# The region enclosed by the given curves is rotated about the specified line. Find the volume of the resulting solid.$$y=1 / x, x=1, x=2, y=0 ; \quad \text { about the } x-axis$$

## $$\frac{\pi}{2}$$

#### Topics

Applications of Integration

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

So what if I volume bounded by the equations? Why it was one over x x equals lawn and X equals two as well as, like, zero. So one over X is that rational function. It's actually both parts of this graph here on the left hand, the right. But because I'm bounded by these vertical and horizontal lines, really, the only area I'm looking at is this part here that I colored in. So once I figure out the area, I need to see where I'm rotating this value, and this volume is gonna rotate around the X axis. So that's right here connected to the volume. Now think about as you spun that around. Is there any missing space? And the answer is no. This whole thing would just spend and the totally filled in. I know that because every single part of that shaded area comes right up against the access of revolution. So this is the disc method, and the disc method is volume equals pi times the integral of r squared and then think about whatever access or parallel to whatever access you're working with here. And since it's the X excess, I want my variables to be exes, including on my limits. Right X one x two. Well, the starting ending points for this area are the givens one and two. So I'm gonna put pie in a girl 1 to 2 of our squared DX, which means to really fill in the rest of this. I just need to know what is that? Capital are so thinking, then back to two dimensions. I want to define that area, which is the radius of the shape I'm spinning, going from the access of revolution to the edge of that area. So thinking their upper minus lower, it's that rational function minus the access which is just zero. So I would fill it in as high times, integral of from 1 to 2 of one over X squared DX. So now this is a great time if I can to go through and solved through the integral. So if I were to solve this Enbrel by hand, it's easy enough to do that quickly. I would think of it as what happens when I square it. Will I get one over X squared and then to do the anti derivative, I need to take one away from the degree and divide by that new exponents. So taking one away from the degree make it just one over X. Dividing by the new exponents makes it the negative going from 2 to 1, and then I need to fill in two and one. So if I feel in the upper bound, they get negative 1/2 minus. If I feel in the lower bound, I actually end up with a positive one. So one minus 1/2 leaves behind 1/2 times pi makes it by halfs. Right? Pi over two is my final answer, but definitely a good moment here. If you have the ability to calculate your integral, totally acceptable to, then calculate that and get back to that final answer.

Oklahoma Baptist University

#### Topics

Applications of Integration

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp