= Solution
For the printed integration formula, $J$ must mean the <Jacobian determinant> of the inverse coordinate map:
$$
\boxed{J=\frac{\partial(x,y,z)}{\partial(u,v,w)}=\det\begin{pmatrix}x_u&x_v&x_w\\y_u&y_v&y_w\\z_u&z_v&z_w\end{pmatrix}}.
$$
If instead one calls $\partial(u,v,w)/\partial(x,y,z)$ the forward Jacobian, its <determinant> is $1/J$ where the derivative is invertible. This convention distinction resolves the direction of the arrow in the printed wording.
To derive the <change of variables formula>, a small rectangular cell of side lengths $du,dv,dw$ maps under the inverse transformation, to first order, to a parallelepiped with edge vectors $\mathbf r_u\,du$, $\mathbf r_v\,dv$, $\mathbf r_w\,dw$. Its volume is the absolute <scalar triple product> of those vectors:
$$
dx\,dy\,dz=|\mathbf r_u\cdot(\mathbf r_v\times\mathbf r_w)|\,du\,dv\,dw=|J|\,du\,dv\,dw.
$$
For a one-to-one continuously differentiable coordinate map with continuously differentiable inverse, summing these local volume contributions and taking the partition limit gives
$$
\boxed{\int_D f(x,y,z)\,dx\,dy\,dz=\int_\Delta f(x(u,v,w),y(u,v,w),z(u,v,w))\,|J|\,du\,dv\,dw}.
$$
The absolute value removes an orientation reversal; injectivity ensures that each volume element is counted once.
For positive semi-axes, take $u=x/a$, $v=y/b$, $w=z/c$. The <solid ellipsoid> becomes the unit ball and the inverse <Jacobian determinant> is $abc$. Thus
$$
\int_Dx^2\,dV=a^3bc\int_{u^2+v^2+w^2\le1}u^2\,du\,dv\,dw.
$$
Rotational symmetry gives equal integrals of $u^2,v^2,w^2$ over the ball. Their sum is the integral of $r^2$, so <spherical coordinates> give $\int u^2\,dV=\frac13\int_0^1 4\pi r^4\,dr=4\pi/15$. The <Cartesian second moment of a solid ellipsoid> is therefore
$$
\boxed{\int_Dx^2\,dx\,dy\,dz=\frac{4\pi}{15}a^3bc}.
$$
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