# Why do both $\iint_R(x^2+y^2)\ dA$ and $\iiint_E 1\ dxdydz$ give the volume of $z=x^2+y^2$? What is the difference between $R$ and $E$?

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?

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

$$E$$ 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 $$z=1$$. 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: $$V = \iiint\limits_E dV = \iiint\limits_E 1 dz dy dx = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{0}^{x^2 +y^2}dz dy dx = \frac{\pi}{2}$$

• What program did you use for that draw? Jun 12, 2022 at 11:19
• @Hans-André-Marie-Stamm, I used Geogebra 3D. 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).