centroid of the region bounded by x = 2 - y^2 and y-axis

[johnq2k7]

Find the centroid of the region bounded by the curve x=2-y^2 and the y-axis

A= integral of (2-y^2)dy from 0 to 1

M_y= (1/2) integral of (2-y^2)^2 dy from 0 to 1

M_x= integral of (y)(2-y^2)dy from 0 to 1

x= (M_y)/A
y= (M_x)/A

centroid is (x,y)

I'm sure I may have made some mistakes in my integration set up for A,M_x,M_y please help

 

[Deleted member 4993]

Re: centroid of the region help needed

Find the centroid of the region bounded by the curve x=2-y^2 and the y-axis


A= integral of (2-y^2)dy from 0 to 1

M_y= (1/2) integral of (2-y^2)^2 dy from 0 to 1

M_x= integral of (y)(2-y^2)dy from 0 to 1


x= (M_y)/A
y= (M_x)/A

centroid is (x,y)

I'm sure I may have made some mistakes in my integration set up for A,M_x,M_y please help

First sketch the given expression and define the region of interest.

You'll see that the region is symmetric about x-axis -> y[sub:liqmxax0]centroid[/sub:liqmxax0] = 0

also for the region

x-intercept = 2

y-intercept = +-sqrt(2)

Now can you set up your limits properly?

 

[johnq2k7]

Re: centroid of the region help needed

Find the centroid of the region bounded by the curve x=2-y^2 and the y-axis


A= integral of (2-y^2)dy from 0 to 1

M_y= (1/2) integral of (2-y^2)^2 dy from 0 to 1

M_x= integral of (y)(2-y^2)dy from 0 to 1


x= (M_y)/A
y= (M_x)/A

centroid is (x,y)

I'm sure I may have made some mistakes in my integration set up for A,M_x,M_y please help

First sketch the given expression and define the region of interest.

You'll see that the region is symmetric about x-axis -> y[sub:eb6mo09w]centroid[/sub:eb6mo09w] = 0

also for the region

x-intercept = 2

y-intercept = +-sqrt(2)

Now can you set up your limits properly?

is it integral of sqrt(2-x) dx from 0 to 2 for A

and then integral of 1/2 * (2-x) from 0 to 2 for M_y

and then integral of (x)(sqrt(2-x) dx from 0 to 2

and then the x-coordinate of the centroid is M_y/A and the y coordinate is M_x/A ... is this the correct approach?

how do u mathematically get y=0 for the y coordinate of the centroid... is the centroid (2,0) with your example... i'm still sort of confused

 

[Deleted member 4993]

Re: centroid of the region help needed

is it integral of sqrt(2-x) dx from 0 to 2 for A - No it should be \(\displaystyle 2\int_0^2 \sqrt{2-x}dx\)

and then integral of 1/2 * (2-x) from 0 to 2 for M_y No - did you draw a sketch before you came to these conclusions?

and then integral of (x)(sqrt(2-x) dx from 0 to 2 - What are you calculating here?

and then the x-coordinate of the centroid is M_y/A and the y coordinate is M_x/A ... is this the correct approach?

how do u mathematically get y=0 for the y coordinate of the centroid... is the centroid (2,0) - No

with your example... i'm still sort of confused

 

[johnq2k7]

Find the centroid of the region bounded by the curve x=2-y^2 and the y-axis

my work is shown below:

A= integral of (2-y^2)dy from 0 to 1

M_y= (1/2) integral of (2-y^2)^2 dy from 0 to 1

M_x= integral of (y)(2-y^2)dy from 0 to 1

x= (M_y)/A
y= (M_x)/A

centroid is (x,y)

I'm sure I may have made some mistakes in my integration set up for A,M_x,M_y please help

i also integrated in terms of x and got:

A= 2 times the integral of sqrt(2-x) dx

is the M_x equal to the integral of (2-x) dx from 0 to 2?

and the M_y equal to the integral of (2)(x)(sqrt(2-x) dx from 0 to 2?

therefore x-coordinate of the centroid is M_y/A

and the y-coordinate of the centroid is M_x/A

therefore centroid is [(M/y/A),(M_x/A)]

is this correct?

then the x-coordinate of the centroid is (M_y / A)

 

[BigGlenntheHeavy]

Re: Centroid Help Needed.. a lot of process is shown

(4/5,0) is the center of mass (the balancing point), or the centroid of the region.

I am giving you the answer, can you take it from here?

 

[Deleted member 4993]

Re: Centroid Help Needed.. a lot of process is shown

Duplicate post:

viewtopic.php?f=3&t=33166&p=128872#p128872

 

[stapel]

Duplicate threads merged.

 

[johnq2k7]

Re: Centroid Help Needed.. a lot of process is shown

(8/5,0) is the center of mass (the balancing point), or the centroid of the region.

I am giving you the answer, can you take it from here?

How did you get the answer (8/5,0).. what is wrong with my process work

how did u get M_x/A is equal to 0 basically

and M_y/A is equal to 8/5

since those represent that y and x coordinates..

i'm confused with that.. is there something wrong with my integral setups.. please help

 

[Deleted member 4993]

I had answered you r questions above - but you started a new thread!

M_y = 0 due to symmetry around x-axis. Each strip of "dA = ydx" has a center of mass(y[sub:aqondauj]1[/sub:aqondauj]) at y[sub:aqondauj]1[/sub:aqondauj] = 0. So when you integrate (y[sub:aqondauj]1[/sub:aqondauj]*dA) it becomes 0. Review the derivation of center of mass of an area carefully.

Show your work completly - so that we can check.

 

[johnq2k7]

I had answered you r questions above - but you started a new thread!

M_y = 0 due to symmetry around x-axis. Each strip of "dA = ydx" has a center of mass(y[sub:357n7d3o]1[/sub:357n7d3o]) at y[sub:357n7d3o]1[/sub:357n7d3o] = 0. So when you integrate (y[sub:357n7d3o]1[/sub:357n7d3o]*dA) it becomes 0. Review the derivation of center of mass of an area carefully.

Show your work completly - so that we can check.

for A= 2 times sqrt (2-x) dx from 0 to 2

by performing integration of A i got (sqrt(2-x))* (2x-4))/3 times 2 from 0 to 2

therefore i got (8*sqrt(2))/3 as my value for A

since Mx equals:

1/2 ? [a to b] [f(x)^2 - g(x)^2] dx

therefore i got the integral

M_x= sqrt(2-x) dx from 0 to 2

when i computed the integration of that i got:

2x- (x^2)/2 from 0 to 2

therefore i got Mx= 2... instead of the value of zero which is needed for the coordinate y to be zero.. since y-coord. is (Mx/A).. i don't know what's wrong with my integral here...

since My is
? [a to b] x [f(x) - g(x)] dx

after performing integration i got: 2 times (sqrt(2-x))*(6x^2-4x-16))/(15) from 0 to 2

therefore i got My equals= (32*(sqrt(2))/5

therefore i got x= 32/5 and y equals= 6/(8*(sqrt(2))

how did u get 8/5 as the x-coordinate and y coordinate equals zero???? and if My is zero.. wouldn't that mean the x- coordinate would be zero

since x= My/A.....please help

 

[Deleted member 4993]

I had answered you r questions above - but you started a new thread!

M_x = 0 due to symmetry around x-axis. Each strip of "dA = ydx" has a center of mass(y[sub:2op1cct9]1[/sub:2op1cct9]) at y[sub:2op1cct9]1[/sub:2op1cct9] = 0. So when you integrate (y[sub:2op1cct9]1[/sub:2op1cct9]*dA) it becomes 0. Review the derivation of center of mass of an area carefully.

Show your work completly - so that we can check.

for A= 2 times sqrt (2-x) dx from 0 to 2

by performing integration of A i got (sqrt(2-x))* (2x-4))/3 times 2 from 0 to 2

therefore i got (8*sqrt(2))/3 as my value for A

since Mx equals:

1/2 ? [a to b] [f(x)^2 - g(x)^2] dx <<<< No - it does not for this problem - look at the derivation carefully

therefore i got the integral

M_x= sqrt(2-x) dx from 0 to 2

when i computed the integration of that i got:

2x- (x^2)/2 from 0 to 2

therefore i got Mx= 2... instead of the value of zero which is needed for the coordinate y to be zero.. since y-coord. is (Mx/A).. i don't know what's wrong with my integral here...

since My is
? [a to b] x [f(x) - g(x)] dx

after performing integration i got: 2 times (sqrt(2-x))*(6x^2-4x-16))/(15) from 0 to 2

therefore i got My equals= (32*(sqrt(2))/5

therefore i got x= 32/5 and y equals= 6/(8*(sqrt(2))

how did u get 8/5 as the x-coordinate and y coordinate equals zero???? and if My is zero.. wouldn't that mean the x- coordinate would be zero

since x= My/A.....please help

I mis-spoke.

M_x = 0 ? because of symmetry around x axis

\(\displaystyle M_x \, = \, \int_0^2 y_1\cdot dA\)

Where
y[sub:2op1cct9]1[/sub:2op1cct9] = distance of center of mass of each elemental strip from the x-axis = 0 ? because each elemental area has their center of mass on x-axis due ti symmetry.

so

\(\displaystyle M_x \, = \, \int_0^2 y_1\cdot dA \, = \int_0^2 0\cdot dA \, = 0\)

 

[johnq2k7]

[quote="Subhotosh Khan":7u9eredt]I had answered you r questions above - but you started a new thread!

M_x = 0 due to symmetry around x-axis. Each strip of "dA = ydx" has a center of mass(y[sub:7u9eredt]1[/sub:7u9eredt]) at y[sub:7u9eredt]1[/sub:7u9eredt] = 0. So when you integrate (y[sub:7u9eredt]1[/sub:7u9eredt]*dA) it becomes 0. Review the derivation of center of mass of an area carefully.

Show your work completly - so that we can check.

for A= 2 times sqrt (2-x) dx from 0 to 2

by performing integration of A i got (sqrt(2-x))* (2x-4))/3 times 2 from 0 to 2

therefore i got (8*sqrt(2))/3 as my value for A

since Mx equals:

1/2 ? [a to b] [f(x)^2 - g(x)^2] dx <<<< No - it does not for this problem - look at the derivation carefully

therefore i got the integral

M_x= sqrt(2-x) dx from 0 to 2

when i computed the integration of that i got:

2x- (x^2)/2 from 0 to 2

therefore i got Mx= 2... instead of the value of zero which is needed for the coordinate y to be zero.. since y-coord. is (Mx/A).. i don't know what's wrong with my integral here...

since My is
? [a to b] x [f(x) - g(x)] dx

after performing integration i got: 2 times (sqrt(2-x))*(6x^2-4x-16))/(15) from 0 to 2

therefore i got My equals= (32*(sqrt(2))/5

therefore i got x= 32/5 and y equals= 6/(8*(sqrt(2))

how did u get 8/5 as the x-coordinate and y coordinate equals zero???? and if My is zero.. wouldn't that mean the x- coordinate would be zero

since x= My/A.....please help

I mis-spoke.

M_x = 0 ? because of symmetry around x axis

\(\displaystyle M_x \, = \, \int_0^2 y_1\cdot dA\)

Where
y[sub:7u9eredt]1[/sub:7u9eredt] = distance of center of mass of each elemental strip from the x-axis = 0 ? because each elemental area has their center of mass on x-axis due ti symmetry.

so

\(\displaystyle M_x \, = \, \int_0^2 y_1\cdot dA \, = \int_0^2 0\cdot dA \, = 0\)[/quote:7u9eredt]


Ok I understand what u got for Mx and how Mx equals zero now.. but how do u find the integral of My then?... if My is not 2 times the integral of (x)(sqrt(2-x) dx from 0 to 2

so far I understand A= 2 times integral of sqrt(2-x) dx from 0 to 2

and Mx is integral of y1 da and as a consequence of the symetry around the x axis.. y1 is equal to zero.. so therefore Mx equals zero.. which now makes the y-coordinate equal to zero since Mx/A is zero 0/A is zero

however for the x coordinate how come 2 times (x)(sqrt(2-x)) dx from 0 to 2 is wrong.. .what should it be?

because I got x= 32/5 instead of someone's reported 8/5 value for x

which means my centroid calc. would be (32/5,0) instead of the reported (8/5,0) please help

 

[soroban]

Hello, johnq2k7!

Let's take it from the top . . .

\(\displaystyle \text{Find the centroid of the region bounded by the curve }x\:=\:2-y^2\text{ and the }y\text{-axis.}\)

            *   |
                *
                |::*
                |::::*
        --------+:o:::*--
                |::::* 2
                |::*
                *
            *   |

Assuming uniform density, the centroid will be on the x-axis,
. . closer to the origin than the vertex (2,0).
We need to determine the x-coordinate of the centroid.

\(\displaystyle \text{The mass is given by the area: }\;A \;=\;2 \times \int^2_0(2-x)^{\frac{1}{2}}dx \;=\;\frac{8\sqrt{2}}{3}\)


\(\displaystyle \text{The moment about the }y\text{-axis is given by: }\;M_y \;=\;2 \times \int^2_0x\sqrt{2-x}\,dx\)

\(\displaystyle \text{Let: }\,u \:=\:\sqrt{2-x} \quad\Rightarrow\quad u^2 \:=\:2-x \quad\Rightarrow\quad x \:=\:2-u^2 \quad\Rightarrow\quad dx \:=\:\text{-}2u\,du\)

\(\displaystyle \text{Substitute: }\;2\int (2-u^2)\!\cdot\!u\!\cdot\!(\text{-}2u\,du) \;=\;4\int(u^4-2u^2)\,du\)

. . . . .\(\displaystyle = \;4\left(\frac{1}{5}u^5 - \frac{2}{3}u^3\right) \;=\;\frac{4}{15}u^3\left(3u^2-10\right)\)

\(\displaystyle \text{Back-substitute: }\;\frac{4}{15}(2-x)^{\frac{3}{2}}\bigg[3(2-x)-10\bigg] \;=\;-\frac{4}{15}(2-x)^{\frac{3}{2}}(3x+4)\,\bigg]^2_0\)

\(\displaystyle \text{Evaluate: }\;-\frac{4}{15}(0) + \frac{4}{15}\cdot2^{\frac{3}{2}}\cdot 4 \;=\;\frac{32\sqrt{2}}{15}\)


\(\displaystyle \text{Therefore: }\:\bar x \;=\;\frac{\frac{32\sqrt{2}}{15}} {\frac{8\sqrt{2}}{3}} \;=\;\frac{4}{5}\)

\(\displaystyle \text{The centroid is: }\:\left(\frac{4}{5},\:0\right)\)

 

[johnq2k7]

thanks for the help... i finally understand

 

[Deleted member 4993]

Another way intercept:

\(\displaystyle y^2 \, = \, x \, - \, 2\)

\(\displaystyle y_{intercept} \, = \, \pm\sqrt{ 2}\)

\(\displaystyle M_y \, = \, \int_{-\sqrt{ 2}}^{\sqrt{ 2}}\frac{x}{2}\cdot dA\)

\(\displaystyle M_y \, = \, \int_{-\sqrt{ 2}}^{\sqrt{ 2}}\frac{x}{2}\cdot x dy\)

\(\displaystyle M_y \, = \, \int_{0}^{\sqrt{ 2}}x^2 dy\)

\(\displaystyle M_y \, = \, \int_{0}^{\sqrt{ 2}}( 2 \, - \, y^2)^2 dy \, = \, [4y \, - \, \frac{4y^3}{3} \, + \, \frac {y^5}{5}]_0^{\sqrt{2}}\)

\(\displaystyle M_y \, = \, \frac{32\sqrt{2}}{15}\)

and so on.....

 

[jimbo57]

Hey guys, sorry to resurrect an old thread but I'm working on the same problem. I am absolutely stuck on how you get the Area to be 2. When I do my integral I get 8*sqrt2/3)... since area is = int[sqrt(2-x)] for x[0,2]

Please set me straight!

 

[Deleted member 4993]

Hey guys, sorry to resurrect an old thread but I'm working on the same problem. I am absolutely stuck on how you get the Area to be 2. When I do my integral I get 8*sqrt2/3)... since area is = int[sqrt(2-x)] for x[0,2]

Please set me straight!

Please start a new thread - re-writing the problem.

Where did you see that we found the area =2?

You have calculated the area correctly .... see soroban's post above.

 

[jimbo57]

Please start a new thread - re-writing the problem.

Where did you see that we found the area =2?

You have calculated the area correctly .... see soroban's post above.

Ah no need, I found the problem. When I saw soroban's reply, all I saw was A = 2 because the type format wasn't loading properly for me. It is now and makes perfect sense.

Thanks!
Jim