### Video Transcript

Find the area of the region that
lies inside the polar curve π squared is equal to eight cos of two π but outside
the polar curve π is equal to two.

In order to find the regions that
lie outside π is equal to two but inside π squared is equal to eight cos of two
π, weβll need to find the points of intersections of the two curves. Therefore, we can start by doing
this. We need to solve π squared is
equal to eight cos of two π and π is equal to two, simultaneously. We can substitute π equals two
into the first equation. This will give us that four is
equal to eight cos of two π. We end up with cos of two π is
equal to one-half. Now, weβre looking for the
solutions for which π is between zero and two π. Therefore, these will be the
solutions for which two π is between zero and four π. These solutions within this range
are that two π is equal to π by three, five π by three, seven π by three, or
11π by three. And so, we find that our points of
intersection are that π is equal to π by six, five π by six, seven π by six, and
11π by six.

Letβs now draw a sketch of π is
equal to two. This will simply be a circle with a
radius of two. Next, we can mark on our points
where the two curves will intersect. The first point is at π by
three. The second is at five π by
three. The third is at seven π by
three. And the fourth intersection point
is at 11π by three. Now, we can attempt to sketch the
curve of π squared is equal to eight cos of two π. We can do this by first rewriting
the curve as π is equal to the square root of eight cos of two π. Next, we can find some points on
the curve by inputting some values of π.

We have that at π is equal to
zero, cos of zero is equal to one. Therefore, π is equal to the
square root of eight, which is also equal to two root two. At π is equal to π by two, we
have cos of two π is equal to cos of π, and cos of π is equal to zero. Therefore, at π is equal to π by
two, π is equal to zero. Continuing this on, at π is equal
to π, we have that π is equal to two root two. And at π is equal to three π by
two, π is equal to zero. And we can add these points to our
graph. We now have seven points of our
curve drawn on our graph. And so, we can draw a rough sketch
for what it should look like. As we can see, it sort of looks
like a figure of eight.

Now, this sketch really helps us to
find the regions of which weβre trying to find their areas. So itβs the regions which lie
outside the curve π is equal to two but inside our curve of π squared is equal to
eight cos of two π. There are, in fact, two regions
which weβre interested in, which is these two regions here. We have a formula for finding the
area of regions between two polar curves. And this formula tells us that the
area is equal to the integral between π one and π two of one-half π one squared
minus π two squared dπ. For the regions which weβre
concerned with, π squared is equal to eight cos of two π is greater than or equal
to π is equal to two. Therefore, π one squared is equal
to eight cos of two π, and π two is equal to two.

We can start by considering the
area on the left of our diagram. This will be equal to the integral
from five π by six to seven π by six of one-half of eight cos of two π minus four
dπ. Next, letβs consider the area of
the region on the right of our graph. This region is between the angles
of 11π by six and π by six. However, if we were to integrate
between 11π by six and π by six, we would be jumping from an angle of two π to an
angle of zero. Which would occur when we cross the
horizontal access. And, of course, we cannot do
this. We can instead change the angle of
11π by six to negative π by six. Since on our graph, negative π by
six is equal to 11π by six.

We obtained that the area on the
right is equal to the integral from negative π by six to π by six of one-half of
eight cos of two π minus four dπ. We can simplify both of these
integrands. And now, weβre ready to
integrate. We have that the integral of four
cos of two π minus two is two sin of two π minus two π. Next, we substitute in our upper
and lower bounds. Now, we can use that sin of seven
π by three and sin of π by three are both root three over two. And that sin of five π by three
and sin of negative π by three are both negative root three over two. Next, we can expand everything
here. And then, for our final step, we
just simplify. Which gives us a solution that the
area of the region that lies inside π squared is equal to eight cos of two π but
outside π is equal to two is four root three minus four π by three.