### Video Transcript

Find the area enclosed by the graph
of π₯ equals nine minus π¦ squared, the π¦-axis, and the lines π¦ equals negative
three and π¦ equals three.

We notice that, in this example,
the area weβve been asked to find is enclosed by a curve with the equation π₯ equals
some function of π¦. The area is also bounded by the
π¦-axis rather than the π₯-axis and two lines with equations of the form π¦ equals
some constant, which are horizontal rather than vertical lines.

Now we could repeat the process
from first principles to see how we can use integration to find this area, rather
than an area bounded by a curve in the form π¦ equals π of π₯ and the π₯-axis. But actually, it is as simple as
swapping π₯ and π¦ around. In order to find the area enclosed
by a curve in the form π₯ equals π of π¦, the π¦-axis, and the two horizontal lines
π¦ equals π and π¦ equals π, we evaluate the definite integral from π to π of π
of π¦ with respect to π¦.

In this case then, weβre evaluating
the definite integral from negative three to three of nine minus π¦ squared dπ¦. Notice that everything in the
integral is in terms of π¦, not π₯. To integrate powers of π¦ not equal
to negative one, we increase the power by one and divide by the new power, giving
nine π¦ minus π¦ cubed over three evaluated between negative three and three. We then substitute our limits,
giving nine multiplied by three minus three cubed over three minus nine multiplied
by negative three minus negative three cubed over three. Thatβs 27 minus nine minus negative
27 plus nine, which is equal to 36.

So we find that the given area is
36 square units. And in this problem, weβve seen
that, in order to find an area enclosed by the graph of π₯ equals some function of
π¦, the π¦-axis, and two horizontal lines, we can just perform a definite integral
with everything in terms of π¦ rather than everything in terms of π₯. Notice also that, in this question,
we werenβt concerned that some of the area lay below the π₯-axis as this time we
were integrating with respect to π¦. And all of the area lay on the same
side of the π¦-axis.