Change of Variables & the Jacobian

The reason is geometric: the map sends a tiny $du \times dv$ rectangle to a tiny parallelogram spanned by the column vectors $(x_u, y_u)$ and $(x_v, y_v)$, and a parallelogram's area is exactly the absolute determinant. Every coordinate system you have used is a special case: polar's factor $r$, cylindrical's $r$, spherical's $\rho^2\sin\phi$ — all Jacobians, all computable in two minutes from the definition rather than memorized. For the linear map $x = 2u + v$, $y = u + 2v$, the columns are $(2, 1)$ and $(1, 2)$ and the determinant is $4 - 1 = 3$: the unit square of the $uv$-plane becomes a parallelogram of area $3$, and every region's area is multiplied by $3$.

Why the reciprocal is allowed: the map $(x, y)\mapsto(u, v)$ and its inverse undo each other, so by the chain rule their Jacobian matrices multiply to the identity, and their determinants are reciprocals. This saves the step of solving for $x$ and $y$ in terms of $u$ and $v$, which is often the hardest part of the problem.

The integrand is simply $u$, so $\iint_R (x + y)\,dA = \int_0^3\!\int_1^3 u\cdot\tfrac13\,du\,dv = \tfrac13\cdot4\cdot3 = 4$. The area is $\int_0^3\!\int_1^3 \tfrac13\,du\,dv = \tfrac13\cdot2\cdot3 = 2$. Check it directly: the corners are $\left(\tfrac23, \tfrac13\right)$, $(2, 1)$, $(3, 0)$ and $\left(\tfrac53, -\tfrac23\right)$, and the shoelace formula also gives $2$.

So $dx\,dy = \dfrac{du\,dv}{2v}$ and the area is $\int_1^4\!\int_1^4 \dfrac{1}{2v}\,du\,dv = 3\cdot\dfrac{\ln4}{2} = 3\ln2 \approx 2.08$. The factor $\tfrac1{2v}$ records that equal squares in the $uv$-plane correspond to smaller pieces of the region near the steep line $y = 4x$ than near $y = x$.

Why a determinant again: the tiny box $du\,dv\,dw$ is carried to a tiny parallelepiped spanned by $\mathbf r_u\,du$, $\mathbf r_v\,dv$, $\mathbf r_w\,dw$, where $\mathbf r_u = (x_u, y_u, z_u)$ and so on. The volume of a parallelepiped is the absolute value of the triple product of its edges, which is the determinant with those edges as columns (see the triple product in the vectors topic).

Expand along the bottom row, which has a zero. The two $2\times2$ minors simplify with $\cos^2\theta + \sin^2\theta = 1$ to $\rho^2\sin\phi\cos\phi$ and $\rho\sin^2\phi$, so the determinant is $\cos\phi\cdot\rho^2\sin\phi\cos\phi + \rho\sin\phi\cdot\rho\sin^2\phi = \rho^2\sin\phi\left(\cos^2\phi + \sin^2\phi\right) = \rho^2\sin\phi$. This is the factor the triple-integrals topic found by measuring the edges of a spherical cell; here it comes from the same machine as every other Jacobian.

So the volume is $24$ times the volume of the unit ball: $24\cdot\tfrac43\pi = 32\pi$. In general the ellipsoid with semi-axes $a, b, c$ has volume $\tfrac43\pi abc$; stretching each direction separately multiplies volume by each stretch factor, and the determinant of a diagonal matrix is just that product.

A linear map multiplies all areas by $|\det J|$: the unit square spanned by $\mathbf e_1, \mathbf e_2$ lands on the parallelogram spanned by the columns of $J$, whose area is $|\det J|$, and general regions are limits of little squares. Summing $f \times (\text{image area})$ over a fine grid gives Riemann sums of $\iint f\,|\det J|\,du\,dv$. $\blacksquare$ SIGNIFICANCE: this is the bridge between linear algebra ("determinant = volume factor") and calculus ("substitution rule") — and in probability it is exactly how densities transform, the change-of-variables formula behind every normalizing flow in machine learning.

The "curvy rectangle" argument from the double-integrals topic was the geometric shadow of this two-line computation — and now spherical's $\rho^2\sin\phi$ is just a bigger determinant, not a new idea to memorize. $\blacksquare$

Step 1 — the region is a diamond, awkward in $x,y$; substitute $u = x + y$, $v = x - y$, which maps it to the square $|u| \le 1$, $|v| \le 1$.

Step 2 — Jacobian: $\dfrac{\partial(u,v)}{\partial(x,y)} = \det\begin{pmatrix}1&1\\1&-1\end{pmatrix} = -2$, so $dx\,dy = \tfrac12\,du\,dv$.

Step 3 — the integrand is $u^2$: $\displaystyle\int_{-1}^1\!\int_{-1}^1 u^2\cdot\tfrac12\,du\,dv = \tfrac12\cdot\tfrac23\cdot2 = \tfrac23$.

Step 4 — check plausibility: the diamond has area $2$ and $(x+y)^2$ averages less than $1$ on it, so an answer below $2$ is right-sized; the exact value $\tfrac23$ took three lines because the substitution matched BOTH region and integrand.

Step 1 — the Jacobian matrix of $(r, \theta, z) \mapsto (x, y, z)$ is $\begin{pmatrix}\cos\theta & -r\sin\theta & 0\\ \sin\theta & r\cos\theta & 0\\ 0 & 0 & 1\end{pmatrix}$.

Step 2 — expand along the last row: the determinant is $1\cdot(r\cos^2\theta + r\sin^2\theta) = r$.

Step 3 — so $dV = |r|\,dr\,d\theta\,dz = r\,dr\,d\theta\,dz$ since $r \ge 0$.

Step 4 — the geometric reading: the $z$-direction is untouched, so the 3D element is just the polar area element $r\,dr\,d\theta$ times $dz$ — the same determinant computed in the Jacobian topic, with one extra dimension riding along for free.