Triple integral - relation to volume

Question

I had to calculate the triple integral $J$ of the function $$F(x,y,z) = z$$ over the body $Q \subseteq \mathbb{R}^3$ defined via the inequality $x^2 + y^2 + z^2 - 2Rz \le 0$,
where $R > 0$ is a constant.

$$J = \iiint_{Q} z \,dx \,dy \,dz$$

This body is the closed ball of radius $R$
centered at the point $M = (0,0,R)$, right?

This is an exercise in using spherical coordinates.

So I figured out that in spherical coordinates this body can be defines as follows:

$\theta \in [0, \pi/2]$
$\phi \in [0, 2\pi]$
$\rho \le 2R\cos\theta$

Here $\theta$ is the inclination (or polar angle) and $\phi$ is the azimuthal angle.

OK... So I calculated $J\ $ by using spherical coordinates and I got this answer.

$$\frac{4\pi R^4}{3}$$

Is my answer correct?
Then I noticed that this is actually the volume of the body multiplied by $R$.

My question is what is the intuition behind this?
There must be a reason why this integral is exactly the volume of the ball multiplied by $R$.

Answer 1

Maybe the symmetry would help our understanding. The ball is symmetric in the plane $z=R$. We separate the ball into upper half $Q_+$ and lower $Q_-$. Also we can pair $(x,y,z)\in Q_-$ with $(x,y,2R-z)\in Q_+$ and vice versa. Then, the function $z$ sums up to a constant $2R$. Therefore, $$J=\int_{Q_+}z+\int_{Q_-}z=\int_{Q_-}2R,$$ that is, $2R$ times the volume of a half-ball. Note that the computation above is justified with the change of variables formula.

Answer 2

Let's introduce a transformation $z^\prime = z - R$ so that the sphere has origin as its centre in the $(x, y, z^\prime)$ coordinates. We then have \begin{equation} J = \iiint z dx dy dz = \iiint (z^\prime + R) dx dy dz^\prime. \end{equation} One can use the usual spherical polar coordinates in the $(x, y z^\prime)$ system to conclude that \begin{equation} \iiint z^\prime dx dy dz^\prime = 0 \end{equation} and \begin{equation} \iiint R dx dy dz^\prime = R \iiint dx dy dz^\prime = R \times \frac{4\pi R^3}{3}. \end{equation} Therefore, we have \begin{equation} J = \frac{4\pi R^4}{3}. \end{equation} Perhaps, this might give you an insight into the origin of the factor $R$. You may also want to note that the dimension of the integrand $J$ is the fourth power of length, which is reflected in the expression of $J$.

Answer 3

A physics point of view helps here. The center of mass of an object or system with density $\rho(\vec r)$ has its center of mass at

$$ \bar r = \frac{\iiint \vec r \rho(\vec r)\, dV }{\iiint \rho(\vec r)\, dV } $$

If we use this sphere and say it has density $1$, the center of mass is

$$ \bar r = \frac{\iiint_Q(x \hat \imath + y \hat \jmath + z \hat k)\, dx\, dy\, dz}{\iiint_Q dx\, dy\, dz} $$

The denominator is simply the volume of the sphere, $\frac{4}{3} \pi R^3$. Since $Q$ is symmetric in $x$ and $y$, the numerator's $\hat \imath$ and $\hat \jmath$ components come out to zero. The fact the numerator equals $\frac{4}{3} \pi R^4 \hat k$ tells us the sphere's center of mass is $(0,0,R)$. Of course, that's the center of the sphere.

For a strictly mathematical argument inspired by all this, we can note that

$$ \iiint_Q (z-R)\, dx\, dy\, dz = 0 $$

by symmetry, so

$$ \iiint_Q z\, dx\, dy\, dz = R \iiint_Q dx\, dy\, dz $$