Question

1. find the area of the part of the surface ( z = x^{2}+y^{2} ) that lies between the cylinders ( x^{2}+y^{2}=4 ) and ( x^{2}+y^{2}=16 ).

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:

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