Mathematical Immaturity

13.25. Miscellaneous review exercises on conic sections

• Show that the area of the region bounded by the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $ab$ times the area of a circle with radius 1. Note: This statement can be proved from general properties of the integral, without performing any integrations.
• 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 $$\int_{a}^{b} f(x)\ dx = \frac{1}{k}\int_{ka}^{kb} f\left(\frac{x}{k}\right)\ dx$$
• We can express a circle of radius 1 centered at the origin as an ellipse with $a = b = 1.$ In other words the standard form of a circle is $x^{2} + y^{2} = 1.$ Expressing this as a function of $x$ gives us the graph of the upper semicircle $y = f(x) = \sqrt{1 - x^{2}}.$ Taking twice the area under the curve from $x = -1$ to $x = 1$ we get $$A_{circle} = 2\int_{-1}^{1}\sqrt{1 - x^{2}} \ dx$$Doing the same for the ellipse with equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ from its left to its right vertices, we see that its area is $$A_{ellipse} = 2|b|\int_{-a}^{a}\sqrt{1 - \left(\frac{x}{a}\right)^2} \ dx$$But we know from Theorem 1.19 (Expansion or Contraction of the Interval of Integration) that this integral is equal to $$2ab\int_{-1}^{1}\sqrt{1 - x^{2}} \ dx = abA_{circle} \quad \blacksquare$$