Mathematical Immaturity

13.25. Miscellaneous review exercises on conic sections

• (a) Show that the volume of the solid of revolution generated by rotating the ellipse $$ \begin{align*} \\ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \end{align*} $$ about its major axis is $ab^{2}$ times the volume of a unit sphere. Note: This statement can be proved from general properties of the integral, without performing any integrations. (b) What is the result if it is rotated about its minor axis?
• The volume of a solid of revolution with cross sections cut along the $x$ axis is $$\int_{a}^{b}\pi f^{2}(x) \ dx$$ To show how the integral scales, use Theorem 1.19 (Expansion or Contraction of the Interval of Integration): If $f$ is integrable on $[a, b],$ then for every real $k\neq0$ we have $$\int_{a}^{b} f(x)\ dx = \frac{1}{k}\int_{ka}^{kb} f\left(\frac{x}{k}\right)\ dx$$
• (a) The volume of a unit sphere can be expressed as the integral $$V_{c} = \pi\int_{-1}^{1} (1 - x^{2})\ dx$$As we saw in problem 1, we can express the upper half of the ellipse defined by the equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ as $y = f(x) = |b|\sqrt{1 - \left(\frac{x}{a}\right)^{2}}.$ Its volume of revolution about the $x$ axis is $$V_{e} = b^{2}\pi\int_{-a}^{a} 1 - \left(\frac{x}{a}\right)^{2}\ dx$$But from Theorem 1.19 we know that this integral is equal to $$ab^{2}\pi\int_{-1}^{1} (1 - x^{2})\ dx$$Which is $ab^{2}$ times the volume of a unit sphere. $\blacksquare$ \begin{align*} \end{align*} (b) The solid obtained by revolving the ellipse around its minor axis (in this case, the $y$ axis) would have a volume $a^{2}b$ times the volume of a unit sphere. The positive half-ellipse can be written as a function of $y$ as follows: $$x = f(y) = |a|\sqrt{1 - \left(\frac{y}{b}\right)^{2}}$$Its volume of revolution is then $$a^{2}\pi\int_{-b}^{b} 1 - \left(\frac{y}{b}\right)^{2}\ dy$$so it follows that its volume is $a^{2}b$ times that of the unit sphere. $\blacksquare$