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Vector Calculus An Introduction

Daniel Daners

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, \(f\text{,}\) defined on the image of a domain \(D\subset \mathbb R^3\) under a map \(\vect g\text{.}\) The domain \(D\) and the deformed domain \(\vect g(D)\) are as shown in Figure 5.15, but in space. We denote the coordinates in \(D\) by \((y_1,y_2,y_3)\text{,}\) and those in its image, \(\vect g(D)\text{,}\) by \((x_1,x_2,x_3)\text{.}\) The map \(\vect g\) has now three components:
\begin{equation*} \vect g(y_1,y_2,y_3) =\bigl(g_1(y_1,y_2,y_3),g_2(y_1,y_2,y_3),g_3(y_1,y_2,y_3)\bigr)\text{.} \end{equation*}
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Now we look at the image of a small rectangular box
\begin{equation*} R=[y_1,y_1+\Delta y_1]\times[y_2,y_2+\Delta y_2] \times[y_3,y_3+\Delta y_3] \end{equation*}
As in two dimensions we find an approximation of the volume of the deformed rectangular box \(\vect g(R)\text{.}\) By similar arguments as used in Section 5.3 we see that the volume of \(\vect g(R)\) is close to the volume of the parallelepiped spanned by
\begin{align*} \vect v_1\amp:=\frac{\partial}{\partial y_1}\vect g(\vect y)\Delta y_1, \amp \vect v_2\amp:=\frac{\partial}{\partial y_2}\vect g(\vect y)\Delta y_2, \amp \vect v_3:=\frac{\partial}{\partial y_3}\vect g(\vect y)\Delta y_3\text{.} \end{align*}
The two dimensional analogue of the present situation is shown in Figure 5.16. Now we know from Theorem 1.29 that the
\begin{equation*} \text{volume of the paralellepiped spanned by }\vect v_1, \vect v_2\text{ and }\vect v_3 = \left|\det \begin{bmatrix} \vect v_1 \amp \vect v_2 \amp \vect v_3 \end{bmatrix} \right|\text{.} \end{equation*}
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Using the definition of \(\vect v_1\text{,}\) \(\vect v_2\text{,}\) \(\vect v_3\) and the properties of the determinant we have
\begin{align*} \det \begin{bmatrix} \vect v_1 \amp \vect v_2 \amp \vect v_3 \end{bmatrix} \amp =\det \begin{bmatrix} \dfrac{\partial g_1}{\partial y_1}(\vect y)\Delta y_1 \amp \dfrac{\partial g_1}{\partial y_2}(\vect y)\Delta y_2 \amp \dfrac{\partial g_1}{\partial y_3}(\vect y)\Delta y_3 \\ \dfrac{\partial g_2}{\partial y_1}(\vect y)\Delta y_1 \amp \dfrac{\partial g_2}{\partial y_2}(\vect y)\Delta y_2 \amp \dfrac{\partial g_2}{\partial y_3}(\vect y)\Delta y_3 \\ \dfrac{\partial g_3}{\partial y_1}(\vect y)\Delta y_1 \amp \dfrac{\partial g_3}{\partial y_2}(\vect y)\Delta y_2 \amp \dfrac{\partial g_3}{\partial y_3}(\vect y)\Delta y_3 \end{bmatrix}\\ \amp =\det \begin{bmatrix} \dfrac{\partial g_1}{\partial y_1}(\vect y) \amp \dfrac{\partial g_1}{\partial y_2}(\vect y) \amp \dfrac{\partial g_1}{\partial y_3}(\vect y) \\ \dfrac{\partial g_2}{\partial y_1}(\vect y) \amp \dfrac{\partial g_2}{\partial y_2}(\vect y) \amp \dfrac{\partial g_2}{\partial y_3}(\vect y) \\ \dfrac{\partial g_3}{\partial y_1}(\vect y) \amp \dfrac{\partial g_3}{\partial y_2}(\vect y) \amp \dfrac{\partial g_3}{\partial y_3}(\vect y) \end{bmatrix}\Delta y_1\Delta y_2\Delta y_3\\ \amp =\det\bigl(J_{\vect g}(\vect y)\bigr)\Delta y_1\Delta y_2\Delta y_3, \end{align*}
where \(J_{\vect g}(\vect y)\) is the Jacobian matrix of \(\vect g\) at \(\vect y\) as introduced in Definition 4.18. Hence the
\begin{align} \text{volume of the parallelepiped spanned by }\amp\vect v_1, \vect v_2\text{ and } \vect v_3\notag\\ \amp=\left|\det\left(J_{\vect g}(\vect y)\right)\right| \Delta y_1\Delta y_2\Delta y_3\text{,}\tag{6.3} \end{align}
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 \(\vect g\text{.}\)
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Remark 6.4.

In the more traditional literature the Jacobian determinant is often denoted by
\begin{equation*} \frac{\partial(g_1,g_2,g_3)}{\partial(y_1,y_2,y_3)} :=\det\bigl(J_{\vect g}(\vect y)\bigr)\text{.} \end{equation*}
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 \(D\) by \(\mathcal R\text{.}\) If \(\vect y\) denotes one corner of each \(R\in\mathcal R\) the sum
\begin{equation*} \sum_{R\in\mathcal R}f(\vect g(\vect y))\volume(\vect g(R)) \end{equation*}
is an approximation for the integral of \(f\) over \(\vect g(D)\text{.}\) If we substitute \(\volume(\vect g(R))\) by (6.3) then the above sum is very close to
\begin{equation*} \sum_{R\in\mathcal R}f(\vect g(\vect y)) \bigl|\det\bigl(J_{\vect g}(\vect y)\bigr)\bigr| \Delta y_1\Delta y_2\Delta y_3\text{.} \end{equation*}
The last expression is a Riemann sum for the function \(\vect y\mapsto f(\vect g(\vect y))\bigl|\det\bigl(J_{\vect g}(\vect y)\bigr)\bigr|\text{.}\) Passing to the limit this suggests that the integral of \(f\) over \(\vect g(D)\) is given by
\begin{equation*} \int_Df(\vect g(\vect y)) \bigl|\det\bigl(J_{\vect g}(\vect y)\bigr)\bigr|\,d\vect y\text{.} \end{equation*}
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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Remark 6.6.

In more traditional notation the transformation formula reads
\begin{align*} \iiint_Df(x_1,x_2,x_3) \amp\Bigl|\frac{\partial(x_1,x_2,x_3)}{\partial(y_1,y_2,y_3)} \Bigr|\,dy_1\,dy_2\,dy_3\\ \amp=\iiint_{\vect g(D)}f(x_1,x_2,x_3)\,dx_1\,dx_2\,dx_3\text{,} \end{align*}
where \(x_1,x_2,x_3\) are considered to be functions of \(y_1,y_2\) and \(y_3\text{.}\)
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Note that the special case considered in Example 5.21 carries over to the present situation.
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