How to find the volume of a solid of revolution? The region bounded by $x^2-y=0$ and $x+y=0$ is rotated around $y=1$
I solved for:
$$y = x^2 $$
and $$y = -x$$
And I then also solved for x for both of those. 
I set up the integral like so:
$$ V = 2\pi\int_{-1}^0 (y-1)(-y-y^{1 \over 2})dy $$
And used $y$ as the variable to integrate by because the functions are being rotated around $y=1$. 
However, I keep getting weird answers like -$17 \over 30$, I'm not sure what I'm doing wrong though.  
 A: You are using the method of cylindrical shells. I will do it more or less like in your approach, though that would not be a first choice.
I imagine you drew a picture. It is correct to integrate with respect to $y$. But the picture makes it clear that $y$ goes from $0$ to $1$.
The radius of the shell at height $y$ is $1-y$. For the 'height" of the cylinder at $y$, note that it is $-y-(-y^{1/2})$, since for the left half of the parabola we have $x=-y^{1/2}$. It follows that the volume is
$$\int_0^1 2\pi (1-y)(y^{1/2}-y)\,dy.$$
If I were using cylindrical shells, I would note that by symmetry the volume is the same as when the region between $y=x^2$ and $y=x$ is rotated about the line $y=1$. Now I would be in my region of comfort, the first quadrant, and writing down the integral for the volume involves less chance of error. Negative numbers are trouble.
We can also do it by slicing perpendicular to the $x$-axis. Again, let's work with the region between $y=x^2$ and $y=x$. The cross-section at $x$ is a circle with outer radius $1-x^2$ with a circular hole of radius $1-x$ cut out. The volume is 
$$\int_0^1 \pi\left((1-x^2)^2-(1-x)^2   \right)\,dx.$$
