The Transformation Formula

Section 6.3 The Transformation Formula

For triple integrals we can derive a transformation formula similar to the one given in Section 5.3. To obtain such a formula we simply generalise the method presented in Section 5.3 to three dimensions. The situation is the following. We want to integrate a function,

f,

defined on the image of a domain

D \subset R^{3}

under a map

g .

The domain

D

and the deformed domain

g (D)

are as shown in Figure 5.15, but in space. We denote the coordinates in

D

(y_{1}, y_{2}, y_{3}),

and those in its image,

g (D),

(x_{1}, x_{2}, x_{3}) .

The map

g

has now three components:

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g (y_{1}, y_{2}, y_{3}) = (g_{1} (y_{1}, y_{2}, y_{3}), g_{2} (y_{1}, y_{2}, y_{3}), g_{3} (y_{1}, y_{2}, y_{3})) .

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Now we look at the image of a small rectangular box

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R = [y_{1}, y_{1} + Δ y_{1}] \times [y_{2}, y_{2} + Δ y_{2}] \times [y_{3}, y_{3} + Δ y_{3}]

As in two dimensions we find an approximation of the volume of the deformed rectangular box

g (R) .

By similar arguments as used in Section 5.3 we see that the volume of

g (R)

is close to the volume of the parallelepiped spanned by

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\begin{aligned} v_{1} & := \frac{\partial}{\partial y_{1}} g (y) Δ y_{1}, & v_{2} & := \frac{\partial}{\partial y_{2}} g (y) Δ y_{2}, & v_{3} := \frac{\partial}{\partial y_{3}} g (y) Δ y_{3} . \end{aligned}

The two dimensional analogue of the present situation is shown in Figure 5.16. Now we know from Theorem 1.29 that the

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volume of the paralellepiped spanned by v_{1}, v_{2} and v_{3} = | det [\begin{matrix} v_{1} & v_{2} & v_{3} \end{matrix}] | .

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Using the definition of

v_{1},

v_{2},

v_{3}

and the properties of the determinant we have

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\begin{aligned} det [\begin{array}{c} v_{1} & v_{2} & v_{3} \end{array}] & = det [\begin{array}{c} \frac{\partial g_{1}}{\partial y_{1}} (y) Δ y_{1} & \frac{\partial g_{1}}{\partial y_{2}} (y) Δ y_{2} & \frac{\partial g_{1}}{\partial y_{3}} (y) Δ y_{3} \\ \frac{\partial g_{2}}{\partial y_{1}} (y) Δ y_{1} & \frac{\partial g_{2}}{\partial y_{2}} (y) Δ y_{2} & \frac{\partial g_{2}}{\partial y_{3}} (y) Δ y_{3} \\ \frac{\partial g_{3}}{\partial y_{1}} (y) Δ y_{1} & \frac{\partial g_{3}}{\partial y_{2}} (y) Δ y_{2} & \frac{\partial g_{3}}{\partial y_{3}} (y) Δ y_{3} \end{array}] \\ = det [\begin{array}{c} \frac{\partial g_{1}}{\partial y_{1}} (y) & \frac{\partial g_{1}}{\partial y_{2}} (y) & \frac{\partial g_{1}}{\partial y_{3}} (y) \\ \frac{\partial g_{2}}{\partial y_{1}} (y) & \frac{\partial g_{2}}{\partial y_{2}} (y) & \frac{\partial g_{2}}{\partial y_{3}} (y) \\ \frac{\partial g_{3}}{\partial y_{1}} (y) & \frac{\partial g_{3}}{\partial y_{2}} (y) & \frac{\partial g_{3}}{\partial y_{3}} (y) \end{array}] Δ y_{1} Δ y_{2} Δ y_{3} \\ = det (J_{g} (y)) Δ y_{1} Δ y_{2} Δ y_{3}, \end{aligned}

where

J_{g} (y)

is the Jacobian matrix of

g

y

as introduced in Definition 4.18. Hence the

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\begin{aligned} volume of the parallelepiped spanned by & v_{1}, v_{2} and v_{3} \\ (6.3) & = | det (J_{g} (y)) | Δ y_{1} Δ y_{2} Δ y_{3}, \end{aligned}

so as with double integrals the Jacobian determinant appears. Again, the Jacobian determinant is the factor by which the volume of a small rectangle is distorted by the map

g .

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Remark 6.4.

In the more traditional literature the Jacobian determinant is often denoted by

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\frac{\partial (g_{1}, g_{2}, g_{3})}{\partial (y_{1}, y_{2}, y_{3})} := det (J_{g} (y)) .

We will occasionally use this notation.

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Now we partition

D

into small rectangular boxes. Let us denote the collection of rectangles covering

D

R .

y

denotes one corner of each

R \in R

the sum

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\sum_{R \in R} f (g (y)) vol (g (R))

is an approximation for the integral of

f

over

g (D) .

If we substitute

vol (g (R))

by (6.3) then the above sum is very close to

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\sum_{R \in R} f (g (y)) | det (J_{g} (y)) | Δ y_{1} Δ y_{2} Δ y_{3} .

The last expression is a Riemann sum for the function

y \mapsto f (g (y)) | det (J_{g} (y)) | .

Passing to the limit this suggests that the integral of

f

over

g (D)

is given by

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\int_{D} f (g (y)) | det (J_{g} (y)) | d y .

The above procedure is not a proof, but with some effort all arguments can be made rigorous. Hence we have the following result, completely analogous to Theorem 5.19.

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