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The region enclosed by the given curves is rotated about the specified line. Find the volume of the resulting solid.

$$x=2 y-y^{2}, x=0 ; \quad \text { about the } y-axis$$

$$

\frac{16 \pi}{15}

$$

Applications of Integration

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Harvey Mudd College

University of Michigan - Ann Arbor

I want to find the value and I'm looking at the equation. X equals two. Why minus y squared as well as X equals euros area. But founded here Is this part of coloring in because X equals you're we know is that why access? But I know it's really tough to graph in X equals equation, especially when maybe I don't have a reference of what it looks like. So I did wanna show over here one way that even just now as I sketched it out, I went through, injured the graph, and I just picked numbers to plug in for why? And then just be really aware, As you're plotting them, you either want to, you know, rewrite it as X comma y as your coordinate or just remember that you're plotting the Y value First is your height and then finding the X. So I just picked easy things a plug in for why and then just kind of quickly calculated that, you know, I fill in zero for both and my answer for X is still zero. If I feel in one for both wise and I just went through and found those points so I can have to help me draw that graph. So talking again about volume I'm looking about that part I shaded in and blue and we're gonna rotate this around the y axis. So the Y axis is that vertical line X equal zero? Well, the Y axis there is no area that I need to take away. There's no volume I need take away as this rotations happening. Every single part of the area comes right up against that access of revolution. So this is a disc set up, and for the disc set up, we know volume is pi times the integral of r squared. And then whatever access you're working with us helping you define that variable in this case D y, or if it was a line parallel to the Y access, it would be D y. And that means my bounds need to also be y values. So as I feel this in, I need to find my y values because I made this t chart over here this table of values I actually know that highest and lowest y value already looking at the table. My wife values that happened on the axis are zero and two. So for those bounds, I'm gonna use zero into So then I just need to figure out how I want to fill in our to complete the integral set up well for capital are we are looking at the radius here as we're rotating it around. So going from the access to the edge, well, going from the access to the edge is that curved graph to why minus y squared and then minus the access is just zero. Because it's that vertical line there. So really, I don't need that minus You're a part of it. I'm just gonna fill in to I minus. Why? Squared asked. The Capitol are so at this point, I could expand that binomial or type this into my calculator. If I type this integral portion into my calculator to evaluate the integral, then it's high time. 16/15 which is 16 high over 15. And that's my finally

Oklahoma Baptist University

Applications of Integration