Show that the volume generated by revolving the region in the first quadrant bounded by the parabolas $y^2=x, y^2=8x, x^2=y, x^2=8y$ about the $x$ axis is $\frac{279\pi}{2}$.
I've used the substitution $y^2=ux, x^2=vy$. The domain I'm supposed to integrate over is something like this leaf shaped region with another leaf cut out from it.
So I've computed the inverse Jacobian to be $\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\frac{\partial v}{\partial x}$ = $-\frac{y^2}{x^2}(-\frac{x^2}{y^2})-\frac{2x}{y}\frac{2y}{x}$=$1-4$= $-3$
So we get the Jacobian of this transformation to be $\frac{-1}{3}$.
The problem is that I don't know what to plug into my limits of integration and the function I have to integrate over. I have the general sense of where I'm supposed to go from here. For instance, I can visualise values of $u$ taking on a family of curves 'fitted' in between $x=y^2$ to $x=8y^2$ and likewise for values of $v$. Increasing values of $u$ and $v$, intutively seems like going along these family of curves until you get to u=8 and v=8 respectively.
But I have no idea how to translate what I know about calculating the volume of a region rotated into this kind of $u$-$v$ 'coordinate system', if that's even a possible way of putting it.
Intutively, it seems like $\int^8_1 \int^8_1 u^2-v^2 (-\frac{1}{3}) du dv$ would work, but it doesn't. Can someone help me from here? Also, the negative value of the Jacobian seems a little fishy to me. I'd be grateful if someone could help me clarify any mistakes/misconceptions.
$y^2 = x\\ y^2 = 8x\\ x^2 = 8y\\ x^2 = y$
$u = \frac {y^2}{x}\\ v = \frac {x^2}{y}$
Applying this transformation to the curves above we get.
$u = 1, u = 8, v=1, v = 8$
The Jacobian is $\frac 13$
The area of the region is $\int_1^8\int_1^8 \frac 13 \ du\ dv = \frac {56}3$
The volume of revolution... in $xy$ space that would be:
$2\pi \int\int y\ dy\ dx$
Converting to $uv$
$y = (u^2v)^\frac 13\\ 2\pi \int_1^8\int_1^8 \frac 13 (u^2v)^\frac 13\ du\ dv$