1
$\begingroup$

Why does: $\iint_D \frac{x^2}{a^2} \:dx\:dy = \iint_D \frac{y^2}{b^2} \:dx\:dy,$ Where: $D:\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1$

1 Answers 1

3

Use the change of variable $ u = \frac{x}{a}, \quad v = \frac{y}{b} $ Then $(u,v)$ belong to the unit disk $D(0,1)$ since $u^2 + v^2 \leq 1$. This leads to $ \int\int_D \frac{x^2}{a^2} dx dy = ab \int \int_{D(0,1)}u^2 du dv$ Then variables $u$ and $v$ can be exchanged because of the symmetry of the problem, i.e. $ \int \int_{D(0,1)}u^2 du dv = \int \int_{D(0,1)}v^2 du dv $ Go back to $(x,y)$ to get the result.