# Do we have to take the absolute value of the jacobian ONLY if it is a number?

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?

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

You have to take the absolute value ALWAYS.

• 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 !!
– MCS
Nov 21 '17 at 22:14
• Then you just take $|-2x +y|$ Nov 22 '17 at 0:24

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 $x = x(y)$.