15. (12 points) Find the area of the figure formed by the set of points on the right-angle coordinate plane $b O a$
$$
S=\left\{(b, a) \mid f(x)=a x^{3}+b x^{2}-3 x\right.
$$

is a monotonic function on $\mathbf{R}$, and $a \geqslant-1\}$.
Three, 15. When $a=0$, by $f(x)$ being monotonic on $\mathbf{R}$, we know $b=0$. $f(x)$ being monotonic on $\mathbf{R}$ $\Leftrightarrow f^{\prime}(x)$ does not change sign on $\mathbf{R}$.
Since $f^{\prime}(x)=3 a x^{2}+2 b x-3$, therefore, by $\Delta=4 b^{2}+36 a \leqslant 0$,
we get, $a \leqslant-\frac{1}{9} b^{2}$.
Thus, the points $(b, a)$ that satisfy this condition form a closed figure in the $b O a$ coordinate plane, bounded by the curve $y=-\frac{1}{9} x^{2}$ and the line $y=-1$, with an area of
$$
6-2 \int_{0}^{3} \frac{1}{9} x^{2} \mathrm{~d} x=4
$$