Find the volume of the solid generated by revolving the region across the $y$-axis bounded by the graphs of the equations: $x=y^2, x=20y-y^2$, the line $x=102$.
I set up an integral from $0$ to $10$ using the equation $\sqrt{x}$ but for the other equation I don't know how to make it usable. Please help.
Put the second parabola into standard form:
$x - 100 = -(y-10)^2$ which has a vertex at $(100, 10)$ and opens to the left. By coincidence, this is also a point on both parabolas. The line $x = 102$ does not come into play (or else there is no region bounded by all three boundaries together).
Because of the form of the two parabolas, it is easier to integrate along the $y-$axis, in which case the washer or disc method is being used. Each disc is an annulus with inner radius $r_i = y^2$, outer radius $r_o = 20y-y^2$, and thickness $dy$. So for the element of volume
$$\mathrm{d}V = \pi(r_o^2 - r_i^2) dy = \pi [(20y - y^2)^2 - (y^2)^2] \, \mathrm{d}y = \pi (400 y^2 -40y) \, \mathrm{d}y$$
So
$$V = \displaystyle\int\limits_{0}^{10}{\pi (400 y^2 -40y) \, \mathrm{d}y}$$
which is straight-forward.