30. Let $a, b \in \mathbf{R}^{+}$, find the minimum value of $y=\frac{a}{\sin ^{3} \theta}+\frac{b}{\cos ^{3} \theta}, \theta \in\left(0, \frac{\pi}{2}\right)$.
30. Introducing a constant $k$, we have $y=\frac{a}{2 \sin ^{3} \theta}+\frac{a}{2 \sin ^{3} \theta}+3 k \sin ^{2} \theta+\frac{b}{2 \cos ^{3} \theta}+\frac{b}{2 \cos ^{3} \theta}+3 k \cos ^{2} \theta-3 k \geqslant$ $5 \sqrt[5]{\left(\frac{a}{2 \sin ^{3} \theta}\right)^{2}\left(k \sin ^{2} \theta\right)^{3}}+5 \sqrt[5]{\left(\frac{b}{2 \cos ^{3} \theta}\right)^{2}\left(k \cos ^{2} \theta\right)^{3}}-3 k=5 \sqrt[5]{a^{2} k^{3}}+5 \sqrt[5]{b^{2} k^{3}}-3 k$, with equality if and only if $\frac{a}{\sin ^{3} \theta}=k \sin ^{2} \theta$ and $\frac{b}{\cos ^{3} \theta}=k \cos ^{2} \theta$, i.e., $\tan ^{5} \theta=\frac{a}{b} \Leftrightarrow \tan \theta=\sqrt[5]{\frac{a}{b}}, \theta=\arctan \sqrt[5]{\frac{a}{b}}$. Therefore, when $\theta=\arctan \sqrt[5]{\frac{a}{b}}$, $y$ attains its minimum value, which is easily found to be $\left(a^{\frac{2}{3}}+b^{\frac{2}{5}}\right)^{\frac{5}{2}}$.