Consider the region bounded by $x + y = 0$ and $x = y^2 + 3y$.
a) With the washer method, set up an integral of the solid that is rotated about the line x = 4
b) with the shell method, set up an integral expression for the solid rotated about the line y = 1
Solution Provided
a)
$V = \pi \int_{-4}^{0} (|y^2 +3y| +4)^2 - (4+y)^2 dy$
www.wolframalpha.com/input/?i=Pi*Integrate[(4+%2B+|y^2+%2B3y|)^2+-++(4%2By)^2%2C{y%2C-4%2C0}]
The solution I thought would be
$V = \pi \int_{-4}^{0} (4 - (y^2 +3y))^2 - (4+y)^2 dy$
http://www.wolframalpha.com/input/?i=Pi*Integrate[%284+-+%28y^2+%2B3y%29%29^2+-++%284%2By%29^2%2C{y%2C-4%2C0}]
Doesn't the absolute value sign in the integral will actually reflect the region to the fourth quadrant and hence make the +4 meangingless?? More importantly, why is my integral wrong?
Solution Provided
b) $V = 2 \pi \int_{-4}^{0} (-y - (y^2 + 3y))(y+1) dy$
I thought it should be $V = 2 \pi \int_{-4}^{0} (-y - (y^2 + 3y))(1-y) dy$
The region is below the x-axis, yet when they have y + 1, wouldn't that give me negative radius?