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The function $A=(\sin(y)\sin(z)+\cos(y)\cos(z))\sin(w)\sin(x)+\cos(w)\cos(x)$, given $w\in[0,\pi], x\in[0,\pi], y\in[0,2\pi], z\in[0,2\pi]$, defines a three-dimensional "surface" in 4D. ($A = f(w,x,y,z)$ represent level sets). How would I calculate the hyper-area of this surface as a function of A?

thank you!

p.s. I don't necessarily need a closed-form solution, I'm going to evaluate the integral numerically, but I don't know what the integral should be.

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Oh, $A$ is not a parametrization. I believe your equation involving $A$ is a level-set for a function of $4$ variables. Is $A$ supposed to describe the level set? i.e. you're solving for $x,y,z,w$ in terms of $A$? My initial answer was assuming you had a parametrization. – Ryan Budney Apr 21 '11 at 22:41
Yes, you are correct. – okj Apr 21 '11 at 22:50

Given a differentiable function $f : U \to \mathbb R^m$ where $U \subset \mathbb R^n$ is open, the $n$-dimensional content/volume of $f(U)$ is traditionally defined as the integral:

$$\mu(f(u)) := \int_U \sqrt{ Det( (f')^t f' ) }$$

where $f'$ is the derivative of $f$, thought of (point-wise) as an $m$-by-$n$ matrix. $(f')^t$ is the transpose of that matrix.

I'll perhaps add to this response in a few moments.


Okay, so although the above reply is technically correct it's not really useful for what you need to do since you don't have a parametrization of your level-set. At least, not yet. But there is a nice parametrization. There's also a pretty cute way to compute the volume without computing a parametrization, maybe I'll describe that later.

Your level set has a lot of symmetry and we can make use of that. Think of $x$, $y$, $z$ and $w$ as being real numbers. Then there is a symmetry $(x,y,z,w) \longmapsto (x,y+a,z+a,w)$ where $a$ is an arbitrary real number.

Since $\sin y \sin z + \cos y \cos z$ is a number between $-1$ and $1$, call it $\alpha$. Similarly, let $\beta = \sin w \sin x + \cos w \cos x$, then this turns the problem into one of solving for

$$\alpha \beta + (1-\alpha) \cos w \cos x = A$$

You can solve for $\beta$, giving

$$\beta = c - \frac{1-\alpha}{\alpha} \cos w \cos x $$

So now you have a parametrization of your domain. Perhaps it's not so clean, especially since the domain of integration is described by all solutions to $-1 \leq c-\frac{1-\alpha}{\alpha} \cos w \cos x \leq 1$.

But if you have a software package where you can specify the domain of integration for a multi-variable function in terms of functional inequalities, this should be fine. Hmm, maybe there's a better way to approach this problem.

** 2nd edit **

And the "cute" way to compute the area would be with the appropriate Cauchy-Crofton theorem. The corresponding theorem for lengths of planar curves is here: but there are analogous formula for $n$-dimensional content of submanifolds of $\mathbb R^m$. The integral is:

$$ \int_{E_{m,n}} |L \cap N| dL $$

here $N \subset \mathbb R^m$ is an $n$-dimensional submanifold of $\mathbb R^n$. $E_{m,n}$ is the space of $(m-n)$-dimensional affine-linear subspaces of $\mathbb R^m$ given a certain natural measure. Given $L \in E_{m,n}$, $L \cap N$ is the intersection of $L$ with $N$, and $|L\cap N| \in [0,\infty]$ is the cardinality of the intersection. The space where $|L \cap N|=\infty$ has measure zero in $E_{m,n}$ so that's not relevant to the integration. Technically there is a constant in front of the above integral and it's generally reasonable to work out. In your case, $m=4$, $n=3$ and $E_{4,3}$ is the tangent bundle of $\mathbb RP^3$ so it has a fairly natural parametrization by $S^3 \times \mathbb R^3$. Approximating the above integral is computationally a little less complicated than rendering (ray-tracing) a small scene.

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Thank you for your prompt response. I think you are right, I've edited the post accordingly. It is a four dimensional surface in 5D. It seems to me that your answer regarding the "content" still applies, am I correct? – okj Apr 21 '11 at 22:48
Let's pull things down two dimensions. Given the equation $A = x^2 + y^2$, I could think of this as asking for (1) the length of a circle of radius $\sqrt{A}$, or (2) I could think of it as asking for the surface area of a paraboloid $z = x^2 + y^2$ in $\mathbb R^3$. If we go back to your question, which version of this are you looking for? – Ryan Budney Apr 21 '11 at 22:52
Essentially I want to know the following: for a given value of A (i.e. for a given level set of $f(w,x,y,z)$), what is the content satisfying $f(w,x,y,z) = A$. Which I believe is situation (1) in your previous post. – okj Apr 21 '11 at 22:57
I'm interested in the "cute" way you mentioned of computing volume without a parametrization, could you elaborate? – okj Apr 22 '11 at 1:51
Thank you. I had never heard of the Cauchy-Crofton Theorem, it's pretty cool. I was wondering, could I go about this problem by generalizing the formula for arc-length? That is, $f(w,x,y,z) = (\sin(y)\sin(z)+\cos(y)\cos(z))\sin(w)\sin(x)+\cos(w)\cos(x)$ is a "surface" in $\Re^5$, which makes the level set $A = (\sin(y)\sin(z)+\cos(y)\cos(z))\sin(w)\sin(x)+\cos(w)\cos(x)$ analogous to a "curve" in $\Re^5$. Can the content of that "curve" be computed simply by: $s = \int\sqrt{{\frac{dw}{dt}}^2+{\frac{dx}{dt}}^2+{\frac{dy}{dt}}^2+{\frac{dz}{dt}}^‌​2}dt$ – okj Apr 22 '11 at 14:29

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