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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.

$$y=1-x^{2}, y=0 ; \quad \text { about the } x-axis$$

$$

\frac{16 \pi}{15}

$$

Applications of Integration

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Numerade Educator

Missouri State University

University of Michigan - Ann Arbor

University of Nottingham

I want to find value Rotated around the X axis here bounded by white was one minus X squared and y equals zero. So first of all, we want to sketch out what this graph is gonna look like one minus x squared. This is a parabola that's been reflected to be facing down. I know that because of the negative, and it's been shifted up one. So I'm gonna make my up one pretty obvious here and then downward facing proble you're going to see on the easier Grafts don't have to be super precise as long as you can see how they're related to each other and how they're bound in. So the Y equals zero is my other boundary. Which means the area that I'm trying to rotate to make a volume is this part that I've shaded in green. So because as I go around the X axis, all of this area comes right up along the X axis, there's no spaces. There's nothing missing. This is a disc method. The volume for your disk method is pi R squared in the integral. You use whatever variable you're rotating around this case DX cause we're going on the X axis, and I need to figure out the first X in the first in the second X that defined this area. So that's gonna be thes points here along the x axis. Pretty easy to solve for, because we just need to let our two equations equal each other. So we're gonna let one minus x squared equal to zero, which means X is plus or minus one. So my boundaries for the integral are gonna be plus and minus one. Now, another option here you definitely don't have to use. You're gonna notice it's symmetric around the XX around the y axis. So because it's symmetric, you could actually just do this half and then double it so I could put a two out front and just go from 01 but not necessary. Okay, just another option. So that's optional. Now, as I'm filling in this integral, the only other thing I need to know is what is this r squared? So R squared is the radius and a radius rumor goes from the center to the edge. The center is whatever you're spinning around, so it's gonna be this y equals zero, the edges made by this parable a shape I like to think of it as upper minus lower. So when this problem is one minus X squared, minus the access, so be subtracting zero, which means I'm just gonna fill in one minus x square. So at this point, we could expand this by no meal and integrated by hand. Or obviously, if you have a way to calculate this, that's a great choice as well. I have the ability to calculate this and just for the sake of time I find this integral is 16/15. So for my final answer, if I keep it exact, I would say 16 high over 15.

Oklahoma Baptist University

Applications of Integration