The question is incomplete. The complete question is :
A plate of uniform areal density [tex]$\rho = 2 \ kg/m^2$[/tex] is bounded by the four curves:
[tex]$y = -x^2+4x-5m$[/tex]
[tex]$y = x^2+4x+6m$[/tex]
[tex]$x=1 \ m$[/tex]
[tex]$x=2 \ m$[/tex]
where x and y are in meters. Point [tex]$P$[/tex] has coordinates [tex]$P_x=1 \ m$[/tex] and [tex]$P_y=-2 \ m$[/tex]. What is the moment of inertia [tex]$I_P$[/tex] of the plate about the point [tex]$P$[/tex] ?
Solution :
Given :
[tex]$y = -x^2+4x-5$[/tex]
[tex]$y = x^2+4x+6$[/tex]
[tex]$x=1 $[/tex]
[tex]$x=2 $[/tex]
and [tex]$\rho = 2 \ kg/m^2$[/tex] , [tex]$P_x=1 \ $[/tex] , [tex]$P_y=-2 \ $[/tex].
So,
[tex]$dI = dmr^2$[/tex]
[tex]$dI = \rho \ dA \ r^2$[/tex] , [tex]$r=\sqrt{(x-1)^2+(y+2)^2}$[/tex]
[tex]$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$[/tex]
[tex]$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$[/tex]
[tex]$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \ dy \ dx$[/tex]
[tex]$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$[/tex]
[tex]$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$[/tex]
[tex]$I=\frac{32027}{21} \times 2$[/tex]
[tex]$= 3050.19 \ kg \ m^2$[/tex]
So the moment of inertia is [tex]$3050.19 \ kg \ m^2$[/tex].
