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in w=F.X, x is displacement of center of mass of the body not the displacement of system. here also force is 'kx' and displacement of c.o.m. is 0.5*x,so work done will be 0.5kx^2.

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This solution uses multiple integrals but you don't need to compute them. The final computation is a single integral. The ball has spherical symmetry so its momentum of inertia is the same with respect to the $x$-axis, the $y$-axis and the $z$-axis $I=I_x=I_y=I_z$ with $$I_z=\iiint \rho(r)\left(x^2+y^2\right)\mathrm dx\mathrm dy\mathrm dz$$ now sum up the ...

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As you can see, $\rho$ only depends on a single variable: $r$. Thus, it should be intuitive that one can do this problem by integrating only over the variable $r$. To see what you are supposed to do, consider what happens if you fix $r$: You obtain a spherical shell (as was pointed out in the comments). The moment of inertia of a spherical shell is quite ...

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The supposed orthogonal functions on $[-l,l]$ are, $$f_n = e^{in\pi x/l}$$ If these are truly orthogonal, then they must satisfy the continuous orthogonality relation, $$\langle f_m \lvert f_n \rangle = \int_{-l}^{l} f^{\dagger}_m f_{n} \, dx = \delta_{mn} = \left\{ \begin{array}{lr} 1 & : m = n\\ 0 & : m \neq n \end{array} ... 0 it looks like the first term (encircled in red) explodes as it goes to +infinity, But it doesn't explode since, for either term, the real part of the exponent \le 0. Write the two numerators as$$e^{(ik -a)x} = e^{ikx}e^{-ax}$$and$$e^{-(ik + a)x} = e^{-ikx}e^{-ax} Over the domain of integration, the numerators are finite.

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The term you have highlighted contains a decaying (real) exponential factor $\exp(-ax)$ which goes to 0 for $x \rightarrow 0$. The other factor in that term is a complex exponential which is non-infinite for all $x$.

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