Question

Find the surface area of the part of the surface 2y+ 4z-x^2= 5 that lies above the triangle with vertices (0,0),(2,0), and (2,4).

Answer 1

$\displaystyle \textsf{If surface}\, S\, \textsf{has equation of the form}\\ z = f(x, y), \, \textsf{where}\, (x, y) \in R\, \textsf{and}\, f\, \textsf{has continuous}\\ \textsf{partial derivatives, the surface area}\\\textsf{is given by}\\ A = \iint_R \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, \mathrm{d}A\\ \begin{aligned} 2y + 4z - x^2 &= 5\\ 4z &=5 + x^2 - 2y\\ z &= \frac{5}{4} + \frac{x^2}{4} - \frac{y}{2} \end{aligned}\\ \frac{\partial z}{\partial x} = \frac{x}{2}, \, \frac{\partial z}{\partial y} = -\frac{1}{2}\\ \therefore A = \iint_R \sqrt{1 + \frac{x^2}{4} + \frac{1}{4}} \, \mathrm{d}A\\ A = \frac{1}{2}\iint_R \sqrt{5 + x^2} \, \mathrm{d}A\\ \textsf{Equation of line AC is}\, y = 2x,\\ x \, \textsf{is integrated from}\, 0 \, \textsf{to}\, 2. \\ \begin{aligned} \therefore A &= \frac{1}{2}\int_0^2 \int_0^{2x} \sqrt{5 + x^2} \,\mathrm{d}y\,\mathrm{d}x\\ &= \frac{1}{2}\int_0^2 y\sqrt{5 + x^2} \,\vert_{y=0}^{y=2x}\mathrm{d}x\\ &= \frac{1}{2}\int_0^2 2x\sqrt{5 + x^2} \mathrm{d}x\\ &=\int_0^2 x\sqrt{5 + x^2} \mathrm{d}x \\ &=\frac{1}{2}\int_0^2 \sqrt{5 + x^2} \,\mathrm{d}(5 + x^2)\\ &=\frac{1}{2}\cdot \frac{2}{3} (5 + x^2)^{\frac{3}{2}}\biggr\vert_0^2\\ &=\frac{(5 + 4)^{\frac{3}{2}} - 5^{\frac{3}{2}}}{3} \\ &= \frac{27 - 5\sqrt{5}}{3} \end{aligned}$