We have four surfaces : $z=0$, $y=0$, $z=1-x$, $y=1-x^2$. That is, the $xy$-plane, the $zx$-plane, a straight line contained in the $zx$-plane, and a parabola contained in the $xy$-plane. One might then determine the volume of this object with a triple integral. But isn't the volume of this object just zero? Meaning these planes do not contain within them any points.
I actually believe this set-up describes a half-cone, where the bottom half circle lies in the $xy$-plane, and its side is a triangle, which lies in the $zx$-plane. Is this the correct conclusion, and if so, how can I justify this from the given equations?
Best Answer
You need to realize that, for instance, $y=1-x^2$ does not represent a line in the $xy$-plane. Since $z$ does not appear in the equation, it can take any value, and the equation actually represents a surface that corresponds to vertically moving the line that you mention. The same goes to $z=1-x$, which represents a plane.
If you want to picture it, draw the set $D = \{(x,y,0): -1 \leq x \leq 1, y\leq 1-x^2\}$ in the $xy$-plane and extend if perpencidularly to the $z$-axis making an infinite cylinder with a funny cross-section ($D$). Now just consider the portion of that cylinder that is above the plane $z=0$ (bottom) and the below the plane $z=1-x$ (top).