Consider the following integral:
$\int_{v} \nabla \cdot V \ dxdydz $
where $v$ is a sphere of radius R so that
$x^2+y^2+z^2 = R^2$
and the vector $V$ is given by
$V = x^5\hat{x}+y^5\hat{y}+z^5\hat{z}$
I attempted to introduce a vector $V_2$ so that:
$V_2 = x^3\hat{x}+y^3\hat{y}+z^3\hat{z}$
and
$\nabla \cdot V_2 = 3x^2+3y^2+3z^2=3R^2$
Using vector identities I would arrive at:
$\frac{5}{9}(\nabla \cdot V_2)^2 = \nabla \cdot V$
This produces $\nabla \cdot V=5R^4$
I then take $R$ as the argument, change the system to an spherical system and do the integral. Obtaining:
$\int_{v} \nabla \cdot V \ dv = \frac{20}{7}\pi R^7$
Yet this is different from the answer one gets from the numerical evaluation of the integral which is just: $\frac{12}{7}\pi R^7$
So what am I doing wrong?
Edit. The Vector Identity result was wrong. So the new questions would be> Is there any vector identity that would help us solve this integral?
It is straightforward to compute the integral as belows,
$$I=\int_{v} \nabla \cdot V \ dxdydz =5\int_{v}(x^4+y^4+z^4)dxdydz$$
The integral is symmetric with respect to $x$, $y$ and $z$, that is,
$$\int_{v}x^4 dxdydz =\int_{v} y^4dxdydz = \int_{v}z^4dxdydz$$
For example, the first equality is obtained by switching $x$ and $y$ in the expression $\int_{v}x^4 dxdydz $.
Thus, simplify the integral and then use the spherical coordinates,
$$I =15\int_{v}z^4dxdydz=15\int_0^R\int_0^{\pi} \int_0^{2\pi}r^6\cos^4\theta\sin\theta drd\theta d\phi$$
$$I =15\int_0^R r^6dr\int_0^{\pi}\cos^4\theta\sin\theta d\theta \int_0^{2\pi} d\phi=\frac{12\pi}{7}R^7$$