The surface area of that portion of the surface \(z=\sqrt{4-x^2}\) that lies above the rectangle R in the xy-plane whose coordinates satisfy 0 ≤ x ≤ 1 and 0 ≤ y ≤ 4 is equal to

\(z=\sqrt{4-x^2}\)

\(\frac{\partial x}{\partial x}=\frac{-x}{\sqrt{4-x^2}},\frac{\partial z}{\partial y}=0\)

Required surface area = \(\displaystyle\int_{x=0}^1\int^4_{y=0}\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}dxdy\)

\(\displaystyle\int_{x=0}^1\int^4_{y=0}\sqrt{\left(\frac{-x}{\sqrt{4-x^2}}\right)^2+0^2+1}\ dxdy\)

\(\displaystyle\int_{x=0}^1\int^4_{y=0}\sqrt{\left(\frac{4}{\sqrt{4-x^2}}\right)}\ dxdy=2\displaystyle\int_{x=0}^1\frac{dx}{\sqrt{4-x^2}}\times\displaystyle\int_{y=0}^4(1)\ dy\)

\(=2\times\left[\sin^{-1}\left(\frac{x}{2}\right)\right]_0^1\times{4}\)

\(=8\times\left[\sin^{-1}\left(\frac{1}{2}\right)-\sin^{-1}(0)\right]\)

\(=8\left(\frac{\pi}{6}-0\right)=\frac{4}{3}.\pi=\frac{4\pi}{3}\)