### Video Transcript

What is the volume of the solid formed when the region bounded by the π¦-axis, π¦ is equal to π to the power of negative π₯, and π¦ is equal to three is rotated around the π¦-axis?

Letβs begin by sketching a diagram of our shape. We know that the curve π¦ equals π to the power of negative π₯ is a reflection of the curve π¦ equals π to the power of π₯ in the π¦-axis. It intersects the π¦-axis at the point with coordinates zero, one. We also have the line π¦ equals three, and itβs bounded by the π¦-axis as well. So, the region shaded is the region weβre going to rotate about the π¦-axis. When we rotate this region about the π¦-axis by 360 degrees, we get a shape that looks a little bit like the one shown. This is called a solid of revolution because itβs a shape thatβs obtained by revolving a region about a line.

In general, we calculate the volume of a solid of revolution formed by rotating a region about the π¦-axis using the formula π is equal to the definite integral between π and π of π΄ of π¦ with respect to π¦, where π΄ of π¦ is the cross-sectional area of the shape perpendicular to the π¦-axis. Now, since the cross-sectional shape is a circle and we know that the area of the circle is ππ squared, we can write this as the integral of π times π₯ squared with respect to π¦. And thatβs because the radius of the circle will be given by the value of π₯ at a given point for π¦. Now in this case, our region is bounded by the horizontal lines π¦ equals three and, actually, π¦ equals one. So, π is equal to three and π is equal to one.

And so, weβre going to integrate between one and three π times π₯ squared with respect to π¦. Now, the problem we have at the moment is that our curve is π¦ equals π to the power of negative π₯. Itβs an equation for π¦ in terms of π₯, so letβs rearrange. We can take the natural log of both sides of this equation. Now, the natural log of π to the power of negative π₯ is just π to the power of negative π₯. So, we find the natural log of π¦ is equal to negative π₯, which means π₯ is equal to negative the natural log of π¦. We then replace π₯ in our integrand with this expression, with negative the natural log of π¦. And we see weβre looking to integrate between one and three π times the negative natural log of π¦ all squared with respect to π¦. But of course, when we square a negative, we get a positive. So, itβs actually π times the natural log of π¦ squared with respect to π¦.

Now, we can actually integrate π times the natural log of π¦ squared with respect to π¦ by using integration by parts more than once. But this question allows the use of a calculator, and we have functions on our calculator that will work this out for us. Typing the definite integral between one and three of π times the natural log of π¦ squared with respect to π¦. Or, alternatively, depending on the functionality of the calculator, we might type the definite integral between one and three of π times the natural log of π₯ squared with respect to π₯. Either way, we get 3.23324 and so on, which, correct to three decimal places, gives us a volume of 3.233 cubic units.