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Let R denote the region bounded by the curves y =x? y =6-x,and y = 0. Find the volume of the solid that results from rotating the region around cach of the followi...

Question

Let R denote the region bounded by the curves y =x? y =6-x,and y = 0. Find the volume of the solid that results from rotating the region around cach of the following: The x-axis:The y-axis:The line xThe line y = -

Let R denote the region bounded by the curves y =x? y =6-x,and y = 0. Find the volume of the solid that results from rotating the region around cach of the following: The x-axis: The y-axis: The line x The line y = -



Answers

Concern the region bounded by $y=x^{2}$, $y=1,$ and the $y$ -axis, for $x \geq 0 .$ Find the volume of the solid. The solid obtained by rotating the region about the line $y=-2$

In the question we have to find the volume of a solid obtained by rotating about by excess. And the region bounded by the curves by school. Three to the power minus x square by schools to zero X is close to zero and excess calls to one. Now moving towards the solution here, rotating about the y axis. The cylindrical shell method will be simpler. So like the height equal to E. To the power minus X. Two minus x square. So the radius will be equal to R. S equals two weeks which goes from 0 to 1. So we will be equal to integral of two by R H. D. X. That is we will be called to to buy into integration from 0 to 1 X into E. To the power minus X square D. X. Now we will be solving this integral. So let you be equal to minus x square and then do you will be equal to minus two X. Dx. That is minus one by two D. You will be equal to x dx. So putting these values you will get your limit going from a equals two minus x square to minus zero square that is zero and we will be from minus one square that is minus. But so the integral will be Dubai into Integration from 0 to -1 -1 by two E. To the power you do you Solving This? You will get -9 E to the power you limit going from 0 to -1. That is minus phi into each of the power minus one minus one. Solving this. You will get your answer approximately equal to 1.986 Thank you.

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

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.

In the question we have to find the volume of a solid obtained by rotating about X. X. S. The region bonded by the cove's vice calls to eat to the power. Ex vice calls to zero access calls to zero and access calls to one. Now moving towards the solution here, we have to rotate the given uh about the X axis. So using that desk method will be simpler option here. So the radius is represented by the function rs equals to eat to the power X. So we will be given by pi into integral from 0 to 1 each of the power X to the power to the of X. Now we will be solving this so we will be equal to pi into integral from 0 to 1. Each of the power two weeks into D. X. Bye Into one x two into it to the power to works limit goes from 0-1. Now putting the limit, you will get by by two into each of the power to -1. So now here the derivative of E. To the power two weeks is too into E. To the power to works. The two coming from the derivative of two weeks because of the chain rule chain rule. Now let you be equal to two weeks. So D. You will be equal to two in two D. X. That is one by two D. You will be equal today affects so integral from 0 to 18 to the power to the to E. To the power to the to works. D. X. Will be equal to integral from A. To B. One by two into E. To the power you do you. That is one by two and two E. To the power you. And limit going from A. To B. So this will be equal to one by two into E. To the power two weeks limit going from 0 to 1, which comes out to be five by two and three to the power to minus one. Thank you.


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