### 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 four
minus π₯ squared in the first quadrant and whose cross-sectional slices are squares
perpendicular to the π₯-axis with one edge in the π₯π¦-plane.

We want to find the volume of a
solid whose base is the region under the parabola π¦ is equal to four minus π₯
squared. Weβre looking in the first
quadrant. And where the solid has
cross-sectional slices which are squares perpendicular to the π₯-axis with one edge
in the π₯π¦-plane. The first thing we can do is draw
our parabola π¦ is equal to four minus π₯ squared. And π¦ is equal to zero, our
equation, gives us π₯ squared is equal to four. So, our parabola crosses the
π₯-axis at π₯ is equal to negative two and two. When π₯ is equal to zero, π¦ is
equal to four. So, our parabola intersects the
π¦-axis at π¦ is equal to four.

The base of our solid is the region
under the parabola in the first quadrant. The cross sections of our solid are
squares which are perpendicular to the π₯-axis with one edge in the π₯π¦-plane. To find our volume, we integrate
the areas of the cross section where the limits of integration are defined by our
parabola, the lower limit is zero, and the upper limit is two.

We know that the area of a square
with side length π is equal to π squared. In our case, the length of the
sides of each square is π¦. And the value of π¦ depends on
where the square meets the π₯-axis. The area of each square is,
therefore, π¦ squared. And since π¦ is equal to four minus
π₯ squared, we have the area of the cross section equal to four minus π₯ squared all
squared. So, we have the area of the cross
section as a function of π₯. Our volume is, then, the integral
between zero and two of four minus π₯ squared all squared with respect to π₯.

To make the integral a little
easier, we can expand the bracket. This gives us the integral between
zero and two of 16 minus eight π₯ squared plus π₯ to the power of four with respect
to π₯. The integral of 16 with respect to
π₯ between zero and two is 16π₯ between zero and two. The integral of negative eight π₯
squared with respect to π₯ is negative eight times π₯ to the power of three divided
by three between zero and two. And the integral of π₯ to the power
of four is π₯ to the five divided by five evaluated between zero and two.

Evaluating this, we have 16 times
two minus eight times eight over three plus two to the five over five minus
zero. Evaluating this gives us 256
divided by 15. The volume of the solid whose base
is the region under the parabola π¦ is equal to four minus π₯ squared in the first
quadrant where the cross-sectional slices are squares perpendicular to the π₯-axis
with one edge in the π₯π¦-plane is equal to 256 divided by 15 units cubed.