in this video we're computing the area bounded by the limasson r of theta equals one plus two
sine theta and we're trying to exclude the area of the inner loop from this so we're just getting
the area inside the big loop outside the little loop and i want to start by trying to figure out
how exactly this thing is traced and the key to this is to think about the origin when r is equal
to 0 i get 0 equals 1 plus 2 sine theta and if i solve for sine theta i get a condition on theta
where my curve is going to be at the origin so i end up at the origin if sine theta equals
negative one-half and it's most convenient to think of the starting point of this big
outer loop where it begins at the origin and so i need a starting angle here that gives
me a sine that's equal to negative one-half i can go ahead and start my curve from negative
pi over six to make this happen this is better than reflexively starting at theta equals zero
because if i set up an area integral starting at theta equals zero i've cut off a tiny slice
of the start of my lymason and i'm going to have to express another integral to capture that
so starting from negative pi over six theta will increase
r will follow the trace of this limits on and then my big loop terminates right here when
the sine of theta is equal to negative one-half once again and that's at an angle of 7 pi over 6.
after that the r values become negative and my inner loop is traced and this terminates
when i get back to this same angle which i'm now going to call pi over 6 less than 2 pi or
11 pi over six so let's check out an animation of this trace and this animation starts at negative
pi over six so we can see in the animation that we're starting with just the big loop
and then that's followed up with the little loop in the middle and then
the animation repeats itself so what do we mean by the area captured by a polar curve
what we mean by the area inside the big loop is something like this
so when we're computing area in polar coordinates we're sweeping out a radius as we go around
our curve and we're adding up the areas of all those little infinitesimal sectors
so what we're looking at here is the area inside the big loop but we need to subtract the area
inside the small loop to find the area inside of both of these curves we're going to use the polar
area formula and so we found that the big loop was traced out i'll call this a sub b starting
from the angle negative pi over 6 and terminating at the angle seven pi over six and our function r
of theta here is one plus two sine theta and we gotta square that one plus two sine theta
and our little loop is traced out as theta goes from seven pi over 6
to 11 pi over 6 and the function in that integrand is the same thing
so our plan here is to just compute these two integrals and then we're going to take the
difference a b minus a s now because each of these integrals has the same integrand
we may as well just compute the anti-derivative once and then to
evaluate the two different integrals we just evaluate across different limits of integration
so we'll start by squaring 1 plus 2 sine theta and i get 1 plus 4 sine theta
plus 4 sine squared theta d theta now we replace sine squared theta with one half
times the quantity one minus cosine two theta and the one half cancels a factor of two there
when i distribute the two and then combine the constants i end up with a 3
plus 4 sine theta and then minus 2 cosine 2 theta d theta
and finally i'm ready to guess antiderivatives and i get one half times the quantity three theta
minus four cosine theta and then this two cosine two theta term is set up really nicely to guess
the antiderivative because the derivative of the interior function is already sitting there out
in front and i get minus sine 2 theta so that's our general anti-derivative and now we're ready
to evaluate our individual loop integrals so the area of the big loop is one half integral
from negative pi over six to seven pi over six of one plus two sine theta l squared d theta
and we already got the anti-derivative and all we have to do is evaluate that across the limits of
integration so we plug in our upper limit and when i take 3 theta i'm going to go ahead and cancel
a factor of 3 from the denominator of 7 pi over 6. so i end up with 7 pi over 2 and then we have
minus 4 times the cosine of theta so the cosine of 7 pi over 6 that's negative root 3 over 2
minus the sine of 2 theta and i'm going to go ahead and cancel a 2 out so that's the sine
of 7 pi over 3. 7 pi over 3 is 6 pi over 3 in other words 2 pi plus an additional pi
over 3 so the sine of that angle is root 3 over 2. then i'm going to subtract what i get when i
put in the lower limit and again 3 theta when i plug in negative pi over 6 that's going to
give me negative pi over 2 then i have a minus 4 cosine of theta where theta is negative pi over 6.
so the cosine of that angle is root 3 over 2. and finally i have a negative sine of 2 theta so
that's the sine of negative pi over 3 which gives me a negative root 3 over 2. now i'll combine my
pi containing terms so i have a 7 pi over 2 and a minus negative pi over 2 that gives me a plus pi
over 2. so i have 8 pi over 2 which reduces to 4 pi and then i'll handle my root 3 containing terms
in two different steps i'm going to combine these into groups first and in the first one i have
a positive 4 root 3 over 2 minus root 3 over 2. that gives me a positive 3 root 3
over 2. in the second one i have a negative 4 root 3 over 2 plus root 3 over 2 inside these
parentheses that gives me negative 3 root 3 over 2 but there's a minus sign distributed
through this whole term so we get another 3 root 3 over 2. when i distribute the one half i get 2 pi
and then my root 3 containing terms if i combine these it gives me 3 root 3 but then
i multiply by a half so i end up with 3 root 3 over 2 and there's the area of my big loop
now the area of the small loop again that terminates at eleven pi over six and again
we already computed the general anti-derivative so all i have to do is evaluate across the endpoints
and when i plug in 11 pi over 6 and get 3 times that angle i'm going to cancel a factor of 3
this gives me 11 pi over 2 then i have minus 4 times the cosine of 11 pi over six again that's
the same as the cosine of negative pi over six just visualizing the angle here and that's a root
three over two and then minus the sine of twice eleven pi over six well that's the sine of eleven
pi over three and eleven pi over 3 is almost 12 pi over 3 in other words it's almost 4 pi it's just
pi over 3 less than that so this is in the fourth quadrant and angle of pi over 3 from the x axis
the sine of that is negative root 3 over 2. then for my lower limit i put in seven pi over six and
three times that gives me seven pi over two then i have minus four times the cosine of seven pi over
six which is negative root three over two and then minus the sine of twice seven pi over six
and twice seven pi over six is seven pi over three that's pi over three greater
than two pi so i'm in the first quadrant and it's the same as the sine of pi over three
which is root three over 2. now i combine my pi containing terms and i have 11 pi over 2
minus 7 pi over 2 which gives me 4 pi over 2 which is 2 pi
and then i look at my root 3 containing terms again i'll split this into a couple steps
and in the first group i have negative 4 root 3 over 2 plus root 3 over 2 that gives me negative 3
root 3 over 2. and in the second group i have a positive 4 root 3 over 2 minus root 3 over
2 which is 3 root 3 over 2 but there's a minus sign to distribute in other words i
have another negative 3 root 3 over 2. and now i distribute the one half and i get pi
and then my root 3 containing terms i end up with negative 3 root 3
inside these brackets but then i have to multiply by 1 half so end up with negative
3 root 3 over 2. all right so there's the area of the inner loop and our plan
is almost complete the area that we want is the bigger loop minus the smaller loop
so i have 2 pi plus 3 root 3 over 2
minus pi minus 3 root 3 over 2 and this gives me pi and then i have a 3 root 3
over 2 minus negative 3 root 3 over 2 in other words 3 root 3 over 2 plus itself
and that gives me a grand total of pi plus 3 root 3.
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