I'm doing the following exercise. However I don't get the book's result:

Consider the transformation $T$ given by the equations: $$x=u+v,\;\;\;\;\;\;y=v-u^2.$$ A triangle Q in the plane $(u,v)$ has vertices in $(0,0)$, $(2,0)$, $(0,2)$. Consider that $T(Q)=D$ in the $xy$ plane. Now compute the area of $D$ through a double integral on $Q$.

We know that the area function is the constant $1$. Therefore, computing the Jacobian which is $2u$: $$a(D)= \int^2_0\int^{2-u}_02u\,dv\,du = \int^2_04u-2u^2 du = \left(2u^2-\frac{2}{3}u^3\right)^2_0=\frac{8}{3}.$$ The correct answer is $\dfrac{14}{3}$. Any thoughts on the problem? Am I missing something concerning the limits of integration?

  • 2
    $\begingroup$ I got $1+2u$ for the Jacobian. $\endgroup$
    – David
    Nov 6, 2014 at 4:02
  • $\begingroup$ Oh silly me. I wrote a + that looked like x. Thanks. $\endgroup$ Nov 6, 2014 at 4:04

1 Answer 1


In summary, $$\int_0^2 dv \int_0^{2-v}(1+2u) du=\int_0^2 dv (6-5v+v^2) = 14/3.$$


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