Compute the volume of the body restricted with $z^2=x^2+y^2; z=0; (x^2+y^2)^2=2xy$ surfaces
I used cylindric coordinates and got:
$z=z; z=0; r=\sqrt{\sin 2t} $
So my problem is that I can’t find out bounds. Could you please explain in detail how can I do that without the picture
First of all, I would say, why try to do it without the picture? Pictures are really important to understand what's going on. In this case, the plane curve $(x^2+y^2)^2 = 2xy$, or $r^2 = \sin 2\theta$, is a lemniscate.
If $D$ is the region enclosed by that lemniscate, the volume you seek is
$$
V = \iint_D \sqrt{x^2+y^2}\,dA
$$
To simplify, first let's just use the symmetry to integrate over the first quadrant and double it. Then we don't have to worry about $\sin 2t$ going negative. We can say $r = \sqrt{\sin 2\theta}$ on this region. So
\begin{align*}
V &= 2\int_{0}^{\pi/2} \int_0^{\sqrt{2\sin 2\theta}}(r)r\,dr\,d\theta
\\&= \frac{2}{3}\int_{0}^{\pi/2}\left.r^3\right|_{r=0}^{r=\sqrt{2\sin 2\theta}} \,d\theta
\\&= \frac{2}{3}\int_{0}^{\pi/2} \sin^{3/2}2\theta\,d\theta
\end{align*}
The integrand cannot be antidifferentiated in elementary terms, however. Wolfram says the answer is about 0.582679.
If the surface under which this solid lies were different, we might have an elementary integral. For instance, if it were $z=x^2+y^2$ instead of $z^2=x^2+y^2$, the sine exponent would be $2$ instead of a fraction, and we could integrate that with a trig identity.