Engineering Statics

F-27


Determine the moment of inertia and radius of gyration of the shaded area with respect to the x axis.

Vertical cuts are used.  Therefore, dA = x dy.
The shape given is an ellipse.  In order to perform the integral x must 
by found as a function of y, or x=f(y).
Since the upper and lower halves of the ellipse are identical, it is easier 
to find the moment of inertia of the upper half and then multiply by 2.
A substitution is made for x.  The two contants a and b are factored out 
of the integral sign.  The integration for the upper half proceeds from 
0 to b.
Trigonometric substitution is used to solve this integral.  The limits of 
integration are changed to <font face=symbol>q</font> variation.
The trigonometric substitution is placed in the integral.

The identity 1-sin2=cos2 is used to clear the square root in the integral." >
The double angle formulas for cos(<font face=symbol>q</font>) and sin(<font face=symbol>q</font>) are used to 
simplify the expression.
The two terms are multiplied.  Note that the multiplication gives the 
difference of two squares.
The identity 1-cos<sup>2</sup>(<font face=symbol>q</font>)=sin<sup>2</sup>(<font face=symbol>q</font>) is used.
The double angle formula is used again.
Constants are factored out of the integral.
The integration is performed.
Limits of integration are substituted.
Doubling the answer gives the moment of inertia about the x axis.
The area of the half ellipse is obtained from tables.
The radius of gyration is given as the square root of the moment of 
inertia divided by the area.
Expressions are substituted for the moment of inertia and the area and the 
expression is simplified.
The square root is taken and the radius of gyration found.