Is it possible to calculate the volume of a general 3D (closed & convex) parametric surface?

Question

I have three parametric equations in two variables that give the coordinates of points on a three-dimensional, closed, convex surface. I want to find the volume enclosed by that surface, but I haven't been able to find a method to do so.

The Cartesian coordinates of a parametric surface can be defined in terms of the curve's support function $h(θ, φ)$ and its partial derivatives $h_{(1, 0)}'(θ, φ)$ and $h_{(0, 1)}'(θ, φ)$ as $$ \begin{alignedat}{1} & x(θ, φ) = \sin(θ) ⋅ \cos(φ) & ,円 ⋅ ,円 h(θ, φ) + \cos(θ) ⋅ \cos(φ) & ⋅ h_{(1, 0)}'(θ, φ) − \frac{\sin(φ)}{\sin(θ)} ⋅ h_{(0, 1)}'(θ, φ), \\ & y(θ, φ) = \sin(θ) ⋅ \sin(φ) & ,円 ⋅ ,円 h(θ, φ) + \cos(θ) ⋅ \sin(φ) & ⋅ h_{(1, 0)}'(θ, φ) + \frac{\cos(φ)}{\sin(θ)} ⋅ h_{(0, 1)}'(θ, φ), \\ & z(θ, φ) = \cos(θ) & ,円 ⋅ ,円 h(θ, φ) − \sin(θ) & ⋅ h_{(1, 0)}'(θ, φ), \end{alignedat}\\ 0 ≤ θ ≤ π \; ∧ \; 0 ≤ φ ≤ 2 π. $$

The support function of a 3D surface describes the (signed) distance between the surface's supporting planes and the origin, where a supporting plane is a plane that touches the surface at one or more points but does not cross the interior of the surface.

I know that the surface area can be found by evaluating the double-integral $$ A = \int_{0}^{2 π}\left(\int_{0}^{π} \bigl|\vec{n}(θ, φ)\bigr| dθ\right) dφ $$ where the normal vector $\vec{n}(θ, φ)$ is the cross product of the two vectors given by differentiating the three parametric equations with respect to $θ$ and $φ$, respectively. However, I have not been able to find a similar integral for the volume.

• Ted Shifrin: Apply the divergence theorem. Do the surface integral, say, of $(x,0,0)\cdot\vec n$ over the closed surface.
• Lawton: @TedShifrin Can you elaborate? I looked up the divergence theorem, and I understand just enough of what I found to think it is indeed what I need, but not enough to figure out what calculations to carry out.
• Ted Shifrin: I told you precisely what calculation to carry out. The flux of $\vec F = (x,0,0)$ gives the triple integral of $\text{div}\vec F=1$ over the inside, i.e., the volume.
• Lawton: @TedShifrin Does that mean I set $\vec{F} = \bigl(x(θ, φ), 0, 0\bigr),ドル take its dot product with $\vec{n},ドル and replace $\bigl|\vec{n}(θ, φ)\bigr|$ in the "surface area" double integral with the result?
• Ted Shifrin: My $\vec n$ stands for the unit normal, so, yes, $$\vec F\cdot \vec n,円dS = \vec F(\vec g(u,v))\cdot (\partial\vec g/\partial u \times \partial \vec g/\partial v)du,円dv,$$ where $\vec g$ is the parametrization.

Answer 1

Given the definitions for the three parametric equations from my question, we have their partial derivatives

$$ \begin{align*} x_{\theta}'(\theta, \varphi) &= \cos(\varphi) \cdot \cos(\theta) \cdot h(\theta, \varphi) + \cos(\varphi) \cdot \cos(\theta) \cdot h_{(2, 0)}'(\theta, \varphi) \\ &\quad + \frac{\sin(\varphi)}{\sin(\theta)} \cdot \frac{\cos(\theta)}{\sin(\theta)} \cdot h_{(0, 1)}'(\theta, \varphi) - \frac{\sin(\varphi)}{\sin(\theta)} \cdot h_{(1, 1)}'(\theta, \varphi), \\ y_{\theta}'(\theta, \varphi) &= \sin(\varphi) \cdot \cos(\theta) \cdot h(\theta, \varphi) + \sin(\varphi) \cdot \cos(\theta) \cdot h_{(2, 0)}'(\theta, \varphi) \\ &\quad - \frac{\cos(\varphi)}{\sin(\theta)} \cdot \frac{\cos(\theta)}{\sin(\theta)} \cdot h_{(0, 1)}'(\theta, \varphi) + \frac{\cos(\varphi)}{\sin(\theta)} \cdot h_{(1, 1)}'(\theta, \varphi), \\ z_{\theta}'(\theta, \varphi) &= -\sin(\theta) \cdot h(\theta, \varphi) - \sin(\theta) \cdot h_{(2, 0)}'(\theta, \varphi), \\ \\ x_{\varphi}'(\theta, \varphi) &= -\sin(\varphi) \cdot \sin(\theta) \cdot h(\theta, \varphi) + \cos(\varphi) \cdot \cos(\theta) \cdot h_{(1, 1)}'(\theta, \varphi) \\ &\quad - \frac{\cos(\theta)}{\sin(\theta)} \cdot \cos(\varphi) \cdot \cos(\theta) \cdot h_{(0, 1)}'(\theta, \varphi) - \frac{\sin(\varphi)}{\sin(\theta)} \cdot h_{(0, 2)}'(\theta, \varphi) \\ &\quad - \sin(\varphi) \cdot \cos(\theta) \cdot h_{(1, 0)}'(\theta, \varphi), \\ y_{\varphi}'(\theta, \varphi) &= \cos(\varphi) \cdot \sin(\theta) \cdot h(\theta, \varphi) + \sin(\varphi) \cdot \cos(\theta) \cdot h_{(1, 1)}'(\theta, \varphi) \\ &\quad - \frac{\cos(\theta)}{\sin(\theta)} \cdot \sin(\varphi) \cdot \cos(\theta) \cdot h_{(0, 1)}'(\theta, \varphi) + \frac{\cos(\varphi)}{\sin(\theta)} \cdot h_{(0, 2)}'(\theta, \varphi) \\ &\quad + \cos(\varphi) \cdot \cos(\theta) \cdot h_{(1, 0)}'(\theta, \varphi), \\ z_{\varphi}'(\theta, \varphi) &= -\sin(\theta) \cdot h_{(1, 1)}'(\theta, \varphi) + \cos(\theta) \cdot h_{(0, 1)}'(\theta, \varphi). \end{align*} $$

The derivatives can be combined into vectors as $$ \vec{r_{θ}}(θ, φ) = \begin{pmatrix} x_{θ}'(θ, φ), \\ y_{θ}'(θ, φ), \\ z_{θ}'(θ, φ) \end{pmatrix} $$ and $$ \vec{r_{φ}}(θ, φ) = \begin{pmatrix} x_{φ}'(θ, φ), \\ y_{φ}'(θ, φ), \\ z_{φ}'(θ, φ) \end{pmatrix}. $$

The cross product of the derivative vectors gives the normal vector to the surface: $$ \begin{align} \vec{n}(θ, φ) &= \vec{r_{θ}}(θ, φ) ×ばつ \vec{r_{φ}}(θ, φ) \\ &= \begin{pmatrix} y_{θ}'(θ, φ) ⋅ z_{φ}'(θ, φ) - z_{θ}'(θ, φ) ⋅ y_{φ}'(θ, φ), \\ z_{θ}'(θ, φ) ⋅ x_{φ}'(θ, φ) - x_{θ}'(θ, φ) ⋅ z_{φ}'(θ, φ), \\ x_{θ}'(θ, φ) ⋅ y_{φ}'(θ, φ) - y_{θ}'(θ, φ) ⋅ x_{φ}'(θ, φ) \end{pmatrix}. \end{align} $$

Based on Ted Shifrin's comments, the following three double integrals should all give the volume of the parametric surface: $$ \int_{0}^{2 π}\left(\int_{0}^{π} \left(\begin{pmatrix}x(θ, φ), \\ 0, \\ 0\end{pmatrix} ⋅ \vec{n}(θ, φ)\right) dθ\right) dφ, \\ \int_{0}^{2 π}\left(\int_{0}^{π} \left(\begin{pmatrix}0, \\ y(θ, φ), \\ 0\end{pmatrix} ⋅ \vec{n}(θ, φ)\right) dθ\right) dφ, \\ \int_{0}^{2 π}\left(\int_{0}^{π} \left(\begin{pmatrix}0, \\ 0, \\ z(θ, φ)\end{pmatrix} ⋅ \vec{n}(θ, φ)\right) dθ\right) dφ. $$

Expanding the expressions inside each double integral to get them all in terms of the support function and its derivatives results in extremely messy forms, but using the Sum Rule of Integration we can add the contents of all the integrals together inside one double integral (remembering to divide it by 3ドル$), which allows many terms to completely cancel out. Taking the simplification as far as I can, I've reached the following form: $$ V = \frac{1}{3} ⋅ \int_{0}^{2 π}\left(\int_{0}^{π} \begin{pmatrix} \sin(θ) ⋅ h(θ, φ)^3 \\ + h(θ, φ)^2 ⋅ \bigl(\cos(θ) ⋅ h_{(1, 0)}'(θ, φ) + \csc(θ) ⋅ h_{(0, 2)}'(θ, φ) + \sin(θ) ⋅ h_{(2, 0)}'(θ, φ)\bigr) \\ - \csc(θ) ⋅ h(θ, φ) ⋅ \bigl(\cot(θ)^2 ⋅ h_{(0, 1)}'(θ, φ) - \csc(θ) ⋅ h(θ, φ) ⋅ h_{(1, 1)}'(θ, φ)\bigr)^2 \\ + h(θ, φ) ⋅ h_{(2, 0)}'(θ, φ) ⋅ \bigl(\csc(θ) ⋅ h_{(0, 2)}'(θ, φ) + \cos(θ) ⋅ h_{(1, 0)}'(θ, φ)\bigr) \end{pmatrix} {dθ}\right) {dφ} $$

It's still quite messy, but evaluating it with sample values of $h(θ, φ)$ produces reasonable results, so I'm reasonably confidant that it works. I'll keep working on trying to simplify it further.

• Ted Shifrin: Without having worked on it at all, my optimistic guess would be that the $z$ integral alone might come out the simplest.
• Lawton: @TedShifrin I double-checked, and just using the $z(θ, φ)$ integral results in a lot of terms not cancelling out (in any way that is obvious to me, at least). It's very possible I'm missing some simplification tricks since I have no formal education in multi-variable calculus or differential geometry, though.
• Ted Shifrin: Your formulas, as we know, are horrendously messy, so I'm not surprised my guess was wrong. This mess has nothing to do with multivariable calculus or differential geometry.