find-the-coordinates-of-the-centroids-of-the-given-figures-each-region-is-covered-by-a-thin-flat-p-5

Name: find the coordinates of the centroids of the given figures each region is covered by a thin flat p 5
Uploaded: 2020-05-18
Duration: 3 min 6 s
Channel: Numerade
Description: Find the coordinates of the centroids of the given figures. Each region is covered by a thin, flat plate. The region bounded by $y^{2}=x, y=2,$ and $x=0$

Question

Find the coordinates of the centroids of the given figures. Each region is covered by a thin, flat plate. The region bounded by $y^{2}=x, y=2,$ and $x=0$

Caelan Thomas · Accepted Answer

we want to find the area of the region bounded by why equals two minus the absolute value of X and y equals X squared. So we can find the area of this piece right here and just double our answer, since the graph is the same on both sides and that will give us the this total shaded area. So it&#x27;s first figure out at what point these curves intersect. So I set X squared. You quit too, minus X. And if we move everything to one side, we get X squared plus X minus two. It was zero. And if we factor, we get X plus two times X minus one equals zero, and so X equals native to and X equals one. And since we&#x27;re talking about this point here in the first quadrant, it&#x27;s the X equals one. So are integral will go from 0 to 1, and it will be to minus X minus X squared. These absolute value signs, uh, can just drop off because we&#x27;re dealing with the positive X is in the first quadrant. So are integral is going to be two x minus 1/2 x squared, minus 1/3. Execute from zero. The one. And if we plug in one, we get to minus 1/2 menace 1/3. And if we put in, Syria will just get zero. So we&#x27;re left with for over two. Mina. Sorry. Uh, six over 12. Over. Six, minus 3/6, minus 2/6, which is 12 minutes. Three is 99 minutes to a 77 over six. That is our area of have of the shaded regions. Our total area bounded by the equations will be 14/6 or 7/3.

Caelan Thomas · Answer

we want to find the area of the region bounded by Y equals X Y equals one over X X equals two and white was era. So it&#x27;s this shaded area down here. I want to do this. They want to divided into two separate areas. One where the top curve is white was X and one where the top curve is why it was one of cracks. So to do this, we first need to figure out where these curves intersect. So we&#x27;ll do that by setting these equal to each other One over X equals X, And if you multiply the X over, we get one equals X squared. So X equals plus or minus one. And since we&#x27;re in the first quadrant, we just have X equals one. So our first integral is gonna be from 0 to 1, and we have the integral of thanks minus zero. So just eggs. And that&#x27;s gonna be 1/2 X squared from 0 to 1. And when we put again, we get 1/2 times one minus one halftime zero, which is just through. So we get 1/2 and for our second area we&#x27;ll do that in a girl from 1 to 2 of one over X and in the growth that is natural log of X from 1 to 2. And so we get on to minus Alan. What and this is equivalent to are that started? The total area is equivalent to 1/2 plus natural out of two minus natural, and the natural arc of one is just zero. Since he did, the zero equals one. So what we&#x27;re left with is 1/2 plus natural out of two, and that&#x27;s the area of our region.

Linh Vu · Answer

So this is the graph. I&#x27;m the on tree functions and we&#x27;re from the graph. We see that intersection here goes under gone, uh, zero and Jew The second point Him? We don&#x27;t know yet. So let&#x27;s What you find is one. And the last one is very clear. That and upon trees, they&#x27;re Oh, so let&#x27;s try to find this intersection here. And it&#x27;s B agenda far minus x square and with the ex crab Bliss Joseph, With them it go. And now we will have we bring everything. One side&#x27;s always again. The Jew ANC square on. I would try to bring the X on one side and the number on the other side. So we have the two x square We could you two so x square court you won so angsty. Could you bless? A man is one so grey and this problem here will be one and three. So now we will try to spin this into grow into the £2. The first part will be here one here and the second branch will be the 2nd 1 hand will be here and we do too up. Damn! So the total area echo Joe the first indica goes from 0 to 1 and it&#x27;s between Duh this girl. Yeah, And this one here. So we have Onda excrement. You is a both that we have the X Square plus to minus the job. Less angst, the angst and for the father&#x27;s on is from the green one and fund a bubble one with the hangers from 1 to 3. And we see that this one here is above this one. So we will have far minus thanks Square minus Juba&#x27;s angst Andi act. And isn&#x27;t his fund up? Burba one. And now we will have a go Jew. And to cover room near to one. Thanks, Crab bliss. Thanks, Ana de act. And I found a second wouldn&#x27;t have from 1 to 3 minus next square. Bless angst. Blessed you do your ex and we get it on the equal to export Treaty of Arbitrary Plus X Square on with you evaluated from 0 to 1. Plus, there&#x27;s only three months expert to develop a tree plus X squared it. Damn it! You just you angst evaluated under one and three. So it we compute this one. Which again? The answer. We co Jew you and that&#x27;s the final Funda. Toto, I uh

Caelan Thomas · Answer

we want to find the area bounded by two x times the square root of one minus X squared and y equals X. That&#x27;s this area right here in the first Squadron. So first we know her left bound zero. We need to find where are right bound Is that where these two curves intersect. So two X times one minus X squared under a square root equals X, so that gives us two times one minus. X squared equals one. A square root of one minus X squared equals 1/2. And if we square both sides, we get one minus. X squared equals 1/4. Then we have negative X squared equals negative 3/4 which means X squared equals 3/4 and X equals route three over to So that is, are right bound with 3/2. So we&#x27;ll set up for in a girl zero to root three over to, and we have two x times two square root of one minus X squared minus X and then integral will be negative. 2/3 one minus X squared to the three halves minus 1/2 x squared. And if you plug in Route 3/2, you&#x27;ll get negative 11/24 and if you plug in zero, you&#x27;ll get negative 2/3. So we have negative 11/24 plus ah 16/24 which leaves us with five over 24. That is our area of the bounded region.

David Marsella · Answer

So here we have a very straightforward one with the lying like ALS to and the curves y equals seeking squared X. We want to find the region from zero to pi over four. So if we look at our general form, the area equals the inter girl from a to B of f of X minus G of X dx. We actually have all of these. We don&#x27;t need to find anything because we&#x27;ve been told to find the area from zero to buy over. Four. So are a zero r B is pi over four were given everything in terms of X. It&#x27;s probably generally that&#x27;s the easiest way to do it with Trig functions. For one, in terms of why we ended with, uh, Arc Trig functions and then things get really messy. So we do a vertical line test. The top of this line is y equals two. So that&#x27;s our f of X. The bottom of this line is R y equals seeking squared X. So that&#x27;s our G of X. Oh, and we don&#x27;t need Teoh actually calculate any of the values. We just plug them in. So here we have the integral from zero two pi over four of two minus seeking squared X DX and remembering that the anti derivative of seeking Squared X is tangent of X are anti derivative is two x minus tangent of X from zero to pi over four. If we perform our substitution ins, it gets us pi over two minus one minus zero minus zero. So the area between the lines and the curves his pi over two minus one and you&#x27;re finished.

Problem

Find the coordinates of the centroids of the give…

Problem 16 Easy Difficulty

Find the coordinates of the centroids of the given figures. Each region is covered by a thin, flat plate. The region bounded by $y^{2}=x, y=2,$ and $x=0$

Answer

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Basic Technical Mathematics with Calculus

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Integration Review - Intro

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

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