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

## $$\frac{16 \pi}{15}$$

#### Topics

Applications of Integration

### Discussion

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##### Top Calculus 2 / BC Educators   ##### Kristen K.

University of Michigan - Ann Arbor

### Video Transcript

I'm gonna find volume of these crops. My A graph I have drawn is X equals Y squared and then X equals one. So the area bounded by these two equations. Is this part here? Once I know the area that I have to decide or I have to be, you know, looking at my prompt to see where the revolution needs to happen. So in this case, let's revolving around X equals one. So because all of that area comes right up against the Axis revolution, there's no empty space between what I shaded and revolved. Then this is a disc set up. So for a disc set up, volume equals pi times the integral of r squared. And then whatever very well defined your access while here X equals one is running along like it's the Y axis. See how there's a parallel. So I'm gonna use why, as my variable again cause it's parallel as my revolution is, X equals one parallel to the y axis. Since I'm using why I also use why, for the limits be really careful. All those terms match so on the limits and the r and as your definition for your integration Very well. So now I need to decide what the highest why value is and what the lowest y value is. We know that at this highest and lowest point for why we know X equals one. So actually we use that X equals one to solve the equation given. Then we confined. Why is cluster minus one member? As you take the square root, you get the plus or minus there. So the lowest. Why here happens at negative one and then the highest. Why? For the upper bound is positive one. So we'll put those endure integral negative 1 to 1. Then I need to define what is our our is that radius. It's the revolution happening. So it goes from the axis revolution all the way to the edge of the area you shaded in. So the access of revolution is one minus. The area shaded in is defined by the proble. So why squared? I do one minus y squared because just like with area in two dimensions, I'm thinking right minus left there. The good news is I'm squaring the whole thing as I filled into my equation. So I should be able to get a positive answer that no matter what, but I like to think right minus left or upper minus lower, as I'm thinking about the two dimensional area for capital are or for lower case R. So now that I filled the whole thing in, I'm ready to calculate this. And as I calculate this in a girl, I get 16/15 multiplying by the pie out front and at 16 pi over 15 as my final answer. Oklahoma Baptist University

#### Topics

Applications of Integration

##### Top Calculus 2 / BC Educators   ##### Kristen K.

University of Michigan - Ann Arbor