Video Transcript
Find the area of the region
enclosed by the inner loop of the polar curve 𝑟 is equal to one plus two sin
𝜃.
Let’s start by drawing a quick
sketch of what this curve would look like. We could do this by using a
graphing calculator, graphing software, or by plotting some points on the curve. Our curve would look something like
this. We can see that the inner loop of
our curve is this little loop at the bottom here. And so, this orange-shaded region
is the area we’re trying to find. We do, in fact, have a formula for
finding areas of regions enclosed by polar curves. This formula tells us that the area
is equal to the integral from 𝜃 one to 𝜃 two of one-half 𝑟 squared d𝜃. We have been given 𝑟 in the
question. And it’s equal to one plus two sin
𝜃. We just need to find the values of
𝜃 one and 𝜃 two.
Now, the values of 𝜃 one and 𝜃
two will occur at the end points of their closed inner loop. Since this is a loop, this will
happen when the curve crosses over itself. Therefore, this is at this blue
point here. We can see from our sketch of the
curve that this occurs when 𝑟 is equal to zero, since this point is on the
origin. And so, we can find 𝜃 one and 𝜃
two by solving 𝑟 is equal to zero. When 𝑟 is equal to zero, one plus
two sin 𝜃 is equal to zero too. We can rearrange this to find that
sin 𝜃 is equal to negative one-half.
We need to find solutions here
within the range, where 𝜃 is between zero and two 𝜋. We can use a graph of sin 𝜃 to
help us find our solutions. We draw in a line of negative
one-half. Our calculator will give us a
solution of negative 𝜋 by six. However, we’re looking for
solutions between zero and two 𝜋. Since sin is periodic in two 𝜋, we
can see from our sketch that one of the solutions will be 𝜋 by six less than two
𝜋, which is 11𝜋 by six. And the other solution will be 𝜋
by six more than 𝜋, which is simply seven 𝜋 by six.
Therefore, we’ve now found our two
values for 𝜃 one and 𝜃 two. We have that 𝜃 one is equal to
seven 𝜋 by six and 𝜃 two is equal to 11𝜋 by six. We’re now ready to use our formula
for finding the area. We have that the area is equal to
the integral from seven 𝜋 by six to 11𝜋 by six of one-half multiplied by one plus
two sin 𝜃 squared d𝜃. We can distribute the square and
multiply through by one-half. Which gives the integral from seven
𝜋 by six to 11𝜋 by six of one-half plus two sin 𝜃 plus two sin squared 𝜃
d𝜃.
Now, we know how to integrate each
of these terms apart from the sin squared term. However, we can rewrite the sin
squared term using the double-angle formula for cos. We have that cos of two 𝜃 is equal
to one minus two sin squared 𝜃. This can be rearranged to make sin
squared 𝜃 the subject. Which gives that sin squared 𝜃 is
equal to one minus cos of two 𝜃 all over two. And this can be substituted back
into our integral. And we can cancel the factor of two
with the factor of two in the denominator. Leaving us with the integral from
seven 𝜋 by six to 11𝜋 by six of one-half plus two sin 𝜃 plus one minus cos of two
𝜃 d𝜃. And we can group together the one
and the one-half. So our integrand ends up looking
like this. Which we’re now ready to
integrate.
Integrating the first term, three
over two, we simply get three 𝜃 over two. When we integrate the second term
of two sin 𝜃, we get negative two cos 𝜃. And for our final term of negative
cos two 𝜃, we get negative one-half sin of two 𝜃. And we mustn’t forget that this is
between the bounds of seven 𝜋 by six and 11𝜋 by six. Now, we start by substituting in
11𝜋 by six, giving us 11𝜋 by four minus two cos of 11𝜋 by six minus one-half sin
of 11𝜋 by three. Next, we substitute in seven 𝜋 by
six. But you mustn’t forget to subtract
this since seven 𝜋 by six is the lower bound. Giving us negative seven 𝜋 by four
minus two cos of seven 𝜋 by six minus one-half sin of seven 𝜋 by three.
Next, we use the fact that cos of
11𝜋 by six is equal to root three over two. Sin of 11𝜋 by three is equal to
negative root three over two. Cos of seven 𝜋 by six is equal to
negative root three over two. And sin of seven 𝜋 by three is
equal to root three over two. We can now multiply through all of
the terms. And we end up with 11𝜋 by four
minus seven 𝜋 by four minus root three plus root three over four minus root three
plus root three over four. Combining these terms together, we
reach our solution. Which is that the area of the
region enclosed by the inner loop of 𝑟 is equal to one plus two sin 𝜃 is equal to
𝜋 minus three root three over two.
