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Given a two-dimensional area $A\subset\mathbb R^2$ lying in the $xy$-plane, i.e. $A:\mathbb R^2\supset U\to A\subset \mathbb R^2, (u,v)\mapsto (x(u,v),y(u,v))$, a straight extrusion along the $z$-axis defines a volume $V = A\times[z_\min,z_\max]$ and integration operator over this volume is simply $$\int_V dV = \int\limits_{z_\min}^{z_\max}dz\int_Adx\,dy$$ while the surface integral operator is $$\int_{\partial V}d\vec n = \vec e_z\left.\int_A dx\,dy\right|_{z=z_\min}^{z_\max} + \int\limits_{z_\min}^{z_\max}dz\oint_{\partial A}d\vec n_A \quad(d\vec n_A = -\vec e_z\times d\vec r) .$$ (Using the Kelvin-Stokes theorem $\oint_{\partial A} d\vec r = \iint_A (d\vec\sigma\times\vec\nabla)$ this can be used to prove the divergence theorem for extruded volumes by the way)

But let's now generalize this process to extrusion along a curved path $\vec\gamma:[0,T]\to \Gamma, t\mapsto \vec\gamma(t)$ defining $A(t):(u,v,t)\mapsto \vec\gamma(t) + \frac{d}{dt}\vec\gamma(t)\times(y(u,v),-x(u,v),0)$ (assuming $|\frac d{dt}\vec\gamma(t)|\equiv 1$) as a translation of $A$ both along and perpendicular to $\vec\gamma(t)$ to obtain the Volume $$V_\gamma = \cup_{t=0}^{T} A(t).$$

As a simple example, imagine a circle of radius $r$ extruded along a bigger (and perpendicular) circle of radius $R>2r$ which would yield a torus, while the extrusion along the $z$-axis would simply yield a cylinder.

What are the volume and surface integral operators now? I imagine it's something like $$\int_V dV = \int_0^T dt D(t)\int_{A(t)}du\,dv$$

And even more generally, what are the expressions if the area is tilted or rotated (torqued) along the way?

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What do you mean by "defining $A(t)$ as a translation of $A$ both along and perpendicular to $\gamma(t)$", can you be a bit more specific? For example, if surface $A$ in the $xy$-plane is defined by the equation $A(u,v) = \left( x(u,v), y(u,v), 0 \right), (u,v) \in [a_1,b_1] \times [a_2, b_2]$, by extrusion $V$ along $\vec{\gamma}(t)$ do you mean $V(t,u,v) = \vec{\gamma}(t) + A(u,v)$ or something else? –  Nikola Milinković Nov 15 '12 at 14:08
    
@nikmil Almost, $A(u,v,t) = \vec r(u,v,t) \perp \vec\gamma(t)$, i.e. the original $x,y(u,v)$ have are rotated into the plane perpendicular to the extrusion path –  Tobias Kienzler Nov 15 '12 at 14:12
    
So if I understood correctly, first we rotate $A(u,v) = (x,y,0)$ to get $A_t (u,v) = \frac{\gamma'(t)}{||\gamma(t)||} \times (y, -x, 0)$ which lies in the binormal plane of $\gamma(t)$ and then we translate it to finally get $V(u,v,t) = \gamma(t) + A_t(u,v)$? –  Nikola Milinković Nov 15 '12 at 14:45
    
@nikmil Precisely, I'll update this into the question –  Tobias Kienzler Nov 15 '12 at 14:57

1 Answer 1

The simplest answer is simply to use the substitution rule: give a parametrization of your volume e.g. $F(x) = (A(x,y),z)$, then compute the Jacobi determinant (in our example $J(F) = J(A)$ and then $\int_V dV = \int_{x,y,z} |J(F)(x,y,z)| dx dy dz$. The Jacobi determinant can be easily computed with a CAS. This is always my way to derive Volume integral identities,

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