WEBVTT

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- [Voiceover] We have
two polar graphs here,

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r is equal to 3 sine
theta and r is equal to

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3 cosine theta and what we want to do

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is find this area shaded in blue.

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That's kind of the overlap
of these two circles.

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So I encourage you to pause

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the video and give it a go.

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All right so I assume
you've tried and what's

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interesting here is we're clearly bounded

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by two different polar graphs.

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And it looks like they
intersect right over here.

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If we eyeball it, it looks
like they're intersecting

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at when theta is equal to pi

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over four, and we can verify that.

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Cosine of pi over four
is the same thing as sine

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of pi over four so it is
indeed the case that these two

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things are going to be
equal to each other.

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Their point of intersection happens at

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theta is equal to pi over four.

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And if that wasn't as
obvious, you'd set these

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two equal to each other
and figure out the thetas

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where this actually happened, but here it

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jumps out at you a little more.

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So this is theta is equal to pi over four.

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And so the key is to
realize is that for theta

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being between zero and pi
over four we're bounded

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by the red circle, we're
bounded by r is equal

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to 3 sine theta and then
as we go from pi over four

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to pi over two we're
bounded by the black circle,

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we're bounded by r is
equal to 3 cosine theta.

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So we can just break up our
area into those two regions.

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So this first area right
over here we already know

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is going to be one half
times the definite integral

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from zero to pi over four
of, what are we bounded by,

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3 sine of theta and we're gonna
square that thing d theta.

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That's the orange region
and then this, I guess you

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could say this blue
region right over here is

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going to be one half times
the definite integral

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and now we're going to go from
pi over four to pi over two

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of 3 cosine theta squared d theta.

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That's this region right over here.

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Now one thing that might jump out to you

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is that they're going to be the same area.

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These two circles they are
symmetric around this line.

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Theta is equal to pi over four.

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So these are going to be the same area.

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So one thing that we
could do is just solve for

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one of these and then
double it and we will get

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the total region that we care about.

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So the total area, and you can

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verify that for yourself if you like,

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but I'm just going to say
the total area, I'm just

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going to double this right over here.

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So the total area if I just double this,

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just the orange expression,
I'm going to get the

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definite integral from
zero to pi over four of...

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nine, three squared is nine
sine squared theta d theta.

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And you could evaluate this by hand,

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you could evaluate this by calculator,

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let's evaluate this analytically.

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So sine square theta is
the same thing as one half

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times one minus cosine of two theta.

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That's a trigonometric
identity that we've seen

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a lot in trigonometry class,

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actually let me just write it up here.

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So sine square theta is
equal to one half times

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one minus cosine of two theta.

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So if we replace this
with this it's going to be

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equal to, let's take the one half out,

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so we're going to get
nine halves times the

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definite integral from
zero to pi over four of

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one minus cosine two theta d theta.

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And so this is going to
be equal to nine halves

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antiderivative of one
is theta and let's see

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cosine two theta, it's
going to be negative sine

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of two theta, negative
sine of two theta over two.

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Actually let me just, negative

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one half sine of two theta.

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And you could do u substitution
and do it like this,

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but this you might be
able to do in your head.

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And you can verify the
derivative of sine two theta

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is going to be two cosine
of two theta and then

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you multiply it times
the negative one half,

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you just get a negative
one right over there.

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And so then we are going to evaluate

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this at pi over four and at zero.

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So if you evaluate it
at, well luckily if you

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evaluate this thing at
zero this whole thing

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is going to be zero so we really just

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have to evaluate it at pi over four.

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So this is going to be
equal to nine halves times

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pi over four minus one
half sine of two times

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pi over four is pi over
two, sine of pi over two.

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Sine of pi over two we
already know is one,

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so it is really pi over
four, so this right over

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here is just going to be equal to one.

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So this is going to be nine halves times,

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we could say pi over
four minus one half or

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we could say pi over
four minus two over four.

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So we could write it like
that or we could multiply

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everything out or we could
say this is going to be

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equal to nine pi minus 18
over eight and we are done.