To start we have two entities:
the vector field $\vec{P}(\vec{r})$ and the positional vector $\vec{r}$ (in cartesian coordinates for example, $\vec{r} = x\hat{x} + y\hat{y} + z\hat{z}$).
My question: Have I simplified the following integral correctly?
$$\int d^3r \space \nabla \times \left(\vec{r} \times \vec{P} \right)$$
My work:
$$\nabla \times \left(\vec{r} \times \vec{P} \right) = \vec{r}\left(\nabla \cdot \vec{P}\right) - \vec{P}\left( \nabla \cdot \vec{r} \right) + (\vec{P} \cdot \nabla)\space \vec{r} \space - (\vec{r} \cdot \nabla) \vec{P}$$
Using $\nabla \cdot \vec{r} = 3$, $(\vec{P} \cdot \nabla)\vec{r} = \vec{P}$, and $(\vec{r} \cdot \nabla) \vec{P} = r \frac{d\vec{P}}{dr} = \frac{d (r\vec{P})}{dr} - \vec{P} \:$ reduces the above equation to:
$$\nabla \times \left(\vec{r} \times \vec{P} \right) = \vec{r}\left(\nabla \cdot \vec{P}\right) - 3\vec{P} + \vec{P} - r \frac{d\vec{P}}{dr} = \vec{r}\left(\nabla \cdot \vec{P}\right) - 2\vec{P} - \frac{d (r\vec{P})}{dr} + \vec{P}$$
$$\nabla \times \left(\vec{r} \times \vec{P} \right) = \vec{r}\left(\nabla \cdot \vec{P}\right) - \vec{P} - \frac{d (r\vec{P})}{dr}$$
$$ \int d^3r \:\: \nabla \times \left(\vec{r} \times \vec{P} \right) = \int d^3r \:\: \vec{r}\left(\nabla \cdot \vec{P}\right) - \int d^3r \:\:\vec{P} - \int d^3r \:\: \frac{d(r\vec{P})}{dr}$$
Working on the left hand side first:
$$ \int d^3r \:\: \nabla \times \left(\vec{r} \times \vec{P} \right) = \int d\vec{S} \:\: \times \left(\vec{r} \times \vec{P} \right) = \int (d\vec{S} \cdot \vec{P}) \vec{r} \:\:- \int (d\vec{S} \cdot \vec{r}) \vec{P}$$
Now working on the right hand side:
Decomposing $\int d^3r \:\: \frac{d(r\vec{P})}{dr}$ in spherical coordinates gives:
$$\int d^3r \:\: \frac{d(r\vec{P})}{dr} = \int\int\int r^2 \sin(\theta)drd\theta d\phi \:\: \frac{d(r\vec{P})}{dr} = \int\int\sin(\theta)d\theta d\phi\int r^2 dr \:\: \frac{d(r\vec{P})}{dr}$$
Using integration by parts on the "r" component of the triple integral yields:
$$ \int r^2 dr \:\: \frac{d(r\vec{P})}{dr} = r^2 r\vec{P} - \int dr \:\: 2r^2 \vec{P}$$
replacing the "r" part of the triple integral with the above gives:
$$\int d^3r \:\:r \frac{d\vec{P}}{dr} = \int\int\int r^2 \sin(\theta)drd\theta d\phi \:\:r \frac{d\vec{P}}{dr} = \int\int\sin(\theta)d\theta d\phi \left(r^2 r\vec{P} - \int dr \:\: 2r^2 \vec{P}\right) = \int dS\left(r\vec{P}\right) - 2\int d^3r \vec{P}$$
So putting everything together gives:
$$ \int (d\vec{S} \cdot \vec{P}) \vec{r} \:\:- \int (d\vec{S} \cdot \vec{r}) \vec{P} = \int d^3r \:\: \vec{r}\left(\nabla \cdot \vec{P}\right) - \int d^3r \:\:\vec{P} - \int dS\left(r\vec{P}\right) + 2\int d^3r \vec{P}$$
$$ \int (d\vec{S} \cdot \vec{P}) \vec{r} \:\:- \int (d\vec{S} \cdot \vec{r}) \vec{P} = \int d^3r \:\: \vec{r}\left(\nabla \cdot \vec{P}\right) + \int d^3r \:\:\vec{P} - \int dS\left(r\vec{P}\right)$$
Now, at this point I wonder whether it is legitimate to say that:
$$ \int (d\vec{S} \cdot \vec{r}) \vec{P} = \int dS\left(r\vec{P}\right) $$
because the left hand side is a formally a dot product between $d\vec{S}$ and $\vec{r}$ but the right hand side has no dot product.
However, if it is true that $\int (d\vec{S} \cdot \vec{r}) \vec{P} = \int dS\left(r\vec{P}\right)$, then I arrive at:
$$ \int (d\vec{S} \cdot \vec{P}) \vec{r} = \int d^3r \:\: \vec{r}\left(\nabla \cdot \vec{P}\right) + \int d^3r \:\:\vec{P} $$