9. find the area of the part of the surface ( z = x^{2}+y^{2} ) that lies between the cylinders (…

Explanation:

Step1: Find partial derivatives

Given (z = x^{2}+y^{2}), then (\frac{\partial z}{\partial x}=2x) and (\frac{\partial z}{\partial y}=2y). The formula for the surface - area (S=\iint_{D}\sqrt{1 + (\frac{\partial z}{\partial x})^{2}+(\frac{\partial z}{\partial y})^{2}}dA). Substitute the partial derivatives: (1+(\frac{\partial z}{\partial x})^{2}+(\frac{\partial z}{\partial y})^{2}=1 + 4x^{2}+4y^{2}).

Step2: Convert to polar coordinates

In polar coordinates, (x = r\cos\theta), (y = r\sin\theta), and (dA=rdrd\theta). Also, (x^{2}+y^{2}=r^{2}). The region (D) is given by (4\leq r^{2}\leq16) (i.e., (2\leq r\leq4)) and (0\leq\theta\leq2\pi). The integrand (\sqrt{1 + 4x^{2}+4y^{2}}=\sqrt{1 + 4r^{2}}). So, (S=\int_{0}^{2\pi}\int_{2}^{4}\sqrt{1 + 4r^{2}}r\ drd\theta).

Step3: Use substitution for the inner - integral

Let (u = 1+4r^{2}), then (du=8r\ dr). When (r = 2), (u=1 + 16=17); when (r = 4), (u=1+64 = 65). (\int_{2}^{4}\sqrt{1 + 4r^{2}}r\ dr=\frac{1}{8}\int_{17}^{65}\sqrt{u}\ du). Using the power rule (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n=\frac{1}{2})), we have (\frac{1}{8}\times\frac{2}{3}u^{\frac{3}{2}}\big|_{17}^{65}=\frac{1}{12}(65^{\frac{3}{2}}-17^{\frac{3}{2}})).

Step4: Evaluate the outer - integral

(S=\int_{0}^{2\pi}\frac{1}{12}(65^{\frac{3}{2}}-17^{\frac{3}{2}})d\theta). Since (\int_{0}^{2\pi}d\theta=2\pi), then (S=\frac{\pi}{6}(65^{\frac{3}{2}}-17^{\frac{3}{2}})).

Answer:

(\frac{\pi}{6}(65^{\frac{3}{2}}-17^{\frac{3}{2}}))