Multiple integrals: Triple integrals
Triple integrals in Cartesian coordinates
Like a double integral, we can define a triple integral of a function \(f(x,y,z)\) in three variables over a three-dimensional bounded area \(R\). The numerical approach consists of the following steps:
- Divide the region of integration \(R\) into small sub-regions \(\dd R\), which are also called volume elements; think, for example, of small beams;
- Choose the volume elements so small that the function above them is almost constant.
- Choose any point \((x,y,z)\) in each volume element \(\dd R\) and calculate \(f(x,y,z)\). Multiply this by the volume \(|\dd R|\) of \(\dd R\).
- Add up all the results obtained in step 3. The sum is then approximately equal to the triple integral. The approximation is better when the volume elements are smaller.
An exact calculation of a triple integral is also a generalisation of the calculation of double integrals. Examples speak for themselves.
\(\displaystyle\iiint_R x^2\,y\,z\,\dd(x,y,z)={}\) \(6\)
The iterated integral can be calculated as follows: \[\begin{aligned}\iiint_R x^2\,y\,z\,\dd(x,y,z)&= \int_{x=0}^{x=2}\left(\int_{y=0}^{y=1}\left(\int_{z=0}^{z=3}x^2\,y\,z\,\dd z\right)\dd y\right)\dd x\\[0.25cm] &= \int_{x=0}^{x=2}\left(\int_{y=0}^{y=1}\bigg[{{1}\over{2}}x^2\,y\,z^2\biggr]_{z=0}^{z=3}\;\dd y\right)\dd x\\[0.25cm] &=\int_{x=0}^{x=2}\left(\int_{y=0}^{y=1}{{9}\over{2}}x^2\,y\,\dd y\right)\dd x\\[0.25cm] &=\int_{x=0}^{x=2}\biggl[{{9}\over{4}}x^2\,y^2\biggr]_{y=0}^{y=1}\;\dd x\\[0.25cm] &= \int_{x=0}^{x=2}{{9}\over{4}}x^{2}\,\dd x\\[0,25cm] &=\biggr[{{3}\over{4}}x^3\biggr]_{x=0}^{x=2}\\[0.25cm] &= 6\end{aligned}\]
The iterated integral can be calculated as follows: \[\begin{aligned}\iiint_R x^2\,y\,z\,\dd(x,y,z)&= \int_{x=0}^{x=2}\left(\int_{y=0}^{y=1}\left(\int_{z=0}^{z=3}x^2\,y\,z\,\dd z\right)\dd y\right)\dd x\\[0.25cm] &= \int_{x=0}^{x=2}\left(\int_{y=0}^{y=1}\bigg[{{1}\over{2}}x^2\,y\,z^2\biggr]_{z=0}^{z=3}\;\dd y\right)\dd x\\[0.25cm] &=\int_{x=0}^{x=2}\left(\int_{y=0}^{y=1}{{9}\over{2}}x^2\,y\,\dd y\right)\dd x\\[0.25cm] &=\int_{x=0}^{x=2}\biggl[{{9}\over{4}}x^2\,y^2\biggr]_{y=0}^{y=1}\;\dd x\\[0.25cm] &= \int_{x=0}^{x=2}{{9}\over{4}}x^{2}\,\dd x\\[0,25cm] &=\biggr[{{3}\over{4}}x^3\biggr]_{x=0}^{x=2}\\[0.25cm] &= 6\end{aligned}\]
The aforementioned properties of double integrals each have their equivalent version in triple integrals. For example, Fubini's theorem says that for iterated integrals with proper limits of integration to specify the region of integration the final result is always the same.
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