### Video Transcript

Use the slicing method to find the volume of the solid whose base is the region inside of the circle π₯ squared plus π¦ squared is equal to one if cross sections taken perpendicular to the π¦-axis are squares.

The question is asking us to use the slicing method to calculate the volume of this solid. And we recall we can use the slicing method to calculate the volume by using the following four steps. First, we need to determine the region which represents the base of our solid. And we can do this by sketching the information given to us in the question. Next, we need to determine the cross sections of our solid, which we will use to calculate the volume. Again, we can do this by sketching the information given to us in the question. Third, we need to find a formula for the areas of our cross sections.

And for this part, we notice the question is telling us to use cross sections which are perpendicular to the π¦-axis. So because our cross sections are perpendicular to the π¦-axis, we want to find our area in terms of π¦. Then, we integrate our area function over an appropriate integral to find the volume of our solid. So letβs start by determining the base of our solid. Weβre told that the region inside of the circle π₯ squared plus π¦ squared is equal to one is the base of our solid. And we know that this represents a circle of radius one centered at the origin. So we found the base of our solid. Itβs a circle of radius one in the π₯π¦-plane centered at the origin.

Now, we want to determine the cross sections of our solid. To do this, weβll start by sketching our base on a three-dimensional set of axes. Next, weβre told that cross sections taken perpendicular to the π¦-axis are squares. And so cross sections taken between a plane perpendicular to the π¦-axis and our solid will give us a square. So we found the shape of our cross section. Next, we need to find a formula for the areas of these cross sections. And since theyβre perpendicular to the π¦-axis, weβll want these in terms of π¦.

To help us find a formula for the area of these cross sections, weβll go back to the sketch of our base. If we sketch our cross section onto the base, we can see that the length of one side of our square can be calculated from the π₯-coordinates. We can find the π₯-coordinate by rearranging the equation of our curve, π₯ squared plus π¦ squared is equal to one. This gives us π₯ is equal to the square root of one minus π¦ squared. And we take the positive square root because this is on the positive side of the π₯-axis. We can do the same on the other side to see that our π₯-coordinate is negative the square root of one minus π¦ squared.

Combining these together, weβve shown that the length of the line of our cross section is two multiplied by the square root of one minus π¦ squared. And since our cross sections are squares, we can calculate the area of our cross sections by squaring this length. This gives us the area is equal to two times the square root of one minus π¦ squared all squared, which we can simplify to give us four minus four π¦ squared. So we found a formula to find the area of our cross sections.

Now all we need to do is integrate this formula for the area over an appropriate interval to find the volume of our solid. And we can see from our sketches that the values of π¦ are varying from negative one to one. So our appropriate interval will be from negative one to one. So we can calculate the volume of our solid as the integral from negative one to one of the areas of our cross sections perpendicular to the π¦-axis, which we showed was four minus four π¦ squared with respect to π¦.

We can evaluate this integral by using the power rule for integration. This gives us four π¦ minus four π¦ cubed over three evaluated at the limits of our integral π¦ is equal to negative one and π¦ is equal to one. Evaluating this at the limits of our integral gives us four times one minus four times one cubed over three minus four multiplied by negative one minus four multiplied by negative one cubed over three. We can evaluate this to give us four minus four-thirds plus four minus four-thirds, which simplifies to give us 16 divided by three.

Therefore, by using the slicing method, we found that the volume of the solid whose base is the region inside of the circle π₯ squared plus π¦ squared is equal to one and whose cross sections perpendicular to the π¦-axis are squares is 16 divided by three.