Change of Variables — Question 9

Question 9 – Solution

Strategy. Scale the ellipse to the unit disk; area determines $ab$, while the $x^2$ moment determines one additional power of $a^2$.

Step 1: Area condition The map $x=au$, $y=bv$ has $dA=ab\,du\,dv$. Since the unit disk has area $\pi$, $$\pi ab=6\pi,\qquad\text{so}\qquad ab=6.$$

Step 2: Moment condition On the unit disk, $$\iint_Du^2\,du\,dv=\frac{\pi}{4}$$ by symmetry or a polar calculation. Therefore $$\iint_{E_{a,b}}x^2\,dA =a^2(ab)\frac{\pi}{4} =\frac{3\pi a^2}{2}.$$ Equating this with $27\pi/2$ gives $a^2=9$. Positivity then yields $$\boxed{a=3},\qquad\boxed{b=2}.$$

Verification The recovered ellipse has area $\pi(3)(2)=6\pi$. Its moment is $a^2(ab)\pi/4=(9)(6)\pi/4=27\pi/2$, so both measurements are satisfied without sign ambiguity.