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

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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.

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