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Let $D$ be the region in the first quadrant ($x>0$, $y>0$) of the $xy$-plane bounded by the curves

  • $y=\sqrt x$,
  • $y=2\sqrt x$,
  • $x^2+y^2=1$,
  • $x^2+y^2=4$.

Using a change of variables, evaluate the double integral
$$\int\int_D\frac{2x^2+y^2}{xy} dA.$$

user327696
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1 Answers1

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First of all , it is clear that u will have to set your u and v in this form :

u = $y/{\sqrt x}$ , v = $x^2 + y^2 $ (Could also consider using u= $y^2 / x $ instead for easier computation but remember to change your values accordingly)

Next , the conventional way is to compute x and y directly in terms of u and v but in this case, it would be too troublesome to do so. Hence, we make use of the fact that det(A) = $\frac{1}{det(B)}$ to calculate the absolute value of the Jacobian instead. It would be just simple integration after that.

mOnku
  • 11