### Video Transcript

Area Bounded by Polar Curves

In this video, we will learn how to
calculate the area of a region enclosed by one or more polar curves. Weβll be looking at a variety of
examples of how we can find integrals to find areas of this form. Letβs now consider the following
polar curve. And we can call this curve π or π
of π. Letβs suppose that we were asked to
find the area bounded by this curve, π one and the π two. Now, typically, when weβre finding
an area in Cartesian coordinates, the area we will be finding here would be the area
between the curve π, the horizontal π₯-axis, and the π₯-values of π one and π
two. So thatβs this area here.

However, this is not a Cartesian
graph. This is a polar graph. And when we say weβre finding the
area between π one and π two, this, in fact, means the area enclosed by the lines
π is equal to π one, π is equal to π two, and π is equal to π of π. Which gives us this shaded area in
blue here. Which, as we can see, is not equal
to the area we would have found using the Cartesian method, which is in yellow. Therefore, we have a new formula
for finding this area. This formula tells us that the area
is equal to the integral from π one to π two of one-half π squared dπ. Letβs now look at an example of how
this formula can be used.

Find the area of the region bounded
by the polar curve π is equal to two cos two π between π is equal to π by two
and π is equal to three π by two.

Now, we have a formula for finding
area bounded by polar curves. And this formula tells us that the
area is equal to the integral from π one to π two of one-half π squared dπ. Now, weβve been given π in the
question, and itβs equal to two cos of two π. And weβve been asked to find the
area between π is equal to π by two and π is equal to three π by two. Since three π by two is greater
than π by two, we can say that π by two must be equal to π one. And three π by two must be equal
to π two. We can substitute these values of
π one, π two, and π into our equation for the area.

We find that the area is equal to
the integral from π by two to three π by two of one-half of two cos of two π
squared dπ. We can start by expanding the
square term. We end up with the integral from π
by two to three π by two of one-half times four cos squared of two π dπ. And we can cancel the half with one
of the twos from the four, which gives us this. Now, we cannot directly integrate a
cos squared term like this on its own. However, there is a way to get rid
of the square. We need to use the double angle
formula for cos. This formula tells us that cos of
two π is equal to two cos squared π minus one. We can rearrange this to make cos
squared π the subject, which gives us that cos squared π is equal to cos of two π
plus one all over two.

Now, weβre nearly ready to
substitute this formula in. However, in our integrand, we, in
fact, have cos squared of two π. And in our formula, we only have
cos squared of π. This can be solved by simply
multiplying each π in our formula by two. Which gives us that cos squared of
two π is equal to cos of four π plus one all over too. And we can substitute this back
into our integral to obtain the integral from π by two to three π by two of two
multiplied by cos of four π plus one all over two dπ. And now, we can cancel out the
factor of two with the two in the denominator. And now, weβre ready to integrate
our function.

Our first term, which is cos of
four π, is a function within a function. The innermost function is four
π. So we start by differentiating four
π with respect to π to give us four. Since four is a constant, weβre
able to do this integration. However, we mustnβt forget to
divide by this four. And so, we start off with one over
four. Then, we integrate the cos. And so, for our first term, we
obtain one over four sin of four π. And our second term is simply
one. Integrating one just gives us
π. And we mustnβt forget that weβre
integrating between π by two and three π by two. So these are the bounds which we
need to substitute in.

Substituting in three π by two, we
obtain one-quarter sin of six π plus three π by two. And next, we substitute in π by
two, but we mustnβt forget to subtract this. Since π by two is our lower
bound. And so, we obtain minus one over
four sin of two π plus π by two. Now, we know that sin of two π is
equal to zero. And we also know that sin is a
periodic function with a periodicity of two π. Since six π is a multiple of two
π, this means that sin of six π will also be equal to zero. Therefore, the two terms which
weβre left with is three π by two minus π by two.

And itβs from here that we reach
our solution. Which is that the area of the
region bounded by the polar curve π is equal to two cos of two π between π is
equal to π by two and π is equal to three π by two is equal to π.

In some questions, we may not be
given the values of π one and π two, for which we need to find the area
between. Sometimes, the region of which we
need to find the area will simply be described. We will see this in the next
example.

Find the area of the region
enclosed by the inner loop of the polar curve π is equal to one plus two sin
π.

Letβs 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 weβre 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 π one to π two of one-half π squared dπ. We have been given π in the
question. And itβs equal to one plus two sin
π. We just need to find the values of
π one and π two.

Now, the values of π one and π
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 π is equal to zero, since this point is on the
origin. And so, we can find π one and π
two by solving π is equal to zero. When π is equal to zero, one plus
two sin π is equal to zero too. We can rearrange this to find that
sin π is equal to negative one-half.

We need to find solutions here
within the range, where π is between zero and two π. We can use a graph of sin π 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, weβre 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, weβve now found our two
values for π one and π two. We have that π one is equal to
seven π by six and π two is equal to 11π by six. Weβre 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 π squared dπ. 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 π plus two sin squared π
dπ.

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 π is equal
to one minus two sin squared π. This can be rearranged to make sin
squared π the subject. Which gives that sin squared π is
equal to one minus cos of two π 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 π plus one minus cos of two
π dπ. And we can group together the one
and the one-half. So our integrand ends up looking
like this. Which weβre now ready to
integrate.

Integrating the first term, three
over two, we simply get three π over two. When we integrate the second term
of two sin π, we get negative two cos π. And for our final term of negative
cos two π, we get negative one-half sin of two π. And we mustnβt 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 mustnβt 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 π is equal to one plus two sin π is equal to
π minus three root three over two.

Next, we will learn how to find the
area enclosed by two polar curves. Letβs consider the following
scenario.

Letβs say weβve been given two
polar curves π one and π two and weβve been asked to find the region enclosed by
π one and π two between π one and π two. So thatβs this shaded region
here.

We can find a formula for this area
by applying the formulas which we already know for finding areas of regions bound by
polar curves. We know that we can find the area
enclosed by π two, π one, and π two, which is this area here. And itβs equal to the integral from
π one to π two of one-half π two squared dπ. We also know how to find the area
enclosed by π one, π one, and π two, which is this area here. We know that itβs equal to the
integral from π one to π two of one-half π one squared dπ. Now, we can see that this yellow
area, which weβre trying to find, must be equal to the area encased by π one
between π one and π two minus the area encased by π two between π one and π
two. So thatβs the integral between π
one and π two of one-half π one squared dπ minus that integral between π one and
π two of one-half π two squared dπ.

Using integral properties, weβre
able to combine these integrals since they have the same bounds, to give the
integral between π one and π two of one-half π one squared minus one-half π two
squared dπ. We can factor out the half to form
our final formula. Which is that the area between two
curves π one and π two between π one and π two is equal to the integral between
π one and π two of one-half π one squared minus π two squared dπ. Letβs note that this only works
when π one is greater than or equal to π two for π-values between π one and π
two.

Now that we formed this new
formula, letβs look at an example of how it works.

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.

We have now seen a variety of
examples and techniques which we can use to find the areas bound by polar
curves. Letβs cover some key points of the
video.

Key Points

The area enclosed by the polar
curve π is equal to π of π, π one, and π two is given by the integral from π
one to π two of one-half π squared dπ. The area enclosed by two polar
curves π one which is equal to π of π and π two which is equal to π of π, π
one, and π two. Where π one is greater than or
equal to π two for π between π one and π two. Is given by the integral from π
one to π two of one-half of π one squared minus π two squared dπ.