How to determine rectangular and polar radii of gyration?

[sagarbhathwar]

Homework Statement

[/B]
Determine the rectangular and polar radii of gyration of the shaded area about the axes shown.

Homework Equations

dI_x = y²dA
I_x = ∫y²dA

dA = ydx
A = ∫(dA

The Attempt at a Solution

dI_x = y²dA
=(x⁶/16) *(x³/4) dx
=(x⁹/64)dx
I_x =(integrating from 1 to 2) ∫dI_x = 1023/640

dA = ydx
A = (integrating from1 to 2)∫(x³/4)dx = 15/16

k_x = √(I_x/A) = 1.705

I am getting the radius of gyration about y-axis correctly but that baout x-axis is wrong.
The actualy answer given is 0.75. I am unable to figure out where I am going wrong.

 

[BvU]

No, my mistake, let me re-think, or rather: think about this one some more.
One at the time: start with I_x:
Yes: work it out dA bit more and you'll see you miss a factor 1/3 in I_x.

By the way, you confuse me with $k_x =\sqrt{I_x/A} = 1.705$; I get $1.306\;$; a typo ?​

 

[sagarbhathwar]

Yes. Sorry I forgot to take the square root so yea a typo.
But to be honest, I don't get what you are saying. Could you be more specific in where I am going wrong?

 

[BvU]

$dI_x = y^2 dA$ needs to be worked out to $dI_x = y^2 dy dx$: it's a double integral. You seem to treat it as a single one.