# Setting up Volume Integrals

• Jan 7th 2010, 09:07 AM
Em Yeu Anh
Setting up Volume Integrals
Q: Set up, but do not evaluate an integral for the volume of the solid obtained by rotating the region bounded by the parabolas $x=8y-2y^2$ and $x=4y-y^2$ about $y=5$ and about $x=-3.$

$8y-2y^2=4y-y^2$
$0=y^2-4y$
$0=y(y-4)$
$y=0, y=4$
So intersection points are $(0,0), (0,4)$

From here on I'm stuck, I'm having difficulty drawing out the graphs and determining which one is outside the other. Err at least if that's even the best approach to take? >.<
• Jan 7th 2010, 09:35 AM
ANDS!
One is two times the other. In that interval all values of f(y) are positive. Therefore. . .
• Jan 7th 2010, 06:30 PM
Em Yeu Anh
So to rotate about x = -3 would I integrate:
${\pi}\int_0^4 ((3+(8y-2y^2))^2-(3+(4y-y^2))^2)dy$?