Answer to Question #95943 in Calculus for Rachel

Question

Evaluate an appropriate double integral (in Cartesian coordinates) to find the area of the ellipse x^2/a^2+y^2/b^2 = 1, a, b >0 What happens to your answer in the limit a→0^+? Does this make sense?

Answer 1

Let us define the area $D$ in Cartesian coordinates:

$y_1=-\dfrac{b}{a}\sqrt{a^2-x^2}\\ y_2=\dfrac{b}{a}\sqrt{a^2-x^2}\\ x_1=-a\\ x_2=a\\$

Then the value of the area of ellipse is given by the integral over this area $D$ from the function $f(x,y)=1:\\$

$A = \iint_D \limits dxdy\\$

Opening the double integral:

$A = \int_{x_1}^{x_2}\limits dx\int_{y_1}^{y_2}\limits dy= \int_{-a}^{a}\limits dx\int_{-\frac{b}{a}\sqrt{a^2-x^2}}^{\frac{b}{a}\sqrt{a^2-x^2}}\limits dy = 2\int_{-a}^{a}\limits \frac{b}{a}\sqrt{a^2-x^2}dx=2\dfrac{b}{a}\int_{-a}^{a}\limits \sqrt{a^2-x^2}dx\\$

The last integral is the half area of circle of radius a, so it is equal to $\pi a^2/2$

$A=2\dfrac{b}{a} \dfrac{\pi a^2}{2}=\pi ab\\$

If $a$ is approaching to 0, the area is approaching to 0 too.