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if we want to evaluate the integration $$I=\int\int(x^3y^3)(x^2+y^2)dA$$ over the region bounded by the curves $$xy=1,\\xy=3,\\x^2-y^2=1,\\x^2-y^2=4$$ I used the transformation $$u=xy,\\v=x^2-y^2$$ I found that the jacobian will be $$J=\frac{1}{-2y^2-2x^2}$$ Do I have to get the absolute value of the jacobian? If I did not take the absolute value , I will get the result of the integration = - 30

if I take the absolute value $$J=\frac{1}{2y^2+2x^2}$$ I will get the result= + 30 .

My friend told me we take absolute value of the jacobian only if it is a number .. if this is right .. why we do not take the absolute value if the jacobian is a function?..I think we are sure here that the jacobian is negative since we have x and y squared , so we have to take the absolute value!

Another question, if we have to take always the absolute value of the jacobian (whether it is a number or function) :

if the jacobian is for example $$J=-2x+y$$ It will be positive for some values of x and y only ! .. how can we apply the absolute value inside the double integration?

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If the Jacobian is negative, then the orientation of the region of integration gets flipped.

You have to take the absolute value ALWAYS.

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    $\begingroup$ what if the jacobian is $$J=-2x+y$$ for example ..so we do not know whether it is positive or negative ! it depends on the values of x and y !! $\endgroup$
    – MCS
    Nov 21 '17 at 22:14
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    $\begingroup$ Then you just take $|-2x +y|$ $\endgroup$ Nov 22 '17 at 0:24
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I'm adding an additional answer, in case the issue is actually with the integration of an absolute value, recall the definition of the absolute value:

$\begin{eqnarray*} \left|-2x+y\right| \quad & = & \quad \begin{cases} -2x+y, & \text{if } -2x+y \geq 0, \quad \text{ (i.e. if the quantity is positive)}\\ -(-2x+y), & \text{if } -2x+y < 0, \quad \text{ (i.e. if the quantity is negative)} \end{cases} \\[8pt] % \quad & = & \quad \begin{cases} -2x +y, & y \geq 2x ; \\ 2x - y, & y < 2x \end{cases} \end{eqnarray*}$

As such, if the domain of integration is in just one of these two halves of the plane, you can just apply the appropriate form of the absolute value; and if the domain of integration is a region containing a part of the line $y = 2x$, you simply break up the integral into multiple parts, where one of the two definitions applies.

N.B. here I have expressed $y = y(x)$, but if you wanted to integrate with respect to $x$ first, you could simply rearrange the expression to have $x = x(y)$.

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