Determine the moment of inertia of the shaded area about the x axis.[/B]
The Attempt at a Solution
Okey so I now get how to do this the standard method. But I want to know if the method I tried is correct aswell or where my mistake lies.
I look at the portion above x-axis and then multiple it with 2.
So I tried to use the method that I have been taught, a small element that touches the graph like in the picture.
dA= (Rcos (a)- X)dady
since x^2+y^2=R^2, x= sqrt (R^2-y^2)
so I(x)= y^2(Rcos(a)-sqrt (R^2 - y^2))dady
First I integrate this to da with the limits 0 to 1/2a (since the angle goes from 0 to 1/2a)
and then that answer I integrate to dy with limits 0 to Rsin(a/2) (since y goes from 0 to Rsin(a/2)
So basically after the first integration I get:
and then when I integrate this I get this very long and quite complicated equation but is it correct?
I know now the simpler way, but I find it important that I understand it.