0
$\begingroup$

I have problem to intrgrate over $D$ a parallelogram restricted by $(-1,-1)$, $(-4,0)$, $(-7,5)$ and $(-10,6)$.

This is my solution, which is wrong. Where did I go wrong?

enter image description here

  • 1
    please type your solution here – 2017-02-11
  • 0
    What is the correct answer supposed to be? Off by a factor 2? – 2017-02-11
  • 0
    Your Jacobian seems wrong. If I well understand your calculation the correct value should be $3/16$. – 2017-02-11
  • 0
    Thank you, i was off by the factor 2. https://i.stack.imgur.com/TXY3p.jpg – 2017-02-11

1 Answers 1

0

Hint:

You have a mistake in the change of coordinates.

The correct values are: $$ \begin{cases} x=\frac{3}{4}u-\frac{1}{2}v\\ y=-\frac{1}{4}u+\frac{1}{2}v \end{cases} $$ so the Jacobian is $J=\frac{1}{4}$, and we have $$ \int_{-4}^{8}\int_{-8}^{-4}\frac{1}{4}u(-v-1)du dv $$