Let \(f(x)=\left\{\begin{aligned}x^{2}&:x<0\\ x&:x>0\end{aligned}\right.\). Area bounded by the curve \(y=f(x),y=0\) and \(x^{2}=9a^{2}\) is \(\frac{9a}{2}\), then \(a=\)?

\(\frac{1}{2}\)

\(\begin{aligned}\text{Area}&=\int_{-3a}^{0}f(x)dx+\int_{0}^{3a}f(x)dx\\ &=\int_{-3a}^{0}x^{2}dx+\int_{0}^{3a}xdx\\ &=9a^{3}+\frac{9a^{2}}{2}\end{aligned}\hspace{5cm}\) Thus, \(\frac{9a}{2}=9a^{3}+\frac{9a^{2}}{2}\). So, \(a=-1,a=\frac{1}{2}\)