Find the area $A$ enclosed by the ellipse $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}=1.$

Figure 8 $\,\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}=1$

Solution Figure 8 illustrates the ellipse. Since the ellipse is symmetric with respect to both the $x$-axis and the $y$-axis, the total area $A$ of the ellipse is four times the shaded area in the first quadrant, where $0 \le x \le 2$ and $0 \le y \le 3.$

We begin by expressing $y$ as a function of $x.$ $$\begin{eqnarray*} \frac{x^{2}}{4}+\frac{y^{2}}{9} &=&1 \\[4.5pt] \frac{y^{2}}{9} &=&1-\frac{x^{2}}{4}=\frac{4-x^{2}}{4} \\[4.5pt] y^{2} &=&\dfrac{9}{4}( 4-x^{2}) \\[4.5pt] y &=&\dfrac{3}{2}\sqrt{4-x^{2}}\qquad {\color{#0066A7}{\hbox{\(y \ge 0\)}}} \end{eqnarray*}$$

So, the area $A$ of the ellipse is four times the area under the graph of $y=\dfrac{3}{2}\sqrt{4-x^{2}},$ $0 \le x \le 2.$ That is, $$A=4 \int_{0}^{2}\dfrac{3}{2}\sqrt{4-x^{2}}\,dx =6\int_{0}^{2}\sqrt{4-x^{2}}\,dx$$

Since the integrand contains a square root of the form $\sqrt{a^{2}-x^{2}}$ with $a=2,$ we use the substitution $x=2\sin \theta$, $-\frac{\pi}{2} \le \theta \le \frac{\pi }{2}$. Then $dx=2\cos \theta \,d\theta$. The new limits of integration are:

Then $$\begin{eqnarray*} A&=&6\int_{0}^{2}\sqrt{4-x^{2}}\,dx=6\int_{0}^{\pi /2}\sqrt{4-4\sin^{2}\theta }\cdot 2\cos \theta \,d\theta\\ &=&6\int_{0}^{\pi /2}2\sqrt{1-\sin^{2}\theta }\cdot 2\cos \theta \,d\theta = 24\int_{0}^{\pi /2}\sqrt{\cos ^{2}\theta }\cdot \cos \theta \,d\theta\nonumber \\ &\underset{\underset{\color{#0066A7}{\hbox{\(\cos \theta \ge 0\)}}}{\color{#0066A7}{\uparrow }}}{=}&24\int_{0}^{\pi /2}\cos^{2}\theta \,d\theta \underset{\underset{\color{#0066A7}{\hbox{\(\cos^{2}\theta =\frac{1+\cos (2\theta )}{2}\)}}}{\color{#0066A7}{\uparrow }}}{=} \frac{24}{2}\int_{0}^{\pi /2}\left[1+\cos (2\theta ) \right] \,d\theta \nonumber \\ & =&12\left[ \theta +\frac{1}{2}\sin (2\theta ) \right]_{0}^{\pi /2}=12\left( \frac{\pi }{2}+0\right) =6\pi \end{eqnarray*}$$

The area of the ellipse is $6\pi$ square units.