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# Worked example: Area between two polar graphs

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.D (LO)
,
CHA‑5.D.1 (EK)
,
CHA‑5.D.2 (EK)

## Video transcript

we have two polar graphs here R is equal to 3 sine theta and R is equal to 3 cosine theta what we want to do is find this area shaded in blue that's kind of the overlap by these two the overlap of these two circles so encourage you to pause the video and give it a go all right so I've assumed I assume you've you've tried and what's interesting here is we're clearly bounded by two different polar graphs and it looks like they intersect right over here if we eyeball it it looks like they're intersecting at when theta is equal to PI over 4 and we can verify that cosine of PI over 4 is the same thing as sine of PI over 4 so it is indeed the case that these two things are going to be equal to each other they're going to be they're going to be their point of intersection happens at theta is equal to PI over 4 and if that was wasn't as obvious you would set these two equal to each other and figure out the Thetas where this actually happened but here it's a jumps out at you a little bit more so this is theta is equal to PI over 4 and so the key is to realize is that for theta theta being between 0 and PI over 4 we're bounded by the red circle we're bounded by 3 R is equal to 3 sine theta and then as we go from PI over 4 to PI over 2 we are bounded by the black circle we're abounded by R is equal to 3 cosine theta so we could just break up our air into those two regions so this first area right over here we already know is going to be 1/2 times the definite integral from 0 to PI over 4 0 to PI over 4 of what are we bounded by 3 sine of theta 3 sine of theta and we're going to square that thing D theta that's the orange region and then this I guess you could say this blue region right over here is going to be 1/2 times the definite integral and now we're going to go from PI over 4 to PI over 2 to PI over 2 three cosine theta squared D theta that's this region right over here now one thing that might jump out you is that they're going to be the same area that these two circles they are i guess you say symmetric around this line theta is equal to pi over four so these are going to be the same area so one thing that we could do is just solve for one of these and then double it and we will get this the total region that we care about so the total area and you can verify that for yourself if you like but I'm just going to say that the total area I'm just going to double this right over here so the total area if I just double this just the the orange expression I'm going to get the definite integral from zero to PI over four of let's see let me just of nine three squared is nine sine squared theta D theta and you could evaluate this by hand you could evaluate this calculator let's how to evaluate this analytically so sine squared theta is the same thing as one half times 1 minus 1 one half times one minus cosine of two theta that's a trigonometric identity that we've seen a lot in trigonometric and AMA tree class actually let me just write it up here so sine squared theta is equal to one half times one minus cosine of two theta so if we replace this with this it's going to be equal to this is going to be equal to let's take the one half out so we're going to get nine halves times the definite integral from zero to PI over four of one minus cosine two theta D theta and so this is going to be equal to nine halves nine halves antiderivative of one is Theta and let's see cosine two theta it's going to be it's going to be negative sine of two theta did I do that negative negative sine of two theta over two over two actually oh maybe so negative 1/2 sine of 2 theta and you could do you substitution and do it like this but this you might be able to do it in your head you can verify the derivative of sine of 2 theta is going to be 2 cosine of 2 theta and then you multiply it times a negative 1/2 you want 5 times negative 1/2 you just get a negative 1 right over there and so then we are going to evaluate this at PI over 4 and at 0 so if you evaluate it at well lucky if we go see if we evaluate this thing at 0 this whole thing is going to be 0 so we really just have to evaluate it at PI over 4 so this is going to be equal to 9 halves times pi over 4 minus minus 1/2 sine of 2 times pi over 4 is PI over 2 sine of PI over 2 sine of PI over 2 we already know is 1 so it is really PI over 4 so this this right over here is just going to be equal to 1 so this is going to be 9 halves times we could say PI over 4 minus 1/2 or we could say PI over 4 minus 2 over 4 so we could write it like that or we could multiply everything out or we could say this is going to be equal to this is going to be equal to 9 PI minus 9 PI minus 18 over over 8 and we are done
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