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This may be a silly question but I have trouble grasping this very basic concept.

In equations, sometimes we have

$$ \int_0^\pi \int_x^\pi \frac{\sin y}{y} dydx$$

function given as f(x,y). And other times it's just double or triple integral with 1 as f(x,y) followed by dydx.

What does this f(x,y) do in getting the volume of the object? Can't we just set f(x,y) = 1 , and get volume by having correct integrals?

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    $\begingroup$ In double integral I guess f(x,y) gives the z-value, which can be useful in calculating volume. But why do we need f(x,y,z) in triple integral? Other than f(x,y,z) being 1? $\endgroup$ – himes Dec 6 '15 at 1:35
  • $\begingroup$ The $f(x,y,z)$ in a triple integral may be a density so the integral is finding mass rather than volume. Or the charge density, leading to total electric charge. Volume is not the only thing found with triple integrals! For that matter, $f(x,y,z)$ could also be the $w$ component of a 4-dimensional figure, and the integral is for 4D volume. $\endgroup$ – Rory Daulton Dec 6 '15 at 1:43
  • $\begingroup$ but isn't $\delta(x,y,z)$ different from $f(x,y,z)$? If I understand correctly, f(x,y,z) in triple integral has no effect whatsoever on volume? $\endgroup$ – himes Dec 6 '15 at 1:49
  • $\begingroup$ The only difference between $\delta(x,y,z)$ and $f(x,y,z)$ is the names. And using $f(x,y,z)$ definitely has an effect on the triple integral, but the integral may not be trying to find a 3-dimensional volume. $\endgroup$ – Rory Daulton Dec 6 '15 at 11:41
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Consider the volume: $V=\displaystyle\int_0^\pi \int_x^\pi \int_0^{\sin(y)/y} 1\operatorname d z\operatorname d y\operatorname d x\\ = \displaystyle\int_0^\pi \int_x^\pi \frac{\sin(y)}{y}\operatorname d y\operatorname d x\\ = \displaystyle\int_0^\pi \operatorname{Si}(\pi)-\operatorname{Si}(x)\operatorname d x\\=2 $

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