Find the volume of $K=\{(x,y,z):|x-z^2|+|y-z^2|+z^2\le1\}$

Question

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

Answer 1

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,

$V=\displaystyle\int_{-1}^1 A(z)\,dz$

$V=2\displaystyle\int_{-1}^1(1-z^2)^2\,dz$

$V=\dfrac{32}{15}$

Answer 2

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, $$ I = 4\cdot 2 \int_0^1 \int_0^{\frac \pi 2}\int_0^\pi r^4 \sin 2\phi \sin^3 \theta\, d\theta\, d\phi\, dr = \frac {32}{15}. $$