# Video: AP Calculus AB Exam 1 β’ Section II β’ Part A β’ Question 1

Let π· be the shaded region bounded by the graph of π¦ = 2 ln 1/π₯ and the line π¦ = 3 β π₯, as shown. i) Find the area of π·. ii) Find the volume of the solid generated when π· is rotated about the horizontal line π¦ = β5. iii) Write, but do not evaluate, an integral expression that can be used to find the volume of the solid generated when π· is rotated about the π¦-axis.

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### Video Transcript

Let π· be the shaded region bounded by the graph of π¦ equals two times the natural log of one over π₯ and the line π¦ equals three minus π₯, as shown. i) Find the area of π·. ii) Find the volume of the solid generated when π· is rotated about the horizontal line π¦ equals negative five. iii) Write, but do not evaluate, an integral expression that can be used to find the volume of the solid generated when π· is rotated about the π¦-axis.

Remember we can find the area π΄ of a region bounded by the curves π¦ equals π of π₯ and π¦ equals π of π₯ and the lines π₯ equals π and π₯ equals π, where π and π are continuous and π of π₯ is greater than or equal to π of π₯ for all π₯ in some closed interval π and π, by evaluating the integral between π and π of π of π₯ minus π of π₯ with respect to π₯.

In our case, π and π will be the π₯-coordinates of the points of intersection of the two lines. And since in the interval weβre looking at the graph of π¦ equals three minus π₯ sits above the graph of π¦ equals two times the natural log of one over π₯, we let π of π₯ be equal to three minus π₯ and π of π₯ be equal to two times the natural log of one over π₯. To find the π₯-coordinates of the points of intersection of our two lines, we need to solve them simultaneously. In other words, weβll find the solutions to the equation three minus π₯ equals two times the natural log of one over π₯.

We could use our graphical calculators to perform this step. Alternatively, we could use our graphical calculators to plot the graphs of π¦ equals three minus π₯ and π¦ equals two times the natural log of one over π₯ and find the π₯-coordinates for where they intersect. Either way, we find the π₯-values weβre interested in are 0.25325 correct to five decimal places and 6.84788. And weβve chosen here five decimal places because we want to be as accurate as possible.

And we now see that we can find the area of π· by evaluating the integral between 0.25325 and 6.84788 of three minus π₯ minus two times the natural log of one over π₯ with respect to π₯. When we evaluate this using our calculators, we get 10.22539 and so on. Correct to three decimal places then, weβll say that the area of π· is 10.225 square units.

For part two, weβre looking to find the volume. This time we use a formula which is sometimes referred to as the washer method. It says that the volume formed by rotating the difference between two functions π of π₯ and π of π₯ about the π₯-axis, where once again π of π₯ is greater than or equal to π of π₯, is the integral evaluated between π and π of π times π of π₯ squared minus π of π₯ squared with respect to π₯. We do need to be a little bit careful here that weβre not rotating π· about the π₯-axis. Instead, weβre rotating it about a line parallel to the π₯-axis, π¦ equals negative five.

In order to deal with this, weβre going to subtract negative five from each of our equations. This is essentially going to shift the lines up by five units. And in doing so, weβll then be actually rotating about the π₯-axis rather than the line π¦ equals negative five. And this makes a lot of sense since when we shift π¦ equals negative five up by five units, we do get the π₯-axis. Since π is a constant, we can take that outside our integral sign and we can see that the volume is going to be given by π times the integral evaluated between 0.25325 and 6.84788 of three minus π₯ minus negative five squared minus two times the natural log of one over π₯ minus negative five squared. If we evaluate this on our graphical calculator, we get 264.19837 and so on. Correct to three decimal places, thatβs 264.198 cubic units.

And for part three, weβre looking to find another volume. This time though instead of rotating about a line parallel to the π₯-axis, weβre rotating about the π¦-axis. In this case then, we interchange the roles of π₯ and π¦. This is sometimes written as the volume is equal to the integral evaluated between π and π, where the limits are now given in terms of π¦ of π times π₯ squared with respect to π¦. Weβre going to rearrange both of our equations so theyβre in terms of π¦.

π¦ equals three minus π₯ can be written as π₯ equals three minus π¦. And when we rearrange π¦ equals two times the natural log of one over π₯, we get π₯ equals π to the negative π¦ over two. And now, we see that the volume of the solid generator when π· is rotated about the π¦-axis will be given by π times the integral evaluated between some new limits π and π of three minus π¦ squared minus π to the negative π¦ over two squared with respect to π¦.

But what are π and π? Weβll use the equation π¦ equals three minus π₯ to establish the corresponding values of π¦ when π₯ is equal to 0.25325 and 6.84788. Our value for π will be given by π¦ equals three minus 0.25325 which is 2.747 correct to three decimal places. Our value of π will be given by three minus 6.84788 which is negative 3.848 correct to three decimal places. And this means we can find the volume of the solid generated when π· is rotated about the π¦-axis by evaluating π times the integral between negative 3.848 and 2.747 of three minus π¦ squared minus π to the negative π¦ over two squared with respect to π¦.

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