Change of variables and Jacobians

A change of variables replaces the pair $(x,y)$ by a new pair $(u,v)$ — typically polar $(r,\theta)$ — chosen because the region or the integrand looks simpler there. The catch is that a small rectangle in the $uv$-plane does not map to a small rectangle of the same area in the $xy$-plane; it maps to a small parallelogram whose area is scaled by a local factor. That factor is exactly $|J|$.

General 2D formula. If $x=x(u,v)$, $y=y(u,v)$ is a smooth, invertible map from region $R'$ in the $uv$-plane onto $R$ in the $xy$-plane, then

where:

is the Jacobian determinant of the transformation — its absolute value is the local area-scaling factor, and $du\,dv$ alone would give the wrong area unless $|J|=1$.

Polar, cylindrical, spherical — the standard results. For $x=r\cos\theta$, $y=r\sin\theta$, the Jacobian works out to $J=r$, so $dA = r\,dr\,d\theta$ (derived in full below). Adding a $z$-axis unchanged gives cylindrical coordinates with the same $J=r$, so $dV = r\,dr\,d\theta\,dz$. For spherical coordinates $x=\rho\sin\phi\cos\theta$, $y=\rho\sin\phi\sin\theta$, $z=\rho\cos\phi$ (with $\rho\ge0$ the radial distance and $\phi$ the polar angle measured from the $z$-axis), the Jacobian is $J=\rho^2\sin\phi$, so $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$. These three are worth memorizing outright, since re-deriving them under exam pressure wastes time — but you must still be able to derive the polar one on demand. A subtlety worth flagging: the change-of-variables formula requires the map $(u,v)\mapsto(x,y)$ to be smooth and invertible (one-to-one) on the region of interest, with $J\ne0$ there — a point where $J=0$ is a place the transformation degenerates, folding or collapsing area rather than merely stretching it. At $r=0$ the polar map does exactly this (every $\theta$ collapses to the single point $x=y=0$), which is why $r=0$ is a single, harmless boundary point rather than a problem in practice.

Triple-integral version. The same idea extends to three variables: for $x=x(u,v,w)$, $y=y(u,v,w)$, $z=z(u,v,w)$,

where: $J$ is now a $3\times3$ determinant of the nine partial derivatives $\partial x/\partial u,\ldots,\partial z/\partial w$. The cylindrical and spherical Jacobians above are just this general formula applied to those two specific substitutions — nothing new is happening in principle, only more bookkeeping.

Problem. Evaluate $\displaystyle\iint_R e^{-(x^2+y^2)}\,dA$ where $R$ is the disc $x^2+y^2\le4$ (radius $2$), using polar coordinates.

Step 1. Recognize that no elementary Cartesian antiderivative exists for $e^{-(x^2+y^2)}$ — this is precisely the situation a coordinate change is built for. In polar coordinates the disc $x^2+y^2\le4$ becomes the simple rectangle $0\le r\le2$, $0\le\theta\le2\pi$, and $x^2+y^2=r^2$, so using $dA=r\,dr\,d\theta$ from the derivation above,

Step 2. Evaluate the inner integral over $r$ by substituting $u=r^2$, $du=2r\,dr$, so $r\,dr=\tfrac12du$, with limits $u=0$ to $u=4$:

Step 3. Evaluate the outer integral over $\theta$. Since the Step 2 result does not depend on $\theta$, integrating over $0$ to $2\pi$ just multiplies by $2\pi$:

Step 4. Compute the numerical value: $\pi(1-e^{-4}) \approx \pi(1-0.0183) \approx \pi(0.9817) \approx 3.084$. Answer: $\displaystyle\iint_R e^{-(x^2+y^2)}\,dA = \pi(1-e^{-4}) \approx 3.084$.

Competitive stretch

The same disc integral generalizes cleanly if the radius is $R$ instead of a fixed $2$: repeating Steps 1–3 with limits $0\le r\le R$ gives $\pi(1-e^{-R^2})$, and letting $R\to\infty$ (a common trick to evaluate the improper Gaussian integral over the whole plane) sends $e^{-R^2}\to0$, so $\iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dA = \pi$. This is the standard route to the classic result $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$, obtained by writing the plane integral as a product of two identical one-dimensional integrals and taking a square root — a favourite competitive-exam trick precisely because it needs polar coordinates and nothing more advanced.

• Forgetting the extra Jacobian factor — writing $dA=dr\,d\theta$ instead of $r\,dr\,d\theta$, or $dV=d\rho\,d\phi\,d\theta$ instead of $\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$.
• Dropping the absolute value $|J|$ in principle, even though the spherical Jacobian $\rho^2\sin\phi$ happens to stay non-negative for $\phi\in[0,\pi]$.
• Mixing up the spherical angle convention — $\phi$ is measured from the $z$-axis here (giving $z=\rho\cos\phi$), but some texts measure it from the $xy$-plane instead; stay consistent within one problem.
• Transforming the integrand but forgetting to re-express the region's boundary curves in the new variables — a circular region must become simple rectangular limits in $(r,\theta)$, not just have its formula substituted.

Ask the tutor: if you keep dropping the Jacobian factor or mis-converting a region's boundary into new coordinates, ask the AI tutor to walk through a full 3D Jacobian derivation for cylindrical or spherical coordinates before you try evaluating a triple integral with them.