Find the volume of the solid that results when the region enclosed by x=y^2, and x=y is revolved about the line x=-1.
most important ... sketch a diagram.
$\displaystyle x = y^2$ and $\displaystyle x = y$ intersect at $\displaystyle y = 0$ and $\displaystyle y = 1$
washers w/r to y ...
$\displaystyle V = \pi \int_c^d [R(y)]^2 - [r(y)]^2 \, dy$
$\displaystyle R(y) = y - (-1) = y + 1$
$\displaystyle r(y) = y^2 - (-1) = y^2 + 1$
$\displaystyle V = \pi \int_0^1 (y+1)^2 - (y^2+1)^2 \, dy$