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consider a circular lamina placed in xy plane whose center lies ot origin...
mass per unit area (d) is given =kx^2
consider a small element of lamina ...let it be a ring of radius x and thickness dx which is co axial to the lamina...
length of this element =dl=2pix
area of this element is 2pixdx...............1
mass of dis element is dm=(area).(mass per unit area)=2pixdx(d)
dm=2pikx^3dx............2
moment of inertia of this element about a axis passing through its center and perpendicular
to plane of element is given by dI=dmx^2
dI =2pikx^5dx.........3
total moment of inertia is equla to integral dI limit from 0 to a
I=2pik (x^6/6) lim 0 to a
I = (pika^6)/3 ..............4
we have eq 2 as dm=2pikx^3dx
if we integrate this eq then it will give the total mass of lamina (M)
M=integral(2pikx^3)dx limit 0 to a
M=(pieka^4)/2...............5
solving 4 and 5
I=2/3 ma^2 ans
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