5

Find the mass of the ellipsoid given by 4x?+4y+22 = 16 Iying above the xy - plane and the density function is given by plx,Y,2)=2.Select one: 163 16T d.87 4t...

Question

Find the mass of the ellipsoid given by 4x?+4y+22 = 16 Iying above the xy - plane and the density function is given by plx,Y,2)=2.Select one: 163 16T d.87 4t

Find the mass of the ellipsoid given by 4x?+4y+22 = 16 Iying above the xy - plane and the density function is given by plx,Y,2)=2. Select one: 16 3 16T d.87 4t



Answers

Find the center of mass of the following solids, assuming a constant density of 1. Sketch the region and indicate the location of the centroid. Use symmetry when possible and choose a convenient coordinate system. The solid bounded by the upper half $(z \geq 0)$ of the ellipsoid $4 x^{2}+4 y^{2}+z^{2}=16$

Were asked to find the mass and center of mass of the solid s because founded by the Arab Lloyd is E equals four expert for a wide squared and the plane Z equals a Where is greater than zero if s has a constant density, which is K Well, first of all, grab a Lloyd Z equals four X squared plus four y squared. This intersects the plane Z equals a and occurred with the equation a equals four x squared plus four y squared. And this can also be rearranged so that X squared plus y squared equals 1/4 a and therefore in cylindrical coordinates, our region E is the set of triples our data z such that data will lie between zero and two pi, as usual are is going to lie between zero. And according to this equation, R squared is less than or equal to 1/4 a or are is less than or equal to one half route A and Z will lie above the parable Lloyd and below the plane. So above deprive await his equation for R squared and the plane. Oh, and so we had the beat mass of this region M, this is going to be the integral from 30 equals 0 to 2 pi, integral from r equals 0 to 1 half square today integral from Z equals four r squared to a of the density. Okay, Times differential are dizzy DRB theta and taking the anti derivative with respect to our and evaluating with Z and evaluating, we get K times integral from 0 to 2 pi integral from 0 to 1 half route a times this is going to be our times a minus for R squared or a R minus four are cube DRG theta taking into their evidence with respect to our you get K times integral from 0 to 2 pi and then of a over to R squared minus part of the fourth from r equals zero to r equals one half route a de Seda and sorry, This is rude over to the theater and evaluating the U K times integral from 0 to 2 pi. And this is, Let's see Avery a over eight minus a squared over 16 that's R squared. That's a squared over eight minus a squared over 16, which reduces to 1 16. He squared the theater, thinking inside derivatives and evaluating we get que times 1 16 a square times two pi or 1/8 they square pie. Okay, this is the mass. Now we assume with against K that the region is homogeneous. And also we know by the way the region was constructed, that it's symmetric about the x axis and the Y axis. Sorry about the y Z plane in the XY plane and therefore follows that moment about the wise You playing in a moment about the XY plane or both. Zero. We have that moment about the X Y plane. Well, this is the integral from Fadi equals 0 to 2 pi, integral from r equals zero to r equals one half fruit A integral from Z equals four r squared to Z equals a of z times the density k times the differential R D CVRD data and take the anti derivative with respect to Z and evaluating. We get que times integral from 0 to 2 pi into girlfriend zero toe, one half route a and then one half a squared or minus. This is going to be one half of 16 are to the fifth or eight hard to the fifth DRD data and taking the anti derivative with respect to our this is K times integral from 0 to 2 pi of 1/4 a squared R squared minus 8/6 Artist six from r equals zero two r equals one half route a The data evaluating we get que times integral from 0 to 2 pi this simplifies to CEO Mhm you It's one 24th a cube, the theater taking anti derivatives and evaluating two simplifies to two times one 24th or 1 12 acute hi times K. So this is the moment about the X Y plane and therefore the Sen Troy or center of Mass for this region X bar y Barzee bar. This is the point with coordinates Moment about the why is he playing over the total mass? This is zero moan about the X the plane from the total mass, which is zero in the moment about the X Y plane over the total mass. This is 1 12 a cube pai que divided by 18 a square to pai que and this simplifies to two thirds a

All right. So we want to find the volume or the mass and they sent out a mass of this solid S. Which is founded by the tabloids as you go four X squared plus four pi squared. And the planes easy but a where A. Is positive. So if you go ahead and draw it just looks like we'll see that. This is just so this is just a parable Lloyd. Mhm. Mhm. That is sort of cut off by me writing colour. So cut off by this plane. So it's sort of like if you draw it out we'll see something like a problem with a flat top. So this top curve is just too easy today and the bottom is Z. Is equal to in this case to be doing cylindrical is just for ar squared. Now we have our Z bounds. We know it's completely symmetrical with data so it's the order to pilot data. And then we just want to find out our balance by setting these two equals so we can see is equal to four R squared A over four is R squared. And then so we're just giving that art is spurred A or two. So we have our three bounds we have Z. It must be between Hey and four square since we know a. is the top cap, we know that Are, you must be between skirt a over two and 0. And then data is completely around security to buy and then we're taking the mask, you know, one of the density which were given which was just a constant rosie quick A. So for all we know our mass is just a triple and a rover S of road E B which is just a triple integral which is one from zero to pi narrow from zero to square a day or two integral from for our square to a of um we have just row tv which is K. R. Dizzy. He's already data going ahead and separating the data out since we know we have no dependence on the inside. Mhm. You can go ahead and write this since we have no Z. We can just go ahead and evaluate this inter interval giving us a minus four R squared D D. R, solving this. Outside integral is two pi. We can pull out the K. As well. Just have this insight in a girl a ar minus for our cube D. R. Now we can plug into pie. I mean we can do our outside to pick a and then to evaluate this integral giving us a R squared over two minus are to the fourth from zero to square day over two. And we'll see that we'll end up with two pi K times. So we have squared a square. That's just a s we have a squared over two times. 1/2 squared is just hate. Oh, so yeah, what half times when have squared is 818? Then we have a squared over 16. So then our final answer for the masses. Just two pi K. He squared over 16. We're just by K. Make sure its capital a squared over eights for the mass. Okay. Now we want to find the center of mass and we can go back to looking at our figure. We know that we are completely symmetrical in data. Okay. And in our so we have no we know that our center of mass since it's a constant density and it's completely symmetrical around the Z axis, We know that the first two coordinates X. Y. must be both zero through the symmetry. And so we want to only find is the Z coordinate of our center of mass, which is just this integral. So then the average value of Z just given by um the integral of rosie tv all divided by the mass. So we know the mass above. So we have to calculate weekends just calculate this integral and then divided by what the masses to find the center of mass. So we have yes, this is equal. We have first song in this interval, we know our balance of the same. And then instead we're going to have K R Z. Is he D already data? Yeah, now we have a Z on the inside. So we first we can still pull out two pi the data And then the square day of the two and then solving the center integral gives us Z squared over two from our bounds for r squared Okay. And so we didn't just pull out the KR and then we'll see that this is just um a squared over two minus 16 arch the fourth over two. And then D. R. Now we're going ahead and solve this outer integral gives us two pi and we just saw this integral here. Okay. Times A it's where it are over two AIDS part to the fourth or are to the fifth. Sorry since we have this extra our factor. Mhm. Yeah. And so we just have still pull out the K. And then so we can just integrate this inside giving us a squared R squared over the floor minus eight. Hard to the 6/6 which is just four thirds arts and six. Yeah. 0 to screwed a over to. Yeah. And so we have two pi K tidies. We'll see that a square gives us A cute over 16. So I screwed I mean let me have four cube four thirds times a cube over 2 to 6. It's just um 64. We'll see that. This uh 16 this four and 64 council. So we have a cube over 16 minus. Um We have 1/3 times a cube over 16. So then this just ends up being two pi OK times. Um We caught a cube and the 16 You get 2/3 here. This is just equal to um We have the four and the 16 cancel to give us high K. A square or a cube all divided by um We have four times three is 12. So this is what the integral evaluates to. And so we just want to divide, we know the z center of mass value is just this pike a cute over 12 divided by our mass which was pi K A squared over AIDS. And then so we can cancel out these pie K's, we can cancel out two of the AIDS. So we're just left with A over 12 times AIDS, which is just equal to To a over three. So let me know our complete tire of mass 00 to a over three. And then this is our see Hello and then we have our maths, the schist pi K K squared upgrades. This is our final I don't answer to both questions.

So in this video, we're given the density function, which is one plus see, divided by nine. And then we're given D where d is domain. Uh, yeah. So are they Darcy we're seeing is between zero and nine. Minus are square and doors ST 03 And we're also told that this is a solid Carrabba Lloyd. So you know that data is from 0 to 2. All right, so the first thing we have to know, So we already know that mass is simply the triple integral off road. You see, where if you use the volume element, so it's not just the volume element into a cylindrical coordinates is just RGC yard data. And now we need to figure out what our limits limits of integration. So we know from here that Z goes from 0 to 9 minus r squared at bar goes from 03 So the 1st 2 are very simple and unlikely said our data goes from 0 to 5 because we have a stolid, uh, probably all right. So now comes the part where the mechanical part where we have to use our integration. So the integral of one pussy divided by nine times ours to C plus square, divided by 18 and are just remains on the outside because it's basically a concert. Let's see their limit limits of integration or from 0 to 9 minus off script. So if we plug in nine minus R squared to get nine minus R squared plus nine minus R squared square, divided by a team and then minus zero plus zero so we don't need to times are New York. All right now, let's try to simplify. So this is nine minus R squared plus 81 minus 18 r squared plus R divided by plus R to the power of four divided by. And if we do further simplifications, we have nine minus R squared. Anyone divided by 18 is just nine divided by two. So if we pull out of nine as a common factor minus 18 r squared, divided by 18 just by this are square plus R to the power for divided like 18 kinds our yard. So now we can distribute This are well, actually first. That's right. A simplified to see. So if we look at these fractions, we have a nine plus nine divided by two that simply are 27 divided by two and then we have negative are square minus are spread. This is negative. Two are square plus r to the power for divided by, so redistribute that for and get 27 r divided by two finest are cute plus r to the power five divided by 18 The art art So now we're gonna take the integral with perspective are so 27 are departed Bite to it Ours just 27 r squared, divided by four Buying this integral of two are cute to our to the power for divided by four which is argued of power for divided by two Boston Integral part of the power 5 18 is part of the power six times 18 times six, which is just 100. There are limits of integration. Zero All right now reporting three we get 27 times three squared, divided before minus treated the power for two plus three to the power of six, divided by 108 minus zero minus Europe 10 So we just forget about now. If we work out the math, we're going to get 27 So this whole chunk right over here. It's simply 27. So we have the equal forms in order to buy 27. 27 is a constant, and the integral of the data is just so that 27 times data where data goes from 0 to 2 pi. The two byline is used to high at 27 times two pi r 54. So our masks off our elliptic off our where our stall it for Rapid Lloyd, it's 54 plus.

Okay, So this form wants you to find the mass and the center of mass of all of a volume s as down and by ah problem and an upward facing crumbling with the equation of Z equals two four times expertos y squared and upward and stealing plane with of the evils eight where he has to be Ah, constant. That is greater than zero All right, to find mass mass eagles to a density times volume. And they told us in the problem that the density of this solid as is K we have over here. So all we have to do is calculate the triple integral. So first we have to do is comfort everything into cover everything into cylindrical coordinates. So let's go ahead and do that. So, for our triple into growing, we have a density off K. So that's there. Que and devi converted into cylindrical coordinates. That becomes our times Easy, D r do you think? And looking at our boundaries we have in this DNC on the way. Looking at our boundary, we have z the upper facing problems of the equals for time extra plus y squared and we know that exports parts where people who are square So we can rewrite this as C equals to four r squared and the ceiling plane. Well, let's just see equals a and cylinder Alcorn itself at stay the same. So now we've got everything into cylindrical coordinates. Now let's figure out the boundaries. So first off, we have Z and we know that Z is founded below by the problem, Lloyd of four r squared it is bounded above by the plane of a Z equals a no, for our are is r is the radius were the maximum Are is when the plane and the probably they intersect. So when for our square you go to a and if you sold for Are you okay? They are value you get in our value off half square root of a So this is our second boundary. Our upward bound e boundary for are 1/2 square root of a And finally, Martina is from 0 to 2 pi because go, uh, nothing interrupts the full rotation of probably. All right, so now we've got our boundaries and got everything into suitable coordinates. So all we have to do now is calculate this triple integral to find the mass. So if we were to from tackling this triple integral who would first he would first integrated would respect to We're personally greeted with respect to Z We would have kr xe that has done it by for r squared in a Let's not forget these pounds euro two pi and square root of a over to All right, so next Oh, and the RD deal. And next if we plugged in the values of a four squared into Z can we get a double integral A double into world off K R A minus four K, aren't you? Do you are dictator All right, now next we have to integrate it with respect to our and it is just appalling on your So that'll be easy. It would just be now an integral from 0 to 2 pi off of 1/2 1/2 k Okay, a r squared minus k r to the fourth. The boundaries for this is square root of a over to your right over two and zero. Do you think? And if we plugged in, swear today over to into our values of our we were gay an integral from 0 to 2 pi of off k. Okay, over 16 times a square. If data, it's pretty nice. Have we integrated this? With respect to data, we would get okay over 16 times a sward time scale with the calories from 0 to 2 pi. So forget to plug in to pi into data. We would get a mass. I would get a mass of pi a squared times K density over eight. So this is your answer for the mass now. Next, The problem wants you to find the said No, not a century. The center of mass off this volume of the solid And if you notice because we're dealing with an upward facing probably probably is cylindrical about the Z axis and has a cost intense to you. Okay, so the X and Y components of the center of mass will become zero as a result because of symmetry. So x bar and why are both equal to zero or the center of mass I can remember is written in terms of expired. Why Barnes er see bar. So all we have to calculate now is Z bar and he calculated Z bar. We just have to. First part calculating is too re Calculate the triple integral with the same boundaries and everything. Except this time. Oh, you got four square for squares. Except this time, instead of integrating okay, are me Multiply it by Z Sore integrating kr KZ are around the square root of a over to So that's DZ tr. Do you think so? We do now is calculate this and we get one step closer to calculating the center of mass. All right, so we integrated this with respect to Z. We get it. Factor out a case that's it's a constant. We get a double integral to die screaming at me over to both zeros on the bottom. We get the double integral off. 1/2 won't have our z squared when the bounds are from or r squared to a DRD data. Okay? And if we plug in these values for Z, you would then get a double integral k times a double integral to find 0 to 2 pi from zero to screw of a over to, we get a double integral off. 1/2 times are times a a squared minus eight times are to the fifth drd data. And if we integrate this with respect to our you would get que times integral from 0 to 2 pi of 1/4 a squared R squared minus 4/3 or 2 to 6. The balance from zero to Brad A over to dif data. If we were to plug these in to our we were gay. Okay, Times the integral from 0 to 2 pi You're on a too high off a cube over 24 de stayed up. If we integrate this with respect data, we will get que a que over 24 times they don't in the bounds are from 0 to 2 pi, so we'll get final value. Final value of final value, love. Sorry. Oh, pie. Okay, a que over 12. So now to calculate Z bar to calculate Z bar, we have to take this value and divided by the mass which we calculated above. So take this value 1 12 times pi Okay, A cube And we divided by the mass and the mass to calculate above is 18 times pi ace work A So that's right 1/8 times pi Okay, a squared. And if we character did all that we would end up with the Z bar off 2/3 a. So our center of mass will be 00 comma 2/3 times a and this is your final answer.


Similar Solved Questions

5 answers
Certain compact disc player draws current of 430 mA at 24.0 V: How much power is requlred to operate the player?Need Help?
certain compact disc player draws current of 430 mA at 24.0 V: How much power is requlred to operate the player? Need Help?...
4 answers
3) Data was collected from 107-Deople and it was found that 43 people were in favor of the new contract: Construct a 95% confidence interval then write the ending statement:
3) Data was collected from 107-Deople and it was found that 43 people were in favor of the new contract: Construct a 95% confidence interval then write the ending statement:...
3 answers
An UTH contains 3 black balls andwhite balls:second urn contains 6 white balls andOne ball is drawn from Ur one and placed in urn (wo- If we now draw ball fromn 8 black balls. urn two what is the probability that the ball is black? (Hint: Use partition result and conditionalprobability formulae)
An UTH contains 3 black balls and white balls: second urn contains 6 white balls and One ball is drawn from Ur one and placed in urn (wo- If we now draw ball fromn 8 black balls. urn two what is the probability that the ball is black? (Hint: Use partition result and conditional probability formulae)...
1 answers
Jmr*~iatn Mattaits Yviktt ir N S /atLt?t
Jmr* ~iatn Mattaits Yviktt ir N S /at Lt?t...
5 answers
Complcte the following Lewis stncture by eddingelectron pirsuppronnaeIncludz all valence lone puirs - your answer; Should you want restart the cxeteise , thc drop-down tenu labcled starting pointsbe used redriw the Stirting molecule on thc skLnemooo3n o DonteShnaralteam
Complcte the following Lewis stncture by edding electron pirs uppronnae Includz all valence lone puirs - your answer; Should you want restart the cxeteise , thc drop-down tenu labcled starting points be used redriw the Stirting molecule on thc sk Lnemooo 3n o Donte Shnaralteam...
1 answers
Show that each sequence is geometric. Then find the common ratio and write out the first four terms. $$ \left\{t_{n}\right\}=\left\{\frac{3^{n-1}}{2^{n}}\right\} $$
Show that each sequence is geometric. Then find the common ratio and write out the first four terms. $$ \left\{t_{n}\right\}=\left\{\frac{3^{n-1}}{2^{n}}\right\} $$...
5 answers
Diagrams showing the appearances of the image = and the object to use for the description of the characteristics of the imageObjectImageCase 2.5 x fCase 2 * fCase L.Sx fThis Image not formed on the screen: Visible looking through the lens Into the light sourceCase 0.5x f
Diagrams showing the appearances of the image = and the object to use for the description of the characteristics of the image Object Image Case 2.5 x f Case 2 * f Case L.Sx f This Image not formed on the screen: Visible looking through the lens Into the light source Case 0.5x f...
5 answers
Question 1Calculole - mMATILt antint of bent Ihat chould [umlrcu contu Lg of water VOt (crnoctlune (25.0 %C} into Ice (at atluxopbctic pressure) Rcler = Folir cuuLuOM sheet for the relevani properties ol wnter: (Nele: I0 } I0')slcet Otk: 490* 10*1 , 1047 -10n] / {0 16i0 J| 0J77*10 ] } 414,0K I0*J}
Question 1 Calculole - mMATILt antint of bent Ihat chould [umlrcu contu Lg of water VOt (crnoctlune (25.0 %C} into Ice (at atluxopbctic pressure) Rcler = Folir cuuLuOM sheet for the relevani properties ol wnter: (Nele: I0 } I0') slcet Otk: 490* 10*1 , 1047 -10n] / {0 16i0 J| 0J77*10 ] } 414,0K ...
5 answers
Applying the first theorem on bounds for real zeros of polynomials, determine the smallest and largest integers that are upper and lower bounds, respectively, for the real solutions of the equation. With the aid of a graphing utility, discuss the validity of the bounds.$$2 x^{3}-5 x^{2}+4 x-8=0$$
Applying the first theorem on bounds for real zeros of polynomials, determine the smallest and largest integers that are upper and lower bounds, respectively, for the real solutions of the equation. With the aid of a graphing utility, discuss the validity of the bounds. $$2 x^{3}-5 x^{2}+4 x-8=0$$...
1 answers
Sketch a right triangle corresponding to the trigonometric function of the acute angle $\theta .$ Then find the exact values of the other five trigonometric functions of $\theta .$ $$\sin \theta=\frac{1}{5}$$
Sketch a right triangle corresponding to the trigonometric function of the acute angle $\theta .$ Then find the exact values of the other five trigonometric functions of $\theta .$ $$\sin \theta=\frac{1}{5}$$...
5 answers
OO J1 dx is convergent zv?TrueFalse
OO J1 dx is convergent zv? True False...
5 answers
(a) Determinc L'Hopital s rule appllespos;Ible that the tunction would ncedretritenFirsttSuec"Y-0*(Sx + 7)Wx50 ec-_Pe"Scccl applicg holtdona nol dpply(6) Use (()9(x) Vx =neai the questlons uclow.Explaln why L'Hopital's Rule appllesI[) 9(*)Now IInd the exact valueg(x)
(a) Determinc L'Hopital s rule applles pos;Ible that the tunction would nced retritenFirstt Suec" Y-0* (Sx + 7)Wx 50 ec-_ Pe "Scccl applicg holtdona nol dpply (6) Use (() 9(x) Vx = neai the questlons uclow. Explaln why L'Hopital's Rule applles I[) 9(*) Now IInd the exact val...
5 answers
Izt 1 6 m < @ And 9 e 4 (Co,(J) Oelermint vhick Ve Ives 1< P, 1 2 # the "eratav To : Lp (Lo,lJ) ~ L9 (Co, | J) bocndt Tglf) = F 9 Aeve V f elp(Co lJ)
Izt 1 6 m < @ And 9 e 4 (Co,(J) Oelermint vhick Ve Ives 1< P, 1 2 # the "eratav To : Lp (Lo,lJ) ~ L9 (Co, | J) bocndt Tglf) = F 9 Aeve V f elp(Co lJ)...
5 answers
Find the 1 1 ofcuQ 1 sin€ for each# 135Accessbrlit Mode
Find the 1 1 ofcuQ 1 sin€ for each# 1 35 Accessbrlit Mode...
5 answers
Problems 5 - 8, use transformations to graph each equation from (~Zx, 2t)Y = 2sinxY =-~sin(i*)Y = 2cos(*)-1Y =-3sin(*) + 2
Problems 5 - 8, use transformations to graph each equation from (~Zx, 2t) Y = 2sinx Y =- ~sin(i*) Y = 2cos(*)-1 Y =-3sin(*) + 2...

-- 0.022142--