Volume Bound by $z+x^2=1; z+y^2=1, x=0;y=0;z=0$

Question

I am trying to produce a triple integral that finds the volume underneath the surraces $x^2+z=1$, $y^2+z=1$, and is bound by the coordinate planes. On the graph, this looks like two parabaloids orthogonal to each other intersecting along the line $x=y$. The bounds I choose are $0<z<1-y^2$, $0<y<x$, $0<x<1$. I then solve this integral and multiply it by two (because of symmetry) and get $5/6$ but apparently the answer is $1/2$ according to the book I'm using. I assume I have my bounds incorrect, could someone explain to me why this is?

Answer 1

You made a mistake in your bounds. The conditions say $$0\leq z\leq\min\{1-x^2,1-y^2\}=1-\max\{x^2,y^2\}\ .$$ In the part $0\leq y\leq x$ of the first quadrant of the $(x,y)$-plane this means $$0\leq z\leq 1-x^2\ .$$ The volume of the body $$B:=\bigl\{(x,y,z)\bigm|x\geq0, \ y\geq0,\ 0\leq z\leq\min\{1-x^2,1-y^2\}\bigr\}$$ therefore is given by $${\rm vol}(B)=2\int_0^1\int_0^x\int_0^{1-x^2}1\>dz\>dy\>dx=2\int_0^1 x(1-x^2)\>dx={1\over2}\ .$$