What is the real-world application of triple integrals?

[Calpalned]

Homework Statement

I know that a single integral can be used to find the area under a y = f(x) curve, but above the x axis. Correct me if this example of a double integral is invalid: If I hold a piece of paper in mid air and it droops, the double integral will give me the volume of the object bounded on the bottom by the ground (flat and even) and on top by the paper (smooth but sloping downwards - like a surface). What then, does the triple integral tell me? Is there a real-world example? Thank you.

I've seen that other people have asked the same question in this forum, but I still couldn't understand the reasoning even after reading them.

Homework Equations

n/a

The Attempt at a Solution

The double and triple integrals both give volume. So is there a difference? What is a real world example of a triple integral graph?

 

[Incand]

It is true you can calculate that volume with both a double and a triple integral.
For the double integral you would have to integrate over the height $f(x,y)$ with $\iint_D f(x,y)dxdy$. However that is equal to the triple integral over the body described by $D$ and the height function. $\iiint_B 1dxdydz$ would do exactly the same thing. Usually you just get the double integral directly when doing this but in some cases you get easier calculations using the triple integral.

Similary when you calculate the area in a plane with a normal integral that is equal to the double integral over 1 for that area $\int_a^b f(x)dx = \iint_D 1dxdy$.

But you don't want just the volume or area or something. The triple integral is very usefull if you are integrating over for example a vector field in 3 dimensions when you wouldn't integrate over 1 but instead the value of the vector field of that point.
Hope this helps you understand it a bit better!

 

[fourier jr]

What then, does the triple integral tell me? Is there a real-world example? Thank you.

If you want the mass of an object you need to know both the volume and the density, so if T is the solid & δ is the density function the mass of the object would be
$$\int\int\int_{T} \delta \,dV$$
If δ=1 you'd just get the volume. The coordinates of the centre of mass are then given by
$$\overline{x} = \frac{1}{m}\int\int\int_{T}x\delta \,dV$$
$$\overline{y} = \frac{1}{m}\int\int\int_{T}y\delta \,dV$$
$$\overline{z} = \frac{1}{m}\int\int\int_{T}z\delta \,dV$$

I suppose 2d objects could be done with double integrals but for more complicated ones you might need a triple integral