### Video Transcript

Use the slicing method to find the
volume of the solid whose base is the region under the parabola π¦ is equal to nine
minus π₯ squared and above the π₯-axis and where slices perpendicular to the π¦-axis
are squares.

Weβre asked to find the volume of a
solid whose base is defined by the parabola π¦ is equal to nine minus π₯ squared and
above the π₯-axis. The slices of the solid are
perpendicular to the π¦-axis and are squares. Letβs first draw the base region of
our solid. Our parabola is π¦ is equal to nine
minus π₯ squared so that when π¦ is equal to zero, π₯ squared is equal to nine so
that π₯ is equal to positive or negative three. When π₯ is equal to zero, π¦ is
equal to nine. And this is where our function
crosses the π¦-axis.

The region under this curve and
above the π₯-axis forms the base of our solid. Our cross sections are squares. And these are perpendicular to the
π¦-axis. The length of the sides of each
square is two π₯. And the value of π₯ depends on
where the square crosses the π¦-axis. The area of a cross-sectional
square is, therefore, two π₯ times two π₯, which is four π₯ squared. The volume of our solid, however,
is an integral along the π¦-axis between π¦ is zero and nine of the area of the
cross-sectional squares as a function of π¦. So, we need to convert our area
into a function of π¦.

We know that π¦ is equal to nine
minus π₯ squared, which gives us π₯ squared equal to nine minus π¦. Our area as a function of π₯ is
four π₯ squared so that four π₯ squared is equal to nine minus π¦ times four. In other words, four π₯ squared is
equal to four times nine minus π¦ so that our area, in fact, is four times nine
minus π¦. And we can now work out our volume,
which is the integral between zero and nine of four times nine minus π¦ with respect
to π¦.

We can take the four outside since
this is a constant multiple. We know that the integral of nine
with respect to π¦ is nine times π¦. And we know that the integral of
negative π¦ with respect to π¦ is negative π¦ squared over two. So, then, our volume is four times
nine π¦ minus π¦ squared over two evaluated between zero and nine. This is four times nine times nine
minus nine squared over two minus zero. That gives us four times 81 minus
40.5, which is 162. The volume of the solid whose base
is the region under the parabola π¦ is nine minus π₯ squared and above the π₯-axis
where slices are perpendicular to the π¦-axis and are squares is, therefore, 162
units cubed.