Lời giải:Diện tích hình phẳng \(\left( H \right)\) là : \(S=\int\limits_{0}^{2}{\left( \sqrt{4-{{x}^{2}}}-2+x \right) \text{d}x}\).
Đặt \(x=2\sin t\)\(\Rightarrow \text{d}x=2\cos t\text{d}t\).
\(\Rightarrow S=\int\limits_{0}^{\frac{\pi }{2}}{\left( 2\cos t-2+2\sin t \right) 2\cos t\text{d}t}=\int\limits_{0}^{\frac{\pi }{2}}{\left( 4{{\cos }^{2}}t-4\cos t+4\sin t\cos t \right) \text{d}t}\)
\(=\int\limits_{0}^{\frac{\pi }{2}}{\left( 2+2\cos 2t-4\cos t+2\sin 2t \right) \text{d}t}=\left. \left( 2t+\sin 2t-4\sin t-\cos 2t \right) \right|_{0}^{\frac{\pi }{2}}=\pi -2\).
\(\Rightarrow a=1, b=-2\)\(\Rightarrow P=2{{a}^{2}}+{{b}^{2}}=2+4=6\).
