Finding Y-Bar for a Constant Density Region

  • Thread starter Thread starter whatlifeforme
  • Start date Start date
  • Tags Tags
    Constant Density
whatlifeforme
Messages
218
Reaction score
0

Homework Statement


constant density. region bounded by y=x-x^2, y=-x.


Homework Equations


this is according to the formula my instructor gave me:

Y(bar) - (1/2)(integral)(a to b)(δ)(f(x) ^2 - g(x)^2) dx
-----------------------------------------------
(integral)(a to b) (δ) (f(x) - g(x)) dx

The Attempt at a Solution



Y(bar) = (1/2) (integral)(0 to 2) [ δ ((x-x^2)^2 - (-x)^2) dx ]
----------------------------------------------------
(integral)(0 to 2) [δ ((x-x^2) - (-x)) dx]

(1/2)(integral) (0 to 2) [δ (-2x^2 + x^4) dx ] = 8δ/15
(integral)(0 to 2) [δ ((x-x^2) - (-x)) dx] = 4δ/3

Y(bar) = 2/5 ; however, the answer is -3/5.
 
Physics news on Phys.org
sorry. solved.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top