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:
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
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 \(\vect g\text{.}\)
Remark6.4.
In the more traditional literature the Jacobian determinant is often denoted by
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
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
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
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.
Theorem6.5.Transformation formula.
Suppose that \(D\subset\mathbb R^3\) is a closed set on which the integral for every continuous function is well defined. Assume that \(\vect g\colon D\to\mathbb R^3\) is a one-to-one function with continuous first order derivatives. Then for every continuous function \(f\colon\vect g(D)\to\mathbb R\) we have