\(f(x)\) defined over the support \(c_1 g(b)\), and thus \([c,d] = [g(b), g(a)].\) Also in this case, \(|g'(u)| = -g'(u)\) for all \(u\). a dignissimos. Generally, the function that we use to change the variables to make the integration simpler is called a transformation or mapping. That is not always the case however. \] again, here we just mean to indicate triple integrals, not necessarily in the given order. Lets start with the \(x\) transformation and plug in the known value of \(u\) for this equation. \det D\mathbf G(\mathbf u) = \frac 1{\det[ D(\mathbf G^{-1})(\mathbf x)] } \qquad We know that, Area of a circle, A = r 2. z Change of variables is an operation that is related to substitution. The third equality holds because, as shown in red on the following graph, for the portion of the function for which \(u(X)\le y\), it is also true that \(X\ge v(Y)\): The fourth equality holds from the rule of complementary events. u = x+y+z, \quad v = x+2y+4z, \quad w = x+2y+8z. So, lets determine the range of \(v\)s we should get. And, let \(Y=u(X)\) be an invertible function of \(X\) with inverse function \(X=v(Y)\). Finally, lets transform \(y = \frac{x}{3} - \frac{4}{3}\). But, because the function is continuous and increasing, an inverse function \(X=v(Y)\) exists. Consequently. Planar Transformations. r\, \cos \theta \\ change of coordinates from polar to Cartesian coordinates, but matrix Use the transformation equations to substitute u and v for x and y. So, the range of \(v\)s for \(u = 4\) must be \( - 1 \le v \le 4\), which nicely matches with what we would expect from the graph of the new region. \iiint_{S}1\, dV = \iiint_{T} (1-z)^2\, dV. \iint_S 1 \, dA, \qquad\quad\text{ for } This explains the presence of \(|\det D\mathbf G(u)|\) in formula \(\eqref{cofv}\).