What integrals in 2D plane and 3D space represent?

Question

We know that if a real valued function $f$ is continuous over an interval $[a,b]$ then the following integral $$\int_a^bf(x)dx$$ represents the area between horizontally the line $y=0$ and the curve of $f$, vertically between the lines $x=a$ and $x=b$. So what represent the following $$\int_{[a,b]\times [c,d]}g(x,y)dxdy$$ and $$\int_{[a,b]\times [c,d]\times [e,f]}h(x,y,z)dxdydz$$ where $g$ and $h$ are two continuous real valued 2d and 3d functions. Thanks

Answer 1

The double integral, represents the sum of the infinite areas under the curves $g(x,y)$ with $x=constant$ defined as: $$A(y)=\int{g(x,y)\,dx}$$ each one between $y$ and $y+dy$, which gives you a volume. (see Fubini 's explanation)

The last integral gives you the infinite sum of volumes $V(z)$ defined as $$V(z)=\int\int{h(x,y,z)\,dxdy}$$ which gives you a hypervolume.

Imagine that if the function $h$ is a sphere the number $\int V(z) \,dz$ will be the sum of all the volumes of the infinite spheres of radius $z$ between the integration interval $[z_1,z_2]$

Answer 2

Two-variable integrations refer to the volume enclosed by the surface, the $xy$-plane and the planes perpendicular to $xy$-plane passing through $x=a$ to $x=b$, and $y=c$ to $y=d$, in the case of:

$$\int_{y=c}^{y=d}\int_{x=a}^{x=b}{g(x,y)dxdy}$$

Going by analogy, the three-variable integrations refer to the hypervolume, or the 4D analog of our usual 3D volume. This is difficult to visualize for human beings as we are only capable of interacting and understanding a 4D world. However, as the volume in the previous case is considered to be enclosed by plane surfaces, the hypervolume may be described as being enclosed by 3D surfaces.

We may consider

$$\int_{z=e}^{z=f}\int_{y=c}^{y=d}\int_{x=a}^{x=b}{h(x,y,z)dxdydz}$$

as the hypervolume enclosed between:

1. The $xyz$-space at $w=0$
2. The 4D curve $w=h(x,y,z)$
3. The $yzw$-spaces at $x=a$ and $x=b$
4. The $xzw$-spaces at $y=c$ and $y=d$
5. The $xyw$-spaces at $z=e$ and $z=f$

Answer 3

karimath, I may contribute.

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1D integral can be used to calculate area. For example, if we want to calculate an area bounded by $y=f(x)$ and $y=0$ in $x \in [a, b]$, then we can do approximate this by :

$$ A \approx \sum_{i=0}^{N} |f(x_{i})-0| \triangle x, \:\:\: \triangle x = \frac{b-a}{N} $$

$N$ is the number of rectangles, $\triangle x$ being the width of each rectangle $i$, and $f(x_{i})$ is the height of the rectangle $i$.

If we take as many rectangles, we get $$ A = \lim_{N \rightarrow \infty} \sum_{i=0}^{N} |f(x_{i})-0| \triangle x, \:\:\: \triangle x = \frac{b-a}{N} $$ which is defined another way by $$ A = \int_{a}^{b} f(x) dx$$

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2D integral is similar. You can approximate the Volume bounded by surface $z=f(x,y)$ and $z=0$, in the region $R : a < x < b, \:\: c < y<d$, by :

$$ V \approx \sum_{j=0}^{N} \sum_{i=0}^{M} |f(x_{i},y_{j})-0| \triangle x \triangle y, \:\:\: \triangle x = \frac{b-a}{M}, \: \: \: \triangle y = \frac{d-c}{N} $$ Notice that $$ |f(x_{i}, y_{j})| \triangle x \triangle y$$ is the volume of the small cuboid with center at position $(x_{i}, y_{j})$, $\triangle x \triangle y$ being the area of the small square as the floor of the cuboid.

Taking as many very-small-cuboids as possible, we get the Volume

$$ V = \int_{c}^{d} \int_{a}^{b} f(x,y) dx dy $$

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For 3D, it is a bit different. The function $h(x,y,z)$ is a quantity that may be measured. The $\triangle x \triangle y \triangle z$ is the small-volume in which a value $h(x_{i},y_{j},z_{k})$ holds. Connect this to @TrevorNorton comment.