Problem Solution
The equation of the ellipse shown above may be written in the form:
\[ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \]Since the ellipse is symmetric with respect to both the \( x \) and \( y \) axes, we can find the area of one-quarter and multiply by 4 in order to obtain the total area.
Solve the above equation for \( y \):
\[ y = \pm b \sqrt{1 - \dfrac{x^2}{a^2}} \]The upper part of the ellipse (\( y \) positive) is given by:
\[ y = b \sqrt{1 - \dfrac{x^2}{a^2}} \]We now use definite integrals to find the area of the upper right quarter of the ellipse:
We now make the trigonometric substitution \( \sin t = \dfrac{x}{a} \), which gives \( x = a \sin t \) and \( dx = a \cos t \, dt \). The integral becomes:
\[ \dfrac{1}{4} \text{Area of ellipse} = \displaystyle \int_{0}^{\dfrac{\pi}{2}} a b \left(\sqrt{1 - \sin^2 t}\right) \cos t \, dt \]Since \( \sqrt{1 - \sin^2 t} = \cos t \) for \( t \) varying from \( 0 \) to \( \dfrac{\pi}{2} \), we have:
\[ \dfrac{1}{4} \text{Area of ellipse} = \displaystyle \int_{0}^{\dfrac{\pi}{2}} a b \cos^2 t \, dt \]Use the power-reduction identity \( \cos^2 t = \dfrac{\cos 2t + 1}{2} \) to linearize the integrand:
\[ \dfrac{1}{4} \text{Area of ellipse} = \displaystyle \int_{0}^{\dfrac{\pi}{2}} a b \left( \dfrac{\cos 2t + 1}{2} \right) dt \]Evaluate the integral:
\[ \dfrac{1}{4} \text{Area of ellipse} = \dfrac{1}{2} ab \left[ \dfrac{1}{2} \sin 2t + t \right]_{0}^{\dfrac{\pi}{2}} = \dfrac{1}{4} \pi ab \]Obtain the total area of the ellipse by multiplying by 4: