Find the volume of the region defined by (x^2 + y^2 + z^2)^2 <= a^3 z, where a > 0.

Step-by-step solution

1. 1

   Translate the inequality into spherical coordinates. With $x^2+y^2+z^2=\rho^2$ and $z=\rho\cos\varphi$, the defining condition becomes

   $\rho^4\le a^3\rho\cos\varphi\;\Longleftrightarrow\;\rho^3\le a^3\cos\varphi\;\Longleftrightarrow\;\rho\le a\left(\cos\varphi\right)^{1/3}$

   One factor of $\rho$ cancels because $\rho>0$ away from the origin. The right-hand side requires $\cos\varphi\ge 0$, so the solid lives entirely in the upper half-space $0\le\varphi\le\tfrac{\pi}{2}$ — expected, since $a^3z\ge 0$ forces $z\ge 0$.

2. 2

   Set up the volume integral. The solid is a surface of revolution about the $z$-axis (no $\theta$ appears), so

   $V=\int_0^{2\pi}\!\!\int_0^{\pi/2}\!\!\int_0^{a(\cos\varphi)^{1/3}}\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$

   The factor $\rho^2\sin\varphi$ is the spherical volume element; omitting it is the standard way this integral goes wrong.

3. 3

   Integrate radially — the cube root disappears. For fixed $\varphi$,

   $\int_0^{a(\cos\varphi)^{1/3}}\rho^2\,d\rho=\frac{\left[a\left(\cos\varphi\right)^{1/3}\right]^3}{3}=\frac{a^3\cos\varphi}{3}$

   This is the point of the whole problem: cubing the radius cancels the exponent $\tfrac13$ exactly, turning an awkward boundary into a plain cosine.

4. 4

   Do the polar-angle integral. What remains is a textbook substitution:

   $\int_0^{\pi/2}\frac{a^3\cos\varphi}{3}\sin\varphi\,d\varphi=\frac{a^3}{3}\left[\frac{\sin^2\varphi}{2}\right]_0^{\pi/2}=\frac{a^3}{3}\cdot\frac12=\frac{a^3}{6}$

5. 5

   Multiply by the angular sweep and check the scaling. The $\theta$-integral contributes $2\pi$:

   $V=2\pi\cdot\frac{a^3}{6}=\frac{\pi a^3}{3}$

   Two checks. Dimensionally the answer must scale like $a^3$, which it does — replacing $a$ by $ka$ scales the solid linearly by $k$. Numerically, taking $a=1$ and estimating the same volume by Monte Carlo sampling of $(\rho^2)^2\le z$ over the box $[-1.2,1.2]^2\times[0,1.2]$ gives about $1.045$, against $\tfrac{\pi}{3}\approx 1.0472$ $\checkmark$.