5
$\begingroup$

In my text book it says that the volume between the some region $R$ in the $xy$ plane and the surface $z=f(x,y)$ can be found by calculating $$\iint_D f(x,y)~dxdy$$

yet in the next page it uses this formula to calculate an area in the $xy$ plane not a volume under a surface why is this the case I don't see how they go from talking about volume to talking about area.

It also says the volume can be calculated as follows:

$$\iiint_S dxdydz $$ but I don't understand why this is the case also I thought we needed $z=f(x,y)$ to calculate the volume not $\omega=f(x,y,z)$ as surely this would be some other quantity in $4$ dimensions, not a volume?

Please help me clear this up thanks.

$\endgroup$

2 Answers 2

5
$\begingroup$

$$\iint _D 1 dxdy$$

I believe writing the number one explicitly will help you understand it. If you integrate number one over a surface, you will find the surface area.

$$\iint _D f(x,y) dxdy$$

If you integrate $z=f(x,y)$ over an area, you will find the volume under the surface defined by function $z=f(x,y)$. However this is a very restrictive method. $f(x,y)$ must be a function, that takes a single value for any pair $(x,y)$

$$\iiint _s 1 dxdydz$$

This will make you find the volume

$$\iiint _s f(x,y,z) dxdydz$$

This will make you find some property of the volume. Consider density. "mass per unit volume". Density of an object can be a function of $(x,y,z)$. If you integrate it over a volume, you will find the total mass. You can consider this to be the 3 dimensional analogue of $\iint _D f(x,y) dxdy$. We called this finding the volume under a surface, and what we just did for density could be called "finding * some 4th dimension quantity* of a volume"

$\endgroup$
6
  • $\begingroup$ Great answer, I think I'm starting to understand it better now many thanks! $\endgroup$
    – Matthew
    Jun 3, 2015 at 23:10
  • $\begingroup$ Yes, it is a good answer. +1. $\endgroup$
    – MPW
    Jun 3, 2015 at 23:11
  • $\begingroup$ Only one part I don't get surely $$\iint_D 1dxdy$$ would be the volume of the region $D$ under the plane $z=1$ and therefore it wouldn't be a area but a volume? $\endgroup$
    – Matthew
    Jun 3, 2015 at 23:12
  • $\begingroup$ $A=\int \int _D 1 dxdy$ A would be the area of the region. Consider a plate with area A and thickness 1. Its volume would be also A. But calling that a volume, i think, would be silly.|| What if you had a length L: that would also be the volume of a plate with length L, width 1 and height 1. $\endgroup$ Jun 3, 2015 at 23:15
  • $\begingroup$ Of course because volume=length x width x height so 1 x length x width=length x width=area. Very good many thanks. Quite silly of me. $\endgroup$
    – Matthew
    Jun 3, 2015 at 23:17
4
$\begingroup$

This is a simple application of Fubbini's theorem, because $$\iiint_S 1 dxdydz = \iint_{D} \int_{0}^{f(x,y)} 1 dz dxdy = \iint_D f(x,y) dx dy$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .