Solid of Revolution Problem (Washer and Shell Method) 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? 
 A: You are correct. 
Consider the crude drawing below:
 
Using the washer method:
A typical washer, generated by revolving the  line segment $\color{orange}{\ell_y}$ about the line $\color{gray}{x=4}$, is shown in gray above. 
The outer radius, $\color{darkgreen}{r_o}$ of this washer is
$$\eqalign{
\color{green}{r_o}&= 4-(\color{maroon}{y^2+3y} )
}
$$
and the inner radius, $\color{darkblue}{r_i}$ is
$$\eqalign{
\color{darkblue}{r_i}&= 4-(\color{pink}{-y})
}
$$
It important to realize that the above expressions work for all washer elements.
The area of the washer element at $y$ is
$$\eqalign{
\pi (r_o^2-r_i^2 ) =\pi\bigl[ \bigl(4-(y^2+3y)\bigr)^2- (4+y )^2 \bigr]
}
$$
Since the washers "start" at $y=-4$ and "end" at $y=0$, the volume of the solid of revolution is
$$
\int_{-4}^0 \pi\bigl[ \bigl(4-(y^2+3y)\bigr)^2- (4+y )^2 \bigr]\, dy,
$$
as you have.
Your solution to part b) is correct as well.
With the shell method, you are revolving the horizontal line segment $\color{orange}{\ell_y}$  about the line $\color{gray}{y=1}$. The length of $\color{orange}{\ell_y}$ is 
$\color{pink}{-y} -(\color{maroon}{y^2+3y})$, and the distance from $\color{orange}{\ell_y}$ to the line $\color{gray}{y=1}$ is  $1-y$.
