14.48 Set up the double integral required to find the moment of inertia I, about the given line, of the lamina bounded by the graphs of the equations. Use a computeralgebra system to evaluate the double integral.x² + y? = b2%3Dp = kline: x = a%3D(a > b)b.I = 2knb2(6²+4a²)r-g²dy dx =4Need Help?Talk to a TutorRead ItWatch ItMaster It

Given:Given that x2+y2=b2and ρ=kThe line x=a a>bTo find:we have to find the moment of Inertia I Here the line x=a and given that a>bIf we draw the circle x2+y2=b2 we get the coordinate (-b,0) and (b,0) on the x-axiswe get the limits of x are x=-b to x=bNow, y2=b2 -x2⇒y=b2 -x2The limits of y are y=0 to y=b2 -x2The moment of inertia isI=2∫-bb∫0b2 -x2 ρ(x-a)2 dy dxBut ρ=kI=2∫-bb∫0b2 -x2 k(x-a)2 dy dx=2k∫-bb (x-a)2∫0b2 -x2 dy dx =2k∫-bb (x-a)2 b2 -x2 dxNow, (x-a)2 b2 -x2 =x2-2ax+a2b2 -x2 =x2b2 -x2-2axb2 -x2+a2b2 -x2=2k∫-bb x2b2 -x2-2axb2 -x2+a2b2 -x2dx=2k ∫-bb x2b2 -x2 dx-2∫-bbaxb2 -x2 dx+∫-bba2b2 -x2dx=2kA-B+C ....(1)A=∫-bb x2b2 -x2 dxput x=b sin v ⇒dx=b cos v dvIf x=b then v=π2If x=-b then v=-π2 A=∫-π2π2 b2 sin2 v b2 -b2 sin2 v b cos v dvA=∫-π2π2 b2 sin2 v b cos v b cos v dvA=∫-π2π2 b4 sin2 v cos2 v dvA=∫-π2π2 b4 1-cos (4v)8 dvA= b4 8v-sin(4v)4-π2π2A= b4 8π2+π2A= π 8b4 B=2∫-bbaxb2 -x2 dxHere axb2 -x2 is odd function we know if f(x) is odd function then ∫-bbf(x) dx=0so, -bbaxb2 -x2 dx=0B=0C=∫-bba2b2 -x2dxC=a2∫-bbb2-x2dxput x=b sin v ⇒dx=b cos v dvIf x=b then v=π2If x=-b then v=-π2 C=a2∫-π2π2 b2 -b2 sin2 v b cos v dvC=a2∫-π2π2 b cos v b cos v dvC=a2b2∫-π2π2 cos2 v dvC=a2b22∫-π2π2 1+ cos(2v)dvC=a2b22 v+sin(2v)-π2π2C=a2b22π substitute the above value in (1)I=2kA-B+CI=2kπ 8b4+a2b22πI=πk4b4+a2b2kπI=b2kπb24+a2Hence, The moment of inertia is I=b2kπb24+a2