- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
13.25. Miscellaneous review exercises on conic sections
- A parabolic arch has base of length $b$ and height $h.$ Determine the area of the region bounded by the arch and the base.
- Recall that the equation for a parabola opening downwards with vertex $(0, h)$ is $$x^{2} = 4c(y - h)\quad (\text{where}\ c < 0)$$If we place its $x$ intercepts (where $y = 0$) at $\frac{-b}{2}$ and $\frac{b}{2},$ then its base will then be of length $b.$ Setting $x = \frac{b}{2}$ and solving for $c$ we get $c = \frac{-b^{2}}{16h}.$
- Let the equation for the parabola with vertex $(0, h)$ and base of length $b$ be the function of $x:$ $$y = f(x) = \frac{x^{2}}{4c} + h\quad (\text{where}\ c = \frac{-b^{2}}{16h})$$Integrating from $x=\frac{-b}{2}$ to $x=\frac{b}{2}$ we get $$\displaylines{\int_{-b/2}^{b/2} \frac{x^{2}}{4c} + h\ dx &= \ \frac{x^{3}}{12c} + hx\ \Biggr|_{-b/2}^{b/2} \\ &=\frac{b^{3}}{48c} + hb}$$Substituting $c = \frac{-b^{2}}{16h}$ we get $$\displaylines{\frac{b^{3}}{48c} + hb = \left(\frac{16h}{-b^{2}}\right)\frac{b^{3}}{48}} + hb \ = \frac{2}{3}hb \quad \blacksquare$$