How to find the volume of the solid

Question

I need some help on this question, I even know nothing about the shape. How can I decide the range of x,y,z? Thanks a lot!

Answer 1

The equation suggests that a cylindrical coordinate system might be useful: let $x = r \cos \theta$, $y = r \sin \theta$, to get $$(r^2 + z^2)^2 = 2z r^2, \quad r \ge 0.$$ Immediately, we see that this surface is independent of the angle $\theta$--it is symmetric through any rotation about the $z$-axis.

But can we do more with this? Indeed, suppose we now write $r = \rho \sin \phi$, $z = \rho \cos \phi$; that is, we consider a spherical coordinate system: then $$\rho = 2 \cos \phi \sin^2 \phi.$$ Note that we must have $\rho \ge 0$, $0 \le \theta \le 2\pi$, and $0 \le \phi \le \pi/2$.

Now can you find the volume as a triple integral in spherical coordinates?

Answer 2

How about convert to sphericals:

$$r^4 = 2 r^3 \cos{\theta} \sin^2{\theta} \implies r = 2 \cos{\theta} \sin^2{\theta}$$

Note that because $z$ should always be positive, $\theta \in [0,\pi/2]$. The volume is then

$$\begin{align}V &=2 \pi \int_0^{\pi/2} d\theta \, \sin{\theta} \, \int_0^{2 \cos{\theta} \sin^2{\theta}} dr \, r^2 \\ &= \frac{16 \pi}{3} \int_0^{\pi/2} d\theta \,\cos^3{\theta} \, \sin^7{\theta}\\ &= \frac{16 \pi}{3} \int_0^{1} du\, (u^7-u^9)\\ &= \frac{16 \pi}{3} \left (\frac18-\frac1{10} \right )\\ &= \frac{2 \pi}{5}\end{align}$$