Find the surface area of the portion inside the cylindrical $x^{2} +y^{2} =1$, where the surface is given by $z\ =\ x^{2} \ +\ y^{2}$
Here is my thinking: \begin{array}{l} \iint _{S}\sqrt{\left(\frac{\partial Z}{\partial x}\right)^{2} +\left(\frac{\partial Z}{\partial y}\right)^{2} +1} \ dxdy\\ =\iint _{S}\sqrt{4x^{2} +4y^{2} +1} \ dxdy\ \ \ ( x=rcos\theta ,\ y=rsin\theta )\\ =\int _{0}^{2\pi }\int _{0}^{1} r\sqrt{4r^{2} +1} \ drd\theta \ \ \ \ \ \ \ \left( Let\ 1+4r^{2} =u,\ du=8rdr\right)\\ =\int _{0}^{2\pi } d\theta \frac{1}{8}\int _{1}^{5}\sqrt{u} du\\ =2\pi \cdot \frac{1}{8}\left(\frac{2}{3} \cdot 5^{\frac{3}{2}} -\frac{2}{3}\right)\\ =2\pi \left(\frac{1}{12} \cdot 5^{\frac{3}{2}} -\frac{1}{12}\right)\\ =\frac{5^{\frac{3}{2}} -1}{6} \pi \end{array}
Is it right? The result seems incorrect (Sorry, I don't have the correct answer). Thanks in advance.
