Calculate volume enclosed by cylinder and paraboloid (integration). I need to calculate the volume enclosed by:
$$x^2 + y^2 = z, \space y = x^2, \space z=0, \space y = 1$$
The shape of the volume I get gets me confused. It is a paraboloid ($x^2 + y^2 = z$) intersected with cylinder ($y = x^2$) and limited by specific $z$ and $y$ plains. When I tried drawing this I saw that the volume is not limited by the "upper" $z$ plain, therefore it seems to be infinite. Did the lecturer provide us with "wrong" conditions, so the volume is infinite?
Am I right? If yes, how can I calculate the volume if I change my previous condition ($z = 0, \space y = 1$) to $0\le z \le 1$? I tried approaching this "updated" problem, but also didn't have any luck.
Any help would be appreciated.
EDIT: The answer including the integral solution was posted - see below. The whole problem was caused by me thinking about the volume "inside" the paraboloid, while the task was to calculate it "outside", enclosed by the cylinder.
 A: Look first at the $xy$-plane (the bottom). The condition limits the area $D$ between $y=x^2$ and $y=1$. It is bounded in $(x,y)$. Now look at what happens along the vertical $z$ axis. It says: take those points $(x,y,z)$ that are between $z=0$ and $z=x^2+y^2$. The set (and the volume) is finite, it is between two surfaces ($xy$ plane and the paraboloid).

Try to split integration as
$$
\iint_D\int_{z=0}^{z=x^2+y^2}\,dz\,dxdy.
$$
A: First of all the volume is not infinite because it is bounded at $z=2$.
Draw the figure on a $xy$-plane at $z=t$ where $0\leq t\leq2 $ you will see that the figure gets closed.
$\int_{0}^{2}S(t)dt$ where $S(t)$ is the area enclosed by $y=1,y=x^2$ and $x^2+y^2=t$.
Keep in mind that you will have to consider two cases separately when $0\leq t\leq1 $ and $1\leq t\leq2 $.
A: Thanks to help from A.Γ. and Chris2006:
It turns out that the volume is enclosed and can be calculated. The answer was accepted. Using the hints given there:
$$\int_{-1}^{1} dx \int_{x^2}^{1} dy \int_{0}^{x^2 + y^2} dz = \int_{-1}^{1} dx \int_{x^2}^{1} (x^2 + y^2) dy = \int_{-1}^{1} (x^2 y + \frac{y^3}{3})_{y =(x^2, 1)} dx= \int_{-1}^{1} (x^2 + \frac{1}{3} - x^4 - \frac{x^6}{3})dx = (\frac{x^3}{3} + \frac{1}{3} x - \frac{x^5}{5} - \frac{x^7}{21})_{x =(-1, 1)} = \frac{88}{105}$$
