Mathematical Immaturity

13.25. Miscellaneous review exercises on conic sections

• Two parabolas having the equations $y^{2} = 2(x - 1)$ and $y^{2} = 4(x - 2)$ enclose a plane region $R.$
  (a) Compute the area of $R$ by integration.
  (b) Find the volume of the solid of revolution generated by revolving $R$ about the $x$-axis.
  (c) Same as (b), but revolve $R$ about the $y$-axis.
• (a) Rearrange the equations for the parabolas as functions of $y:$ $$y^{2} = 2(x - 1) \quad \rightarrow \quad x = f(y) = \frac{y^{2}}{2} + 1$$ $$y^{2} = 4(x - 2) \quad \rightarrow \quad x = f(y) = \frac{y^{2}}{4} + 2$$ To find the limits of integration, find the points at which the two functions intersect, or in other words where: $\frac{y^{2}}{2} + 1 = \frac{y^{2}}{4} + 2.$
  (b) Take $y$ as a function of $x$ to give $y^{2} = f^{2}(x)$ for each parabola and then use $f^{2}(x)$ to calculate the solid of revolution about the $x$-axis: $$\int_{a}^{b}\pi f^{2}(x) \ dx$$To find the corresponding limits of integration, note that both the functions for $y^{2}$ are symmetric about the $x$-axis (an exercise for the reader: why is this the case?). Thus, for each integral, we only need to find the points at which the graphs intersect eachother and the $x$-axis.
  (c) The cross-sectional areas are now functions of $y,$ and thus the solid of revolution about the $y$-axis is $$\int_{a}^{b}\pi f^{2}(y) \ dy$$To find the limits of integration, find the points at which the two graphs intersect. Note: these are the same points of intersection from (a).
• (a) To find the area of the region $R$ bounded by the graphs of the two functions, first we must find the areas underneath each graph from their points of intersection. $$\displaylines{A_{0} &= \int_{-2}^{2} \frac{y^{2}}{2} + 1 \ dy\\ &= \frac{y^{3}}{6} + y\ \Biggr|_{-2}^{2}\\ &= \frac{80}{12}}$$ $$\displaylines{A_{1} &= \int_{-2}^{2} \frac{y^{2}}{4} + 2 \ dy\\ &= \frac{y^{3}}{12} + 2y\ \Biggr|_{-2}^{2}\\ &= \frac{112}{12}}$$Subtracting $A_{0}$ from $A_{1}$ we get $\frac{32}{12}$ or $\frac{8}{3}$ $\quad \blacksquare$ \begin{align*} \end{align*} (b) To find the volume $V$ of the solid obtained by rotating the region $R$ about the $x$-axis, we find the difference between the volumes obtained by rotating each parabola $y^{2}$ about the $x$-axis. We denote these volumes by $V_{0}$ and $V_{1},$ respectively. Let: $$\displaylines{V_{0} &= \pi\int_{1}^{3}2(x - 1) \ dx\\&=2\pi(\frac{x^{2}}{2} - x)\Biggr|_{1}^{3}\\&=4\pi}$$ $$\displaylines{V_{1} &= \pi\int_{2}^{3}4(x - 2) \ dx\\&=4\pi(\frac{x^{2}}{2} - 2x)\Biggr|_{2}^{3}\\&=2\pi}$$Subtracting the smaller volume $V_{1}$ from the larger $V_{0}$ we get $V = 2\pi\quad\blacksquare$ \begin{align*} \end{align*} (c) We know from (a) that the parabolas taken as functions of $y$ are $$x = \frac{y^{2}}{2} + 1; \quad x = \frac{y^{2}}{4} + 2$$The volume of their solids of revolution $V_{0}$ and $V_{1}$ are $$\displaylines{V_{0}&=\pi \int_{-2}^{2}\frac{y^{4}}{4} + y^{2} + 1 \ dy\\&=\pi \left(\frac{y^{5}}{20} + \frac{y^{3}}{3} + y\right) \Biggr|_{-2}^{2}\\&= \pi \left(\frac{16}{5} + \frac{16}{3} + 4\right)}$$ $$\displaylines{V_{1}&=\pi \int_{-2}^{2}\frac{y^{4}}{16} + y^{2} + 4 \ dy\\&=\pi \left(\frac{y^{5}}{80} + \frac{y^{3}}{3} + 4y\right) \Biggr|_{-2}^{2}\\&= \pi \left(\frac{4}{5} + \frac{16}{3} + 16\right)}$$Subtracting $V_{0}$ from $V_{1}$ we get $V = \frac{48 \pi}{5}\quad\blacksquare$