I can't understand how do we calculate volume with triple integral. for example $z =x^2+y^2$, we can calculate its volume with both: $$\iint\limits_R (x^2+y^2)\ dA$$ and $$\iiint\limits_E 1\ dxdydz$$ what is difference between $E$ and $R$ here?

  • 2
    $\begingroup$ Could you say what E and R are first? $\endgroup$
    – Paul
    Jun 9, 2020 at 10:02
  • 1
    $\begingroup$ Using the disk method, we can even do a single integral: $\int_0^H \pi z \,dz.$ $\endgroup$
    – David K
    Jun 9, 2020 at 12:11

2 Answers 2


$E$ and $R$ are both domains of integration. In the double integral case, $R$ represents the region of area under the surface where you're calculating the volume. In the triple integral case, $E$ represents the entire region of volume that we want to find.

Intuitively you can think of the double integral as the natural extension of a one-dimensional integral. Since $\int_{I} f(x) \ dx$ gives you the area under the function $f(x)$ over the interval $I \subseteq \mathbb{R}$, the analogous $\iint\limits_R f(x,y) \ dydx$ gives you the volume under the function $f(x,y)$ over the region $R \subseteq \mathbb{R}^2$.

For the intuition of the triple integral, you can think of this as dividing up the volume into tiny blocks "$dV$" and then you add the volumes of these blocks to find the value of your original volume. Here $E$ denotes the region where these "$dV$ pieces" are in $\mathbb{R}^3$.

For example, let's say that you want to find the volume below $z=x^2 +y^2$ delimited by $z=1$. enter image description here If we were to use a double integral to solve this, we need to find the region of area $R$ that's being enclosed by the blue circle on the $xy$ plane since this is the area under the volume of integration we're interested in. Using this we get that: $$ R = \left \{(x,y) \in \mathbb{R}^2 : -1 \le x \le 1, \ -\sqrt{1-x^2} \le y \le\sqrt{1-x^2} \right \} $$ which means we can compute the volume as: $$ V = \iint\limits_R (x^2 +y^2) dA = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(x^2 +y^2) dy dx = \frac{\pi}{2} $$

Alternatively, if we want to use a triple integral to find the volume, we need to first find the region $E$ made up of all the points in the volume we want to find. We then see that the volume is delimited by:

$$ E = \left \{(x,y,z) \in \mathbb{R}^3 : -1 \le x \le 1, \ -\sqrt{1-x^2} \le y \le\sqrt{1-x^2}, \ 0 \le z \le x^2 +y^2 \right \} $$ so we can find the volume by doing: $$ V = \iiint\limits_E dV = \iiint\limits_E 1 dz dy dx = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{0}^{x^2 +y^2}dz dy dx = \frac{\pi}{2} $$

  • $\begingroup$ What program did you use for that draw? $\endgroup$
    – Heidegger
    Jun 12, 2022 at 11:19
  • $\begingroup$ @Hans-André-Marie-Stamm, I used Geogebra 3D. $\endgroup$
    – Robert Lee
    Jun 12, 2022 at 11:20

In the first case you obtain the volume as "volume = height $\times$ area".

This is useful when you wan to calculate the volume contained within a certain surface $z = f(x,y)$ and the plane $z=0$. The "base" of this volume is $R$, a region of the $(x,y)$ plane. You can regard $f(x,y)$ as the local height and $dA(x,y)$ as a small local area around $(x,y)$, so that the volume is (formally) obtained as the sum

$$ V = \sum_{(x,y) \in R} f(x,y) \, dA(x,y) $$

In the second case you are calculating the volume as "volume = sum over small local volumes in a region of space". In this case you have a region $E \subset \mathbb{R}^3$ and, from the "philosophical" point of view,

$$ V = \sum_{(x,y,z) \in E} dV(x,y,z) = \sum_{(x,y,z) \in E} 1 \, \, dV(x,y,z) $$

where $dV(x,y,z)$ is a small volume around the point $(x,y,z)$.

Usually you write $dA(x,y) = dx dy $ and $dV(x,y,z) = dx dy dz$ but this may not be always the case (it depends on the volume form you are using, but this is a more advanced topic in differential geometry).


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