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# Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.$$y=x, y=\sqrt{x} ; \quad \text { about } y=1$$

## $$V=\frac{\pi}{6}$$

#### Topics

Applications of Integration

### Discussion

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##### Top Calculus 2 / BC Educators ##### Catherine R.

Missouri State University  ##### Kristen K.

University of Michigan - Ann Arbor ##### Michael J.

Idaho State University

### Video Transcript

so I'm gonna find the volume. If I'm rotating around, why equals one? And I've already sketched out the graph for y equals X and then the graph for why I equals the square root of X. So that means the area bounded by. These is this part here. And if we want to take it around the line y equals one. They were thinking about this horizontal line y equals one that were rotating around retaining towards. So because I see that there's this big, empty space that I'm not using here, that means I'm considering this A washer method for the volume and for that washer method we know volume is pi times the integral of capital R squared minus lower case R squared. Now it is still with respect to X, the d X. I use here because why equals one is moving like it's the X axis. So that means I'm using X is my variable. So now I need to decide where did the X values happen? What is the first in the last x for the area and then set up capital R and lower case R. So first I like to look at the X values. And if you know a little bit about these graphs, then you should know hopefully where X and square root of X intersect. If not, you can solve for where these two things intersect. But it happens where X equals your own X equals one. So I know my limits for the integral are zero and one now for the capital are you're trying to think about? What would this volume be if I didn't have any missing pieces? Well, I didn't have anything missing at all. This volume would go all the way up to that line. Y equals one. See, high filled in with no space now well, that would be defined. Upper minus lower as one minus the straight line and straight line is X so capital are gonna call one minus X for lower case R. You're thinking about what am I trying to take away? So all this part here in between that I just colored in is what I want to take away. Well, that goes from one down to the square root graph. So one minus the square Durex. So these are the two things I want to fill in for Capital R and for lower case R on this problem. If I were to go through and then fill it in my hand, I would need to expand thes by no meals. Remember, since its squared your actual think of it like it's multiplying times itself. That would take a little time to multiply out. If you have the option to use a calculator, that's definitely the fastest choice here. If I type this whole thing in for my inner girl, I get 1/6. So my final answer is pi over six for the volume. Oklahoma Baptist University

#### Topics

Applications of Integration

##### Top Calculus 2 / BC Educators ##### Catherine R.

Missouri State University  ##### Kristen K.

University of Michigan - Ann Arbor ##### Michael J.

Idaho State University