Find the volume of $K=\{(x,y,z):|x-z^2|+|y-z^2|+z^2\le1\}$ How can I find the volume of $K=\{(x,y,z):|x-z^2|+|y-z^2|+z^2\le1\}$?
First, $z^2\le1$ since $|x-z^2|+|y-z^2|\ge0$
This would give me $$\int \limits_{-1}^1 \left(\iint\limits_K \,dx\,dy\right)\,dz$$
where $K=\{(x,y):|x-z^2|+|y-z^2|\le1-z^2\}$
Now I don't recognize this form of $K$, though $i$ does look look like a square if you let the absolute values represent intervals with respect to the $x$- and $y$ axis. How can I calculate the double integral?
Thanks! Alexander
 A: Switch to "spherical" coordinates
\begin{align}
\left |x-z^2 \right | &= r^2\sin^2\theta \sin^2 \phi \\
\left |y-z^2 \right | &= r\sin^2 \theta \cos^2 \phi \\
z &= r \cos \theta
\end{align}
or after some manipulation, you can get completely parametrized version of the region
\begin{align}
x &= r^2 \left(\pm \sin^2\theta\sin^2\phi + \cos^2\theta \right )\\
y &= r^2 \left(\pm \sin^2\theta\cos^2\phi + \cos^2\theta \right )\\
z &= r\cos \theta
\end{align}
I leave showing that $\phi \in [0, \frac \pi 2]$, $\theta \in [0, \pi]$  and $r \in [0, 1]$ to you.
Now, you can easily plot that region

I'm not sure if you can see it, but it has sail-shape.
After that, the whole integration should be reduced to
$$
I = \left | \int_0^1 \int_0^{\frac \pi 2} \int_0^\pi J (\phi, \theta, r)\, d\theta\, d\phi\, dr \right |
$$
where $J(\phi, \theta, r)$ is the Jacobian of the transformation.
$$
J = \left |\begin{array}{ccc}
x_\phi & x_\theta & x_r \\
y_\phi & y_\theta & y_r \\
z_\phi & z_\theta & z_r \\
\end{array}\right |
$$
Again, I'll leave its calculation to you as an exercise. Answer is below
$$
J(\phi, \theta, r) = \pm 2r^4 \sin 2\phi \sin^3 \theta
$$
You have four regions with different signs of Jacobian (recall expressions for $x$ and $y$), and in each region Jacobian has constant sign. Due to the symmetry, volume of each region is the same, so you can calculate only one of them. Finally,
$$
I = 4\cdot 2 \int_0^1 \int_0^{\frac \pi 2}\int_0^\pi r^4 \sin 2\phi \sin^3 \theta\, d\theta\, d\phi\, dr = \frac {32}{15}.
$$
A: Let $k=z^2$, then $0\leq k\leq1$.
Now, rewriting the set K, 
$$K=\{(x,y);|x-k|+|y-k|\leq 1-k\}$$
We have to consider four  cases:
1.$x,y\geq k$, then the inequality is just,  $x+y\leq1+k$
2.$x,y\leq$, then the inequality is just, $x+y\geq3k-1$
3.$x\geq k,y\leq k$, then the inequality is just, $x-y\leq1-k$
4.$x\leq k,y\geq k$, then the inequality is just, $y-x\leq1-k$
When $k$ attains its minimum value ($k=0$), then $K$ is a square of side 2, other hand, when $k$ attains its maximum value ($k=1$), $K$ is a degenerate square.
The point is, how estimate the square's area variation? 
For this, note that the side of a square is just the distance between the lines $x+y=1+k$ and $x+y=3k-1$
This way if the square has side iqual $\ell$, then
$\ell=\sqrt{2}(1-k)$
Hence
$A(k)=2(1-k)^2$ where $0\leq k\leq1$
Or
$\displaystyle\iint\limits_K \,dx\,dy=2(1-z^2)^2$
Now,
$V=\displaystyle\int_{-1}^1 A(z)\,dz$
$V=2\displaystyle\int_{-1}^1(1-z^2)^2\,dz$
$V=\dfrac{32}{15}$
