### Video Transcript

The region π in the figure shown is bounded by π¦ is equal to negative sec π₯ and π¦ is negative three. What is the volume of the solid formed when π is rotated about the π₯-axis?

Weβre asked to find the volume of the solid formed when we rotate our region π about the π₯-axis. In order to do this, we need to define the boundaries of our region, sketch the solid so we have a graphical representation of what weβre looking for, sketch and determine the cross-sectional area. And finally, for the volume, we integrate the cross-sectional area between the boundaries. To find the boundaries, letβs look at our region π. Our region is bounded by the function π¦ is equal to negative sec π₯ and the line π¦ is negative three. Remember that sec π₯ is the secant of π₯, which is one of the cos of π₯.

Where our function negative sec π₯ crosses the π¦-axis, π₯ is equal to zero. And when π₯ is zero, negative sec of zero is negative one over cos of zero which is negative one. So our function negative sec π₯ meets the π¦-axis at negative one. We need also to find the π₯-values where our function negative sec π₯ intersects the line π¦ is at negative three. So we need to solve the equation negative sec π₯ is negative three. The negatives cancel each other out so that sec π₯ is equal to three. This means that the inverse sec of three is equal to π₯.

We can either solve this on our calculator or note that one of our cos π₯ is equal to three, which means that cos π₯ is equal to one over three so that the inverse cos of one over three is equal to π₯. And taking the inverse cos of one over three on our calculator, we get π₯ is equal to 1.23096 to five decimal places. Since our function is symmetric about the line π₯ is equal to zero, which is the π¦-axis, then our boundary to the left of π₯ is equal to zero is negative 1.23096. When we come to find our volume via integration, these will be our limit of integration.

Our next step is to try and sketch the solid formed when this region is rotated about the π₯-axis. Our initial region is in the π₯π¦-plane, bounded by the function π¦ is equal to negative sec π₯ and the line π¦ is negative three. And we want to rotate this area about the π₯-axis. If we continue in this way, we end up with a solid that resembles a tyre. Our next step is to look at our cross sections. If we slice vertically through the solid, the cross section is in this shape of a washer. We use the washer method to calculate the volume of the solid when the axis of rotation is not part of the boundary of our initial region. In our case, the axis of rotation is the π₯-axis which is not part of our region π.

Our cross section has an inner radius π
in which is the radius of the hollow centre circle. We also have an outer radius, which is the radius from the axis of rotation to the outer circumference of the solid. To calculate the volume, we move through the solid summing the area of each of the infinitesimally thin cross sections via integration. The area of each cross section is π times the outer radius squared minus the inner radius squared. If we look again at our region π, the outer radius is the distance from the π₯-axis to π¦ is negative three, which is the outside edge of the solid. And thatβs equal to three.

Our inner radius, on the other hand, depends on π₯ because this is the distance from the π₯-axis to the function π¦ is negative sec π₯. So our inner radius is sec π₯. For example, if we take π₯ equal to zero, the inner radius is one because itβs the distance from zero to negative one. Thatβs the absolute value of negative sec π₯ at π₯ is equal to zero. Remember that radius is a distance and distance is always positive. So we can now work out the area of our cross section. Our area, which is a function of π₯, is π times the outer radius squared, which is three squared, minus the inner radius squared, which is sec squared π₯, which is π times nine minus sec squared π₯.

And finally, to find the volume of our solid, we integrate the area between the boundaries of our region. Since π is a constant, we can take this outside. So the volume of our solid is π times the integral between negative 1.23096 and positive 1.23096 of nine minus sec squared π₯ with respect to π₯. Before we integrate, itβs worth noting that if the axis of rotation was vertical, then our area would be a function of π¦. The limit of integration would then be π¦-values. In our case, however, the axis of rotation is the π₯-axis. So our limits are π₯-values.

Since our region is symmetric about π₯ is equal to zero, we can actually take our lower limit as π₯ is equal to zero and multiply the integral by two. This gives us a volume of two π times the integral between zero and 1.23096 of nine minus sec squared π₯ with respect to π₯. We can do this integral using a calculator or by hand. If we perform the integral by hand, since integration is additive, our volume is then two π times the integral between zero and 1.23096 of nine with respect to π₯ minus two π times the integral between zero and 1.23096 of sec squared π₯ with respect to π₯.

Our first integral gives us two π multiplied by nine π₯ evaluated between zero and 1.23096. And for our second integral, we can use the standard integral. The integral of sec squared π₯ with respect to π₯ is tan π₯ plus constant so that our second integral is two π times tan π₯ evaluated between zero and 1.23096. This gives us two π times nine times 1.23096 minus two π times the tan of 1.23096 minus the tan of zero. Nine multiplied by 1.23096 is equal to 11.078635. And the tan of 1.23096 radiance is 2.828427. The tan of zero is equal to zero. Evaluating this gives us a volume of 51.837 to three decimal places. And remember that volume is measured in units cubed.