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September 21, 2014

September 21, 2014

Posted by **COFFEE** on Sunday, June 17, 2007 at 6:15pm.

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Is this basically 1/4 of an oval/ellipse? If so then the area would be: pi*9*3, correct?

So the X coordinate would equal: 1/Area * Integral from 0 to 9 of (x*f(x))*dx

Which equals: (4/(27*pi))*[(2/5)(x^(5/2))] evaluated at 9 and 0 which equals: 4.584?

The Y coordinate would equal: 1/Area * Integral from 0 to 3 of (1/2)*[f(x)]^2*dx

Which equals: (4/(27*pi))*(x^2)/4 evaluated at 3 and 0 which equals: 0.955

Am I using the wrong equation for area?

If you mean the area bordered by y = sqrt x, y=0 and x=9.

The value of that area is

INTEGRAL OF: sqrt (x) dx

0 to 9

= x^(3/2)/(3/2) @ x=9 - x^(3/2)/(3/2) @ x=0

= (2/3)*27 - 0 = 18

===============================

The x-centroid Xc is

[INTEGRAL OF: x*sqrt (x) dx]/(area)

0 to 9

= [x^(5/2)/(5/2)@x=9]/ 18

Xc = (2/5)(243)/18 = 5.4

===============================

The y-centroid Yc is

[INTEGRAL OF: y*sqrt (x) dx]/(area)

0 to 9

= [INTEGRAL OF: x dx]/(area)

0 to 9

= [x^2/2]/18 @ x=9

= 2.25

- Calculus -
**Gary**, Monday, September 22, 2008 at 6:43pmThe answer to the Y-centroid is incorrect.

The answer for that part is 1.125 or (81/72)

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