Calc 3 question. Please help

Question

anonymous · Answer

A wire of constant mass density (assume to be 1) is bent into the shape of the curve $$y^2=\frac{ 4 }{ 9 }x^3 for   \left| y \right|\le 16/3 $$ and $$0\le x \le4.$$ Set up integrals to find the length of wire and its moments about the x- and y-axes. (This curve has the parameterization $$x(t) = t^2 $$ and $$y(t) = \frac{ 2 }{ 3 }t^3 for.. -2\le t \le2$$

amistre64 · Answer

length is simple enough:$$\int~ds$$
$$\int_{at}^{bt}~\sqrt{(x')^2+(y')^2}~dt$$

amistre64 · Answer

moments i seem to have forgotten the most about

amistre64 · Answer

x' = 2t
y' = 2t^2

integrated from -2 to 2 it seems

anonymous · Answer

I got length, but the moments is what I was really looking for help with. I was kinda assuming it was $$M_y = \int\limits_{x=0}^{x=4}xds$$ but that's a complete guess based off of what I have learned about moments in regular coords.

anonymous · Answer

and a similar equation for M_x just substitute y for x

amistre64 · Answer

is rho(x,y) $$\rho=\frac{ 4 }{ 9 }x^3-y^2~:~ for   \left| y \right|\le 16/3$$???

amistre64 · Answer

rho is 1, bent into the shape of ....

amistre64 · Answer

mass = density times length in this case ...  which is just the length since density is 1

anonymous · Answer

yeah rho is 1 and f(x,y) (the shape is $$y^2 = (4/9) x^3$$ im pretty sure

amistre64 · Answer

does it look like this?
|dw:1384463312444:dw|

amistre64 · Answer

our mass is about 13.57

the symmetry about the x axis would suggest that y=0 would balance it out;  if we balance out the top half; then we could determine the x= axis ....

integrate 2/3 x^(3/2), from 0 to b such that it equals 1/4 of the weight(length)

$$\frac13(\frac25b^{5/2}-0)$$
$$b^{5/2}=\frac31\frac52\frac{W}{4}$$
$$b^{5}=(\frac31\frac52\frac{W}{4})^2$$
$$b=\sqrt[5]{(\frac31\frac52\frac{W}{4})^2}$$

is what im thinking of