Expressing the integrand under a change of variables

Question

I have a question which asks me to compute the double integral $$\iint_By^2-x^2\,dA$$ where B is the region enclosed by $$y=x,y=x+2,y=\frac{2}{x},y=\frac{2}{x}$$I made a change of variables by letting $$u=xy \qquad \text{and}\qquad v=y-x$$which gives me a very nice region in the $uv$-plane $$1\le u\le 2\qquad\text{and}\qquad 0\le v\le 2$$ However i am having a hard time representing the integrand in terms of $u$ and $v$ $$y^2-x^2=(y-x)(y+x)=v(y+x)$$ Is it possible to express this in terms of $u$ and $v$ or am i wasting my time?

Answer 1

I'm assuming your bounds were actually

$$y=x \hspace{15 pt} y=x+2 \hspace{15 pt} y = \frac{1}{x} \hspace{15 pt} y = \frac{2}{x}$$

It is completely possible to find $x+y$, the trick is noticing the degree of the terms. $u$ is quadratic in $xy$ while $v$ is linear, so we should expect another linear term to involve a square root of a $u$ and $v^2$ somehow. For example

$$v^2 = y^2-2xy+x^2$$

$$v^2+4u = y^2+2xy+x^2 $$

$$\sqrt{v^2+4u} =x+y$$

Now that you have formulas for $y-x$ and $y+x$ you maybe tempted to invert and solve for the Jacobian, but that would be a waste of your time. It's simpler derivative wise to compute the inverse Jacobian instead

$$J^{-1} = \left|\begin{vmatrix}y & x \\ -1 & 1\end{vmatrix}\right| = y+x \implies J = \frac{1}{J^{-1}} = \frac{1}{x+y}$$

In other words finding the formula for $x+y$ was unnecessary because the Jacobian would have canceled it out. I showed the trick for solving for $x+y$ because it can still be useful for other problems. This means our integral is

$$\int_0^2\int_1^2 v\:dvdu$$

Answer 2

Hint:

You may use the following identity

$$(y+x)^2-(y-x)^2=4xy$$