= Solution
The PDF diagram includes a solid base of thickness $d$, occupying $-d\leq z\leq0$ and $0\leq r\leq R$. Above it, $0\leq z\leq h$, the metal lies between the two specified curved surfaces. In <cylindrical coordinates>, the <Jacobian determinant> contributes a factor $r$, so the <mass> is
$$
\begin{aligned}
M={}&\rho\int_{-d}^0\int_0^{2\pi}\int_0^Rr\,dr\,d\phi\,dz\\
&+\rho\int_0^h\int_0^{2\pi}\int_{(9/10)R(1+z^2/h^2)}^{R(1+z^2/h^2)}r\,dr\,d\phi\,dz.
\end{aligned}
$$
The base contributes $\pi\rho R^2d$. The wall contributes
$$
\pi\rho R^2\left(1-\frac{81}{100}\right)\int_0^h\left(1+\frac{z^2}{h^2}\right)^2dz=\pi\rho R^2\frac{19}{100}h\left(1+\frac23+\frac15\right)=\frac{133}{375}\pi\rho R^2h.
$$
Consequently
$$
\boxed{M=\pi\rho R^2\left(d+\frac{133h}{375}\right).}
$$
The base thickness $d$ comes from the diagram; omitting it would omit part of the cup's <volume>.
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