00:01
The given curve is y is equal to x cube and x equal to negative 2 and x's is the given region is bonded by the y is equal to x cube x equal to and x axis so graph of the reason is the graph of the given curve is so this is our given graph and this is our x equal to negative 2 so this is our area of the bonded region and this coordinate is negative negative 2 comma negative weight so we need to find the center of mass of the given result the center of mass that is x c comma y c is equal to m m y moment inertia along the y xx is over m negative momentum awareness here about xx is over m so now the first of all m y can be calculated at the double integration over the is an r of x multiple to dy.
01:23
This is equal to x varied from negative 2 to 0 and y varied from x cube to 0 that is the x dy.
01:33
So here will be d a d a, d y, so integrate with respect to y, it will give negative 2, that is a y, here will be x, lower limit x to 0 multiple d x.
01:50
So to the upper limit in place of y it gives 0.
01:55
It gives 0.
01:55
Negative x cube multiply d x this is equal to negative of negative 2 to 0 x to the power 4 multiply to d x so integrate with respect to x is equal to negative x to the power 5 over 5 limit the right from negative 2 to 0 this equal to negative 0 negative of negative to over 5 over 5 so this is equal to negative 32 over 5 similarly we need to find the mass m is equal to the well integral over the nr of d y d x it is equal to negative 2 to 0 x cube to 0 d x integrate with respect to y negative 2 to 0 that is 0 negative x cube multiply to d x this is equal to negative integrate with respect to x it gives x to power 4 over 4 limit vary from negative to 0 this is equal to negative 0 negative 16 over 4 so this is equal to negative negative negative positive 4 so now again we need to find the moment of inertia about x x is equal to double integral 4 zion r of x multiply to sorry here will be y multiply to d x is equal to negative to 0 here x cube to 0 that is a y d .y, dx.
03:40
Integrate with respect to y.
03:42
This is equal to y square over 2.
03:45
Limit vary from, that is x cube to 0, multiply to dx.
03:50
Sub to the upper limit in place of y, it gives 0 negative x to 2 x to the 5 over 2, multiply to dx...