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I just started learning about the substitution rule for multiple integrals and I decided to give myself an example problem:

Calculate $\iint_R{(x^2 + y^2)dA}$ with $R = \{(x, y) \in \Bbb{R} \ |\ 0 \le x + 2y \le 2 \land 0 \le x - y \le 1\}$

To calculate this, I decided to substitute $x = \frac{1}{3}(u+2v)$ and $y = \frac{1}{3}(u-v)$, (which is the same as $u = x + 2y$ and $v = x - y$). Under this substitution, the region of integral becomes $S = \{(u, v) \in \Bbb{R} \ |\ 0 \le u \le 2 \land 0 \le v \le 1\}$, which we can integrate directly. Now our integral is $$ \begin{align} \iint_R{(x^2+y^2)dA} & = \int_0^1\int_0^2{\left[\frac{1}{9}(u+2v)^2 + \frac{1}{9}(u-v)^2\right]\left|\frac{\partial{(x, y)}}{\partial{(u, v)}}\right|dudv} \\ & = -\frac{1}{27}\int_0^1\int_0^2{\left[(u+2v)^2 + (u-v)^2\right]dudv}. \end{align} $$ Because $\left|\frac{\partial{(x, y)}}{\partial{(u, v)}}\right| = -\frac{1}{3}$.

Here is when I run into a problem. The original integrand is positive everywhere, so the volume is positive as well. But the transformed integrand is positive everywhere as well, so when multiplied by a negative constant it must mean that the volume is negative. How can the volume be both positive and negative at the same time? Where was my error?

user3002473
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    A negative sign says that your substitution reversed the orientation of the region $R$. This is a 2-dimensional analogue of what you have seen in univariate case, where you need to swap the boundaries, when you make a substitution with a negative derivative, i.e. use a decreasing function. If left unfixed that will reverse the sign. The remedy is easy: negate the result. BTW Well spotted! If only all the students were equally observant! – Jyrki Lahtonen Oct 12 '14 at 13:27
  • You should be taking $\left |-\dfrac1 3\right|$. See this and this. – Git Gud Oct 12 '14 at 13:30
  • Oh, I thought the vertical lines around the jacobian meant you take the determinant of it, not the absolute value of the determinant. Thanks guys for pointing that out! – user3002473 Oct 12 '14 at 13:44
  • Great question. – mookid Oct 12 '14 at 14:21
  • @user3002473 And it can be determinant also. This depends on the text. – Git Gud Oct 12 '14 at 14:43

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