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

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

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Video Transcript

Let 𝐷 be the shaded region bounded by the graph of 𝑦 equals two times the natural log of one over π‘₯ and the line 𝑦 equals three minus π‘₯, as shown. i) Find the area of 𝐷. ii) Find the volume of the solid generated when 𝐷 is rotated about the horizontal line 𝑦 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 𝐷 is rotated about the 𝑦-axis.

Remember we can find the area 𝐴 of a region bounded by the curves 𝑦 equals 𝑓 of π‘₯ and 𝑦 equals 𝑔 of π‘₯ and the lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏, where 𝑓 and 𝑔 are continuous and 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ for all π‘₯ in some closed interval π‘Ž and 𝑏, by evaluating the integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ minus 𝑔 of π‘₯ with respect to π‘₯.

In our case, π‘Ž and 𝑏 will be the π‘₯-coordinates of the points of intersection of the two lines. And since in the interval we’re looking at the graph of 𝑦 equals three minus π‘₯ sits above the graph of 𝑦 equals two times the natural log of one over π‘₯, we let 𝑓 of π‘₯ be equal to three minus π‘₯ and 𝑔 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, we’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 𝑦 equals three minus π‘₯ and 𝑦 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 we’re interested in are 0.25325 correct to five decimal places and 6.84788. And we’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 𝐷 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, we’ll say that the area of 𝐷 is 10.225 square units.

For part two, we’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 𝑓 of π‘₯ and 𝑔 of π‘₯ about the π‘₯-axis, where once again 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯, is the integral evaluated between π‘Ž and 𝑏 of πœ‹ times 𝑓 of π‘₯ squared minus 𝑔 of π‘₯ squared with respect to π‘₯. We do need to be a little bit careful here that we’re not rotating 𝐷 about the π‘₯-axis. Instead, we’re rotating it about a line parallel to the π‘₯-axis, 𝑦 equals negative five.

In order to deal with this, we’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, we’ll then be actually rotating about the π‘₯-axis rather than the line 𝑦 equals negative five. And this makes a lot of sense since when we shift 𝑦 equals negative five up by five units, we do get the π‘₯-axis. Since πœ‹ 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œ‹ 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, that’s 264.198 cubic units.

And for part three, we’re looking to find another volume. This time though instead of rotating about a line parallel to the π‘₯-axis, we’re rotating about the 𝑦-axis. In this case then, we interchange the roles of π‘₯ and 𝑦. This is sometimes written as the volume is equal to the integral evaluated between 𝑐 and 𝑑, where the limits are now given in terms of 𝑦 of πœ‹ times π‘₯ squared with respect to 𝑦. We’re going to rearrange both of our equations so they’re in terms of 𝑦.

𝑦 equals three minus π‘₯ can be written as π‘₯ equals three minus 𝑦. And when we rearrange 𝑦 equals two times the natural log of one over π‘₯, we get π‘₯ equals 𝑒 to the negative 𝑦 over two. And now, we see that the volume of the solid generator when 𝐷 is rotated about the 𝑦-axis will be given by πœ‹ times the integral evaluated between some new limits 𝑐 and 𝑑 of three minus 𝑦 squared minus 𝑒 to the negative 𝑦 over two squared with respect to 𝑦.

But what are 𝑐 and 𝑑? We’ll use the equation 𝑦 equals three minus π‘₯ to establish the corresponding values of 𝑦 when π‘₯ is equal to 0.25325 and 6.84788. Our value for 𝑑 will be given by 𝑦 equals three minus 0.25325 which is 2.747 correct to three decimal places. Our value of 𝑐 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 𝐷 is rotated about the 𝑦-axis by evaluating πœ‹ times the integral between negative 3.848 and 2.747 of three minus 𝑦 squared minus 𝑒 to the negative 𝑦 over two squared with respect to 𝑦.

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