and saw for the height h So now we could find Sinan co sign. So from the previous line of integration, we had our split over too. You minus sign you coz I knew and we can evaluate these terms now we have three terms to deal with the first term issue. You could solve that from this equation up here by taking our co sign on both sides Coasts in verse, XO over our. So that gives us you and then sign a co sign. We can get right from the picture here. So coming back to our inner rule, we're ready to go back in terms of the variable X co sign inverse X over our minus. Now we have scientists go side. So sign is h over our some You put that down here that's radical over our and then we have co sign which is X over our Now finally, let's go ahead and distribute this r squared over to this has been hanging in front of the integral for awhile now. So push this toe both terms and notice the cancellation of the are squares on the second term. And also we can go back to our original limits are co sign in our So let's go ahead and plug this in. So first we'll plug in capital or the small are into the ex. When we do that, we're going to close Inverse of Erin Grant. So let's start writing this over here. Some plugging small. Our first are over. Our is one. So we have co sign inverse of one. And we could evaluate this. This is your zero. And then we have a minus. Sign here. I should be careful. I should have actually distributed already. So when I distributed the negative r squared over two, I got a positive. On the second term, he r squared has canceled and I just had a one half and then I have X r squared minus explorer. So if I plug in, little are and for X here, I'm going to get radical of zero, which is zero. So they're gonna zeros her. So it looks like both terms go to zero and we plug it in R for X. Now we'LL go ahead and plug in our coastline, so I have to subtract some running out of room here so we subtract. The first term is negative. So we have a negative r squared over two coastline in verse. Our co signed data over our so that that is a negative out here in front of the So that's a denominator. And then we do have a negative there from of the R squared inside apprentices and then the second became a plus after we simplified. And then we have one half our co signed data square root R squared minus R squared coastline data. So this is all from plugging in our co signed data for X. So let's go ahead into the next page and evaluate this. We're almost done. So we have r squared over two cosign verse. We cancel those ours. We just can't co sign data and then we have one half our course idea and it's simplifying inside the radical R squared science where are split over too cosign hearse. Of course, San is just coast. It was just the angle data it is. And then we have one half our times are R squared, coz i data scientist, but you might recognise the second term. This was also the area of the triangle that we computed on the first page. So now is to find the final area a equals area of the triangle, plus the area of P Q. R. Well, from the first page, we have one half our square scientist coastline data and then we have this expression up here R squared over to data minus one half R squared sine data Cho Santa. And then we can cancel thes one half r squared sine cosine terms. And we're just left over with one half r squared data. And who's the formula that we intended to prove? And there it is.

In mathematics, integration is one of the two main operations in calculus, w…

In grammar, determiners are a class of words that are used in front of nouns…

Prove that the area of a sector of a circle of radius $r$ associated with a …

PROOF Prove that the area of a circular sector of radius $r$ wi…

Geometry Show that the area of a circular sector of radius $r$ with central …

Area of a circular sector Prove that the area of a sector of a circle of rad…

(a) Show that $r=2 \sin \theta+2 \cos \theta$ is a circle.(b) Find the a…

(a) Given a circular sector with radius $L$ and central angle $\theta$ (see …

Find the area of the indicated region.Inside of $r=2$ and outside of bot…

Group Activity Area of a Sector A sector of $a$ circle (shaded in the figure…

Areas of circles Use integration to show that the circles $r=2 a \cos \theta…

Find the area of the indicated region.Inside of $r=2 \sin 2 \theta$ and …