Htydjtk

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.

I'm not speaking about

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

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?

I guess it depends in how the function is defined.

f(x,y,z) dosent have to be something with density.

If you had a density distrupution f(x,y,z) where we would assume that f(x,y,z) is measured in $kg/m^3$ then to find the complete masse of the object you would have to take the integraf over x,y and z weighet by the density "distrubution". Given that the density changes as you move around in the 3d volume.

$m = int rho cdot dv$

If as you normaly do, have a density that dosent depend on position, $rho$ is a constant then it can be moved outside.

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

Another way of thinking about it is:

Say you look at a very small volume where the density is constant, then to find the volume of that small volume would be to multiply its volume dv with its density:

dm = $rho cdot dV$

To find the masse of the whole object you would have to add up all the small incriments:

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

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

Hope I answered what you asked:)

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

Analogously:

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

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

If you consider the integral $$intintint_D dxdydz$$ 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 $$intintint_D rho dx dy dz$$

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

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

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: $$intint_D rho (x,y) f(x) dx dy = rho (x,y)_{avg} intint_D f(x,y)dxdy$$

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