Calculate the area between $x^2 = py$,$ x^2 = qy$, and $y = ax$, $y = bx$! So how do you calculate the area between $x^2 = py$,$ x^2 = qy$, and $y = ax$, $y = bx$?
I have a solution, it might be bad, so that is why I'm asking. 
So I drew the area, supposing that $b>a$ and $p>q$. 
The points of intersections in the area we are looking for are: $ap$, $bp$, $aq$,$bq$. 
Let $I$ be the area. So I suppose that 
$$I=\int_{ap}^{bp}\int_{bx}^{\frac{1}{p}x^2}\,\mathrm{d}y\,\mathrm{d}x+\int_{bp}^{aq}\int_{ax}^{bx}\,\mathrm{d}y\,\mathrm{d}x+\int_{aq}^{bq}\int_{\frac{1}{q}x}^{bx}\,\mathrm{d}y\,\mathrm{d}x$$
Is this right?
Thanks in advance!!
 A: Let's assume that $ b > a > 0 $ and  $ p > q > 0 $. Solve all with respect to $ y $ and you get : $ y=ax, y=bx, y=(1/q)x^2, y=(1/p)x^2 $. By drawing up these functions with the way we defined the constants, the calculation of the area will go like that : 
1) Find the intersections :
$ ax=(1/q)x^2 \Leftrightarrow x=aq $ (calculating the others) : $x=aq, x=ap, x=bp, x= bq $ 
2) Looking carefully on the plot and making sub-areas, the total area is : 
$\int_{aq}^{bq}((1/q)x^2 - ax) dx + \int_{bq}^{bp} (bx - (1/p)x^2)  dx - \int_{bq}^{ap} (ax - (1/p)x^2)  dx $
3) The  condition is that $ ap > bq $ . Otherwise, the solution changes but is in the same way of thinking.
The exercise is demanding because you need to see and subtract a small area between $ax $ and $ (1/p)x^2 $. It took some time to make a good plot and notice the small things. I hope I haven't done any silly mistake with the limits of integration. Feel free to ask me anything that you may want to ask or point out something that doesn't seem correct.
