Change of variable in integral using metric

What is the formula to evaluate a multi-integral by a change of coordinates using the squareroot of the metric instead of the determinate of the Jacobian? Thanks.

 

So check my work, let's see if I understand this correctly,

Suppose we want to evaluate a triple integral by going from coordinates (x,y,z) to coordinates (u,v,w), then we start from

[tex]\int_X {f(\vec x)d\vec x\, = \,\int_{{z_1}}^{{z_2}} {\int_{{y_1}}^{{y_2}} {\int_{{x_1}}^{{x_2}} {f(x,y,z)dxdydz} } } } [/tex]

and this equals

[tex] \int_{Q(X)} {f(\vec x(\vec q))\sqrt {g(\vec q)} d\vec q\, = \,} \int_{{w_1}}^{{w_2}} {\int_{{v_1}}^{{v_2}} {\int_{{u_1}}^{{u_2}} {f(x(u,v,w),y(u,v,w),z(u,v,w))\sqrt {g(u,v,w)} dudvdw} } } [/tex]

with

[tex]\int_{Q(X)} {f(\vec x(\vec q))\sqrt {g(\vec q)} d\vec q = \,} \int_{{w_1}}^{{w_2}} {\int_{{v_1}}^{{v_2}} {\int_{{u_1}}^{{u_2}} {f(u,v,w)\left| {\frac{{\partial (x,y,z)}}{{\partial (u,v,w)}}} \right|dudvdw} } } [/tex]

and with

[tex]\sqrt {g(u,v,w)} \, = \,\left| {\frac{{\partial (x,y,z)}}{{\partial (u,v,w)}}} \right|\, = \,\left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}&{\frac{{\partial x}}{{\partial w}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}&{\frac{{\partial y}}{{\partial w}}}\\
{\frac{{\partial z}}{{\partial u}}}&{\frac{{\partial z}}{{\partial v}}}&{\frac{{\partial z}}{{\partial w}}}
\end{array}} \right|[/tex]

This last term is the determinant of the Jacobian, right?

 

And then if we make another change of coordinates from (u,v,w) to coordinates (r,s,t) we would get,

[tex]\int_{{w_1}}^{{w_2}} {\int_{{v_1}}^{{v_2}} {\int_{{u_1}}^{{u_2}} {f(u,v,w)\left| {\frac{{\partial (x,y,z)}}{{\partial (u,v,w)}}} \right|dudvdw} } } = \int_{t({u_1},{v_1},{w_2})}^{t({u_2},{v_2},{w_2})} {\int_{s({u_1},{v_1},w)}^{s({u_2},{v_2},w)} {\int_{r({u_1},v,w)}^{r({u_2},v,w)} {f(u(r,s,t),v(r,s,t),w(r,s,t))\left| {\frac{{\partial (u,v,w)}}{{\partial (r,s,t)}}} \right|drdsdt} } } [/tex]
Is this right? Please check the limits of the last integral using the limits in the first integral.

Is there then a relationship between the determinant in the first integral and the determinant of the last integral? Would there be a tensor transformation between the two determinants?

 

Ahh yes, the determinant from xyz to uvw is part of the integrand that is being converted, right? If that determinant were not included in the left hand side, it would not be needed on the right hand side.

 

And can we just as easily use,
[tex]\int_X {f(\vec x)d\vec x\,} = \int_{Q(X)} {f(\vec x(\vec q))\sqrt {g(\vec q)} d\vec q\,\,} [/tex]
where the metric, g(q), is defined in post 3, if we transfrom from flat space to curved space?

I'm thinking of a transformation from a disk to a hemisphere directly above it. We should be able to map every point on the disk to points of the hemisphere (except for the equator). We can draw lines in flat space on the disk and map them to lines on the hemisphere. However, the metric on the disk will be different from the metric on the hemisphere, and so the length of the line on each will be different. At this point, I'm not sure that the integral on the disk will be equal to the integral on the hemisphere. Any insight into all this would be appreciated. Thanks.