Volume of the solid between $2z+x^2-y^2=0$ and the region bounded by $y=x$ and $y=x^2$ I want to compute the volume of the solid between the paraboloid
$$2z+x^2-y^2=0$$
and the region in the $xy$-plane bounded by $y=x$ and $y=x^2$.
By the equation of the paraboloid, $z={1\over 2}(y^2-x^2)$ and the region over which I have to integrate is $x\in[0,1], y\in[x^2,x]$. Thus,
$$V=\int_0^1\int_{x^2}^x\;{1\over 2}(y^2-x^2)\;dy dx.$$
The computation of this double integral (triple-checked with software) gives as result
$$\int_0^1\int_{x^2}^x\;{1\over 2}(y^2-x^2)\;dy dx={1\over 2}\int_0^1 {-2\over 3}x^3-{1\over 3}x^6+x^4\;dx={1\over 2}\left(-{1\over 6}-{1\over 21}+{1\over 5}\right)={-1\over 140}$$
which is not the value that one would expect for a volume...
Can anyone spot what I'm missing here? Can this volume be negative? What would be the reason?
Thanks a lot.
 A: As comments already stated, you have $x\in[0,1]$ and $y\in[x^2,x]$ so $y\le x$ and therefore $z=\tfrac12(y^2-x^2)<0$. You are integrating over a bunch of negative values.
The dobule integral over a function which is always positive gives the volume its surface makes with the $z=0$ plane. For a function which is always negative, you get the negated volume. And for a function which can change its signs, you get the sum of the volumes above the plane minus the sum of the volumes below. Which might indeed be zero if they happen to cancel out.
Or in other words, if you really want the volume, you have to take the absolute value of the function you're iterating over, which usually means you'll have to do case distinctions and integrate over each component by itself. Luckily it's all negative (or rather non-positive) in your case, so you can just flip the sign to get the absolute value.
Responding to your comment: negative volumes, just like negative areas in the plane, go against convention. There are many situations where dealing with these beasts makes a lot of sense. But then it's usually best to explicitely refer to them as signed volumes, and explain the meaning of the sign unless obvious from context. Otherwise people used to the more common convention of unsigned values will just get confused.
