Volume between $x=0,y=0,z=1,z=\sqrt{x},z=\sqrt{y-x^2}$ I am trying to calculate the volume trapped by the surfaces $x=0,y=0,z=1,z=\sqrt{x},z=\sqrt{y-x^2}$. But when I draw the graph at a graphing calculator, I can't find an area that's bounded by theses surfaces! (I've been using this utility: http://calculator.runiter.com/graphing-calculator/online-graphing-calculator.htm)
Can anyone help unconfuse me?
Thanks!
Edit: Thanks for all the great(!!) comments! I went to sleep after posting this so I couldn't respond immediately.
 A: The first four surfaces, $x=0$, $y=0$, $z=1$, and $z=\sqrt{x}$ bound a region that looks like a routed edge extending from 0 to infinity in the y-direction. You might use such an object in a woodworking project.

(Imagine turning the piece of wood upside down and ignore all but the curving part)
The question is, does the 5'th constraint ($z=\sqrt{y-x^2}$) limit how far the piece can go in the positive y-direction? The answer is yes - if you rearrange the equation a little you get $y=z^2+x^2$, a paraboloid touching the origin and opening out in the y-direction.
I can't say for sure, but my guess is that the graphing software you're using is trying to find the value of z when $y-x^2$ is negative, and failing because this is imaginary (ie, there is no point on the paraboloid with those x and y coordinates - visualize a vertical line that doesn't pass through a paraboloid that opens horizontally).
A: First find the region $R_1$ consisting of all points $\langle x,y,z\rangle$ such that $x\ge 0$, $y\ge 0$, and $\sqrt x\le z\le 1$. This has the same cross-section perpendicular to the $y$-axis for each $y\ge 0$; that cross-section projects to the curvilinear triangle in the $xz$-plane bounded by $z=\sqrt x$, $x=0$, and $z=1$, so it lies above the curve $z=\sqrt x$.
Now revolve the parabola $y=x^2$ about the $y$-axis to produce a paraboloid of revolution; its cross-sections perpendicular to the $y$-axis have the form $x^2+z^2=y$, so $z=\sqrt{y-x^2}$ describes the part of this paraboloid lying on and above the $xy$-plane. Let $R_2$ be the region consisting of all points $\langle x,y,z\rangle$ such that $x\ge 0$, $z\ge 0$, and $0\le y\le x^2+y^2$; $R_2$ is the set of points in the first octant between the paraboloid and the $xz$-plane.
Finally, let $R=R_1\cap R_2$; then $R$ is a finite volume bounded by $x=0$, $y=0$, $z=1$, and $z=\sqrt{y-x^2}$ and actually having each of these four surfaces as part of its boundary.
