centroid problem please help !

[sjay88]

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

 

[daon2]

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

Where are you stuck, what don't you understand?

 

[sjay88]

Where are you stuck, what don't you understand?

first of all when calculating the area im nt sure why it is 2 times ∫(√(2-x) dx

also when calculting the x coordinate of the centroid i got (3/16) integral 0,2 x(√(2-x) dx

is this correct ?

 

[daon2]

first of all when calculating the area im nt sure why it is 2 times ∫(√(2-x) dx

also when calculting the x coordinate of the centroid i got (3/16) integral 0,2 x(√(2-x) dx

is this correct ?

The solution you are looking at may have used the property of symmetry. The graph of this function is a horizontal parabola opening to the left, and is symmetric about the x-axis. Instead of finding the entire area, they find the area above the x-axis (which is half the area), and multiply by two.

You can find the area in different ways:

(1) integrate with respect to y.

\(\displaystyle A = \displaystyle \int_{-\sqrt{2}}^\sqrt{2} 2-y^2 dy\)

(2) integrate with respect to x and use symmetry. First, solve for y: \(\displaystyle y = \pm\sqrt{2-x}\). Choose either the positive function or the negative function.

\(\displaystyle A = \displaystyle 2\int_{0}^2 \sqrt{2-x} dx\)

For the centroid, what you have looks odd (maybe a mistake on finding the area?)

\(\displaystyle \displaystyle \bar x = \dfrac{1}{A}\int_0^2 x\cdot \left[(\sqrt{2-x}) - (-\sqrt{2-x})\right] dx = \dfrac{2}{A} \int_0^2 x\sqrt{2-x} dx\)

 

[sjay88]

centroid problem please help

The solution you are looking at may have used the property of symmetry. The graph of this function is a horizontal parabola opening to the left, and is symmetric about the x-axis. Instead of finding the entire area, they find the area above the x-axis (which is half the area), and multiply by two.

You can find the area in different ways:

(1) integrate with respect to y.

\(\displaystyle A = \displaystyle \int_{-\sqrt{2}}^\sqrt{2} 2-y^2 dy\)

(2) integrate with respect to x and use symmetry. First, solve for y: \(\displaystyle y = \pm\sqrt{2-x}\). Choose either the positive function or the negative function.

\(\displaystyle A = \displaystyle 2\int_{0}^2 \sqrt{2-x} dx\)

For the centroid, what you have looks odd (maybe a mistake on finding the area?)

\(\displaystyle \displaystyle \bar x = \dfrac{1}{A}\int_0^2 x\cdot \left[(\sqrt{2-x}) - (-\sqrt{2-x})\right] dx = \dfrac{2}{A} \int_0^2 x\sqrt{2-x} dx\)

okay got it! that makes much more sense now thank you, so for the Area i used that expression and now got A= 16/3 can you please verify that my value is correct ?

 

[daon2]

The area is not rational.

\(\displaystyle \displaystyle 2\int_0^2 \sqrt{2-x} dx = 4\dfrac{2^{3/2}}{3}=\dfrac{8\sqrt{2}}{3}\)

 

[sjay88]

centroid problem please help

The area is not rational.

\(\displaystyle \displaystyle 2\int_0^2 \sqrt{2-x} dx = 4\dfrac{2^{3/2}}{3}=\dfrac{8\sqrt{2}}{3}\)

okay for the x coordinate of the centroid im kinda stuck at -3 ((4/3) ^3/2 + (4/5)^5/2) for an answer is this correct and how can it be simplified ? Also is the y coordinate of the centroid 0 ?[FONT=MathJax_Main][/FONT]

 

[daon2]

okay for the x coordinate of the centroid im kinda stuck at -3 ((4/3) ^3/2 + (4/5)^5/2) for an answer is this correct and how can it be simplified ? Also is the y coordinate of the centroid 0 ?

Rather than me do it all, can you post your work? The coordinates must, of course, be within the region, so I'm not at all certain how you're arriving at this [negative] number outside of it. But yes, the y coordinate is zero, just by the symmetry of the region.

 

[sjay88]

centroid problem please help

Rather than me do it all, can you post your work? The coordinates must, of course, be within the region, so I'm not at all certain how you're arriving at this [negative] number outside of it. But yes, the y coordinate is zero, just by the symmetry of the region.

yeah ill post my work.



x= 2/A integral (from 0 to 2) xsqrt(2-x) dx
x= 2/ (8sqrt2/3) integral (from 0 to 2) xsqrt(2-x) dx
x= 6/8sqrt2 integral (from 0 to 2) xsqrt(2-x) dx

u= 2-x
du=-dx

x = 6/8sqrt2 integral (from 0 to 2) (-2+u) sqrtu du
x= 6/8sqrt2 integral (from 0 to 2) (-2sqrtu+u^3/2)du
x=6/8sqrt2 (-2) integral (from 0 to 2) (sqrtu du)+integral (from 0 to 2) (u^3/2)du
x=-3 ( 2/3) (2-x) ^(3/2)+ (2(2-x)^(5/2))/5

then i got to -3(4/3)^3/2 + (4/5)^5/2

i agree that i shouldnt be getting a negative value but can u please tell me where i went wrong ?
thanks

 

[DrPhil]

yeah ill post my work.



x= 2/A integral (from 0 to 2) xsqrt(2-x) dx
x= 2/ (8sqrt2/3) integral (from 0 to 2) xsqrt(2-x) dx
x= 6/8sqrt2 integral (from 0 to 2) xsqrt(2-x) dx

u= 2-x
du=-dx
NOTE limits of integration: x=0-->u=2, x=2-->u=0
Integration is from 2 to 0, NOT 0 to 2
This will flip the sign of the integral.

x = 6/8sqrt2 integral (from 0 to 2) (-2+u) sqrtu du
x= 6/8sqrt2 integral (from 0 to 2) (-2sqrtu+u^3/2)du
x=6/8sqrt2 (-2) integral (from 0 to 2) (sqrtu du)+integral (from 0 to 2) (u^3/2)du
x=-3 ( 2/3) (2-x) ^(3/2)+ (2(2-x)^(5/2))/5

then i got to -3(4/3)^3/2 + (4/5)^5/2

i agree that i shouldnt be getting a negative value but can u please tell me where i went wrong ?
thanks

When you substitute the variable of a definite integral, you have to substitute the limits as well.