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Concept Version 8
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Volumes

Volumes of complicated shapes can be calculated using integral calculus if a formula exists for the shape's boundary.

Learning Objective

  • Calculate the volume of a shape by using the triple integral of the constant function 1


Key Points

    • Volume is the quantity of three-dimensional space enclosed by some closed boundary—for example, the space that a substance or shape occupies or contains.
    • Volumes of some simple shapes, such as regular, straight-edged, and circular shapes can be easily calculated using arithmetic formulas.
    • Volumes of complicated shapes can be calculated using a triple integral of the constant function $1$: $\text{volume}(D)=\int\int\int\limits_D dx\,dy\,dz$.

Terms

  • volume

    a unit of three-dimensional measure of space that comprises a length, a width and a height; measured in units of cubic centimeters in metric, cubic inches, or cubic feet in English measurement

  • integral

    also sometimes called antiderivative; the limit of the sums computed in a process in which the domain of a function is divided into small subsets and a possibly nominal value of the function on each subset is multiplied by the measure of that subset, all these products then being summed

  • cuboid

    a parallelepiped having six rectangular faces


Full Text

Volume is the quantity of three-dimensional space enclosed by some closed boundary—for example, the space that a substance or shape occupies or contains. Three dimensional mathematical shapes are also assigned volumes. Volumes of some simple shapes, such as regular, straight-edged, and circular shapes can be easily calculated using arithmetic formulas. The volumes of more complicated shapes can be calculated using integral calculus if a formula exists for the shape's boundary. One-dimensional figures (such as lines) and two-dimensional shapes (such as squares) are assigned zero volume in three-dimensional space.

A volume integral is a triple integral of the constant function $1$, which gives the volume of the region $D$. That is to say:

$\displaystyle{\text{volume}(D)=\int\int\int\limits_D dx\,dy\,dz}$

It can also mean a triple integral within a region $D$ in $R^3$ of a function $f(x,y,z)$, and is usually written as:

$\displaystyle{\iiint\limits_D f(x,y,z)\,dx\,dy\,dz}$

Fig 1

Triple integral of a constant function $1$ over the shaded region gives the volume.

Example

The volume of the cuboid with side lengths 4, 5, and 6 may be obtained in either of two ways.

Method 1

Using the triple integral given above, the volume is equal to:

$\displaystyle{\iiint_\mathrm{cuboid} 1 \, dx\, dy\, dz}$

of the constant function $1$ calculated on the cuboid itself. This yields:

 $\displaystyle{\int_{z=0}^{z=5} \int_{y=0}^{z=6} \int_{x=0}^{x=4} 1 \, dx\, dy\, dz = 120}$

Method 2

Alternatively, we can use the double integral:

$\displaystyle{\iint_D 5 \ dx\, dy}$ 

of the function $z = f(x, y) = 5$ calculated in the region $D$ in the $xy$-plane, which is the base of the cuboid. For example, if a rectangular base of such a cuboid is given via the $xy$ inequalities $3 \leq x \leq 7$, $4 \leq y \leq 10$, our above double integral now reads:

$\displaystyle{\int_4^{10}\left( \int_3^7 \ 5 \ dx\right) dy =120}$

This is the volume under the surface.

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