Answer to Question #142087 in Calculus for Promise Omiponle

Question

Find the surface area of the part of the circular paraboloid z=x^2+y^2 that lies inside the cylinder x^2+y^2=25.

Answer 1

$\displaystyle \textsf{The area of a surface,}\,z, \, \textsf{above}\\\textsf{a region}\,R\, \textsf{of the Cartesian}\\ \textsf{plane is given by}\\ S = \iint_R \sqrt{\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 + 1}\, \mathrm{d}A\\ z = x^2 + y^2, \frac{\partial z}{\partial x} = 2x, \frac{\partial z}{\partial y} = 2y\\ \textsf{The surface area over the region}\\ \textsf{defined by}\, x^2 + y^2 = 25\, \textsf{is given by}\\ S = \iint_R \sqrt{1 + 4x^2 + 4y^2}\, \mathrm{d}x\,\mathrm{d}y\\ \textsf{Converting to Polar coordinates}\\ \begin{aligned} S &= \iint_R \sqrt{1 + 4x^2 + 4y^2}\, \mathrm{d}x\,\mathrm{d}y\\ &= \int_0^{2\pi}\int_0^5\sqrt{1 + 4(r\cos\theta)^2 + 4(r\sin\theta)^2}\, r\mathrm{d}r\mathrm{d}\theta \\ &= \int_0^{2\pi}\int_0^5\sqrt{1 + 4r^2(\cos^2\theta + \sin^2\theta})\, r\mathrm{d}r\mathrm{d}\theta \\ &= \int_0^{2\pi}\int_0^5\sqrt{1 + 4r^2}\, r\mathrm{d}r\mathrm{d}\theta \\ &= \int_0^{2\pi}\frac{1}{12}(1 + 4r^2)^{\frac{3}{2}}\biggr\vert_0^5 \,\mathrm{d}\theta \\ &= \int_0^{2\pi}\frac{1}{12}\left(101^{\frac{3}{2}} - 1\right)\, \mathrm{d}\theta\\ &= \frac{1}{12}\left(101\sqrt{101} - 1\right)\int_0^{2\pi}\mathrm{d}\theta \\ &= \frac{1}{12}\left(101\sqrt{101} - 1\right)\cdot\theta\vert_0^{2\pi} = \frac{1}{12}\left(101\sqrt{101} - 1\right)\cdot 2\pi \\ &= \frac{\pi}{6}\left(101\sqrt{101} - 1\right) \end{aligned}$