# Higher dimensional volume using triple integral

As a normal single variable integral is used to find an area under a certain region below a 2d-curve and double integral are used to find the volume under a 3d-curve, I used to think triple integrals can be used to find the volume under a 4d-graph.

$$\int \int \int \,dx\,dy\,dz$$

but this,

$$\int \int \int f(x,y,z) \,dx\,dy\,dz$$

The former makes complete sense to me as of why it calculates volume. But doesn't latter the one should calculate the volume under a 4d-graph.

Just like double integral,

$$\int \int f(x,y,z) \,dx\,dy$$ to find volume and $$\int \int \,dx\,dy$$ to find area.

But from my understanding in the $$\int \int \int f(x,y,z) \,dx\,dy\,dz$$ the $$f(x,y,z)$$ is called density function and the integral is used to find the mass.

Why density has anything to do with $$f(x,y,z)$$?

can someone explain to me this?

• The formula $\int_I fdxdydz$ would give the mass in a region $I$ of density $f$, but that's not the only possible interpretation of it. – J.G. Feb 17 at 14:08
• buy why f is considered as density? – Vignesh War Feb 17 at 14:11
• to get the units right.... – Jasser Feb 17 at 14:15
• @Jasser Yes at first I thought just like you said. But I'm kinda wondering whether there is a more intuitive explanation of why f can be considered as a density? – Vignesh War Feb 17 at 14:17

One could describe the triple integral $$\int \int \int f(x,y,z) \,\mathrm{d}x\mathrm{d}y\mathrm{d}z$$ as a method for finding a $$4$$-volume. But, generally, people don't visualize four dimensional regions very well, so this description has very little explanatory power.

People do have an understanding of volumetric density and do understand integrating the density to get a total quantity. (Could be mass density giving mass, charge density giving charge, probability density giving probability, or any of several more varieties of densities.) So describing the triple integral using density has more explanatory power.

But, you are right: the $$n$$-fold integral can be described as the integration of an $$n$$-dimensional density or as the (signed) $$n+1$$-volume between the graph of the integrand and the (hyper-)plane of the independent variables.

The curve $$S$$ defined by $$y=f(x)$$ bounds $$w\ge 0$$ measure $$\int_I y dx$$ (the $$2$$-dimensional kind, i.e. area) under $$S$$ over some region $$I\in\Bbb R$$ of values for $$x$$.

Analogously:

The surface $$S$$ defined by $$w=f(x,\,y,\,z)$$ bounds $$w\ge 0$$ measure $$\int_I w dxdydz$$ (the $$4$$-dimensional kind, but let's call it "volume" as you did) under $$S$$ over some region $$I\in\Bbb R^3$$ of values for $$(x,\,y,\,z)$$.

I guess it depends on how the function is defined.

$$f(x,y,z)$$ doesn't need to have anything to do with density.

If you have a density distribution $$f(x,y,z)$$ that is measured in $$kg/m^3$$, then to find the mass of the entire object, you would have to take the integral over $$x$$, $$y$$, and $$z$$, weighted by the density distribution, given that the density changes as you move around in the 3d volume.

$$m = \int \rho \cdot dv$$

If your density function is a constant, then you can simply move $$\rho$$ outside of the integral, as if you were multiplying a the volume of a container of water by its density.

$$m = \rho \int 1 \; dv = \rho \cdot V$$

Another way of thinking about it is:

Say you look at a very small volume where the density is constant, then to find the mass of that small volume, you would multiply the volume $$dV$$ with the density:

$$dm$$ = $$\rho \cdot dV$$

To find the mass of the whole object you would have to add up all the small increments:

$$dm_1 + dm_2 + dm_3 +... = \rho_1 \cdot dV + \rho_2 \cdot dV + \rho_3 \cdot dV +... = \int \rho(x,y,z) \;dV$$

But this is excatly what an integral does, but now $$\rho$$ depends on position and plays the role of $$f(x,y,z)$$ here.

I'd like to provide some intuition for this, at least.

If you consider the integral $$\int\int\int_D dxdydz$$ over some region D, what you're doing is summing up all the volume elements of the constant function, $$1$$, such that the value of each volume unit is the same.

Now, if you consider a constant density function, $$\rho$$, by computing $$\int\int\int_D \rho dx dy dz$$

You're simply scaling the volume up by the $$\rho$$, much like a material whose density is $$4\frac{kg}{m^3}$$ weighs $$4$$ times its volume.

Now, let's consider the function $$f(x,y) = x$$. By taking the double integral $$\int\int_D f(x,y)dxdy$$

You get the volume under the triangular region. By introducing a non-constant density function, $$\rho (x,y)$$, we will scale each volume element $$dV$$ by the average value of the density function over the region such that: $$\int\int_D \rho (x,y) f(x) dx dy = \rho (x,y)_{avg} \int\int_D f(x,y)dxdy$$

Of course, if you want to actually compute $$\rho (x,y)_{avg}$$, you need to evaluate the whole integral first without making it constant.