I am required to calculate the volume bound by the following surfaces:
$$z=x+y$$ $$xy=1$$ $$xy=2$$ $$y=x$$ $$y=2x$$ $$z=0$$
where $x, y>0$
So my first instinct was to make a subtitution:
$$x+y-z=u$$ $$xy=v$$ $$\frac{y}{x} = w$$
I did this substitution because there are a lot of surfaces and I had to employ the help of a 3d plotter to help me plot the function.
Now here's the problem. While I can figure out the boundaries for $v$ and $w$, I can't do the same for $u$. I might have chosen a bad substitution but I can't see any other.
What should I do next?
Update: I tried putting $u=z$ where $u$ would range from $ [0, \sqrt{vw} + \sqrt{\frac{v}{w}}]$ but my answer still isn't the same as in the workbook. I got the upper bound by expressing $x+y$ via $u,v$
Yes change of variable should work. But it can also be set up in polar coordinates.
We focus on the projection in XY-plane as the bound of $z$ is straightforward.
Using change of variable as you did -
$u = xy, v = \dfrac{y}{x}$.
Then $1 \leq u \leq 2, 1 \leq v \leq 2$
$J^{-1} = \dfrac{2y}{x} \implies |J| = \dfrac{1}{2v}$
$0 \leq z \leq x + y = \sqrt{\dfrac{u}{v}} + \sqrt{uv}$
That leads to integral,
$\displaystyle \int_1^2 \int_1^2 \dfrac{1}{ 2v} \left(\sqrt{uv} + \sqrt{\dfrac{u}{v}}\right) \ du \ dv$
In polar coordinates, $x = r \cos\theta, y = r\sin\theta$
$y = x, y = 2x$ lead to bounds of $\theta$ as,
$\frac{\pi}{4} \leq \theta \leq \arctan(2)$.
$xy = 1, xy = 2$ lead to bounds of $r$ as,
$\sqrt{2\csc (2\theta)} \leq r \leq \sqrt{4\csc (2\theta)}$
$0 \leq z \leq r(\cos\theta+\sin\theta)$