Ring of non-uniform density

according to the problem the density is given by

$\lambda =\delta x$ where x is the distance from the end of the rod till the point when we bent the rod we have two parts of the rod with same mass and inertia (like we have divided the ring into two equaled parts ) We calculate the momentum of inertia in one part the multiply it by 2 when we move to polar coordinates the x transform into $R\theta$ we know that $J=\int { { r }^{ 2 }dm }$ here $m=\lambda R \theta$ that yields $dm=\lambda Rd\theta$ and $J=\int { { (\overrightarrow { R } +\overrightarrow { R } ) }^{ 2 }dm }$ then $J=\int { 2R^ 2(1-cos\theta )\lambda Rd\theta }$ then

$J=\int _{ 0 }^{ \pi }{ 2{ R }^{ 2 }(1-cos\theta )\lambda Rd\theta }$ putting $\lambda=\delta R\theta$ then $J=\int _{ 0 }^{ \pi }{ 2\delta { R }^{ 2 }(1-cos\theta ){ R }^{ 2 }\theta d\theta }$ Dont forget to multiply by 2 because you are calculating for only one part of the ring

$J=\int _{ 0 }^{ \pi }{ 4\delta{ R }^{ 4 }(1-cos\theta )\theta d\theta } =1386960.2$

hope my solution had done its' purpose.