= Solution
At height $z$, the <circular sector> has <radius> $r(z)=Re^{-z/H}$ and <angle> $\Theta(z)=2\pi-\pi z/H$. Its <area> is $\Theta(z)r(z)^2/2$, so slicing by horizontal cross-sections gives the <volume integral>
$$
V=\pi R^2\int_0^H\left(1-\frac z{2H}\right)e^{-2z/H}\,dz.
$$
Use the <substitution rule> $t=z/H$. Then
$$
V=\pi R^2H\int_0^1(1-t/2)e^{-2t}\,dt.
$$
An <integration by parts> gives the <antiderivative> $(t/4-3/8)e^{-2t}$, whose endpoint difference is $(3-e^{-2})/8$. Therefore
$$
\boxed{V=\frac{\pi R^2H}{8}(3-e^{-2}).}
$$
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