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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*}
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*}
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 \(\vect g\text{.}\)

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.
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.

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 \(y_3\text{.}\)
Note that the special case considered in Example 5.21 carries over to the present situation.