
STATICS  EXAMPLE

Area Between Curve and x and yaxis


Example 1


Find moment of inertia of the shaded area about
a) x axis
b)
y axis




Solution (a)


Recall, the moment of inertia is the second moment of the area about a given axis or line.
For part a) of this problem, the moment of inertia is about the xaxis. The differential element, dA, is usually broken into two parts, dx and dy (dA = dx dy), which makes integration easier. This also requires the integral be split into integration along the x direction (dx) and along the y direction (dy). The order of integration, dx or dy, is optional, but usually there is an easy way, and a more difficult way. 



Crosssection Area 

For this problem, the integration will be done first along the y direction, and then along the x direction. This order is easier since the curve function is given as y is equal to a function of x. The diagram at the left shows the dy going from 0 to the curve, or just y. Thus the limits of integration is 0 to y. The next integration along the x direction goes from 0 to 4. The final integration from is
Expanding the bracket by using the formula,
(ab)^{3} = a^{3}  3 a^{2} b + 3 a b^{2}  b^{3}






Solution (b)



Similar to the previous solution is part a), the moment of inertia is the second moment of the area about a given axis or line. But in this case, it is about the yaxis, or




Crosssection Area 

The integral is still split into integration along the x direction (dx) and along the y direction (dy). Again, the integration will be done first along the y direction, and then along the x direction. The diagram at the left shows the dy going from 0 to the curve, or just y. Thus the limits of integration is 0 to y. The next integration along the x direction goes from 0 to 4. The final integration from is






Comment 


The area is more closely distributed about the yaxis than xaxis. Thus, the moment of inertia of the shaded region is less about the yaxis as compared to xaxis. 






Example 2


Determine the moment of inertia of y = 2  2x^{2} about the x axis. Calculate the moment of inertia in two different ways. First, (a) by taking a differential element, having a thickness dx and second, (b) by using a horizontal element with a thickness, dy. 



Solution






a) The area of the differential element parallel to y axis is dA = ydx. The distance from x axis to the center of the element is namedy.
y = y/2
Using the parallel axis theorem, the moment of inertia of this element about x axis is
For a rectangular shape, I is bh^{3}/12. Substituting I_{x}, dA, and y gives,
Performing the integration, gives,






(b) First, the function should be rewritten in terms of y as the independent variable. Due to the x^{2} term, there is a positive and negative form and it can be expressed as two similar functions mirrored about y axis. The function on the right side of the axis can be expressed as
The area of the differential element parallel to x axis is
Performing the integration gives,
Performing a numerical integration on calculator or by taking t = 2(2  y) the above integration can be found as,
As expected, both methods (a) and (b) provide the same answer. 



