Exercises — Change of variables — general Jacobian

The four partials are $x_u=3,x_v=0,y_u=0,y_v=5$. $J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}3& 0\\ 0& 5\end{array}\right)=3\cdot 5-0\cdot 0=15.$ What it looks like: a unit square in $uv$ becomes a $3\times 5$ rectangle in $xy$ — area $15$ times bigger. The scaling factor is $|J|=|15|=\boxed{15}$.

The polar Jacobian is $r$ (derived in the parent note). At $r=2$ it equals $\boxed{2}$: a polar cell at radius $2$ has twice the area of the same $drd\theta$ box at radius $1$. The farther out, the fatter the wedge — see the Polar Coordinates picture.

(a) $J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)=-2\ne 0$ everywhere — valid. (b) $J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}2u& 0\\ 0& 1\end{array}\right)=2u$, which is $0$ at $u=0$. The map also fails to be one-to-one there ($u$ and $-u$ give the same $x$). Invalid at $u=0$. A zero Jacobian means the little cell got squashed flat — see the Inverse Function Theorem.

$J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}\cos v& -u\sin v\\ \sin v& u\cos v\end{array}\right)=u{\cos }^{2}v+u{\sin }^{2}v=u.$ Why the collapse: ${\cos }^{2}v+{\sin }^{2}v=1$, so everything folds to $u$. Hence $dxdy=ududv$, exactly the $rdrd\theta$ of the parent note.

First the forward Jacobian (easy, since $u,v$ are given as functions of $x,y$): $\frac{\partial (u,v)}{\partial (x,y)}=\det \left(\begin{array}{cc}2x& -2y\\ 2y& 2x\end{array}\right)=4x^2+4y^2.$ Why reciprocate: the theorem needs old area per new cell, i.e. $\partial (x,y)/\partial (u,v)$, and the two Jacobians are inverse matrices so their determinants multiply to $1$. Therefore $J=\frac{\partial (x,y)}{\partial (u,v)}=\frac{1}{4(x^2+y^2)}.$ At $(1,1)$: $\frac{1}{4(1+1)}=\frac{1}{8}=\boxed{0.125}$.

$J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)=6-1=5.$ The unit $uv$-square maps onto $R$, so $\text{Area}={\int }_0^1{\int }_0^1|J|dudv={\int }_0^1{\int }_0^{1}5dudv=5.$ This is exactly the Determinant as Volume Scaling statement: the parallelogram's area equals $|\det |$ of the edge vectors. Answer: $\boxed{5}$.

The four corners map to $(0,0),(1,1),(2,0),(1,-1)$ — the parallelogram in the figure below. $J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)=-2,|J|=2.$ The parallelogram's edge vectors are $\mathbf{a}=(1,1)$ (image of the $u$-edge) and $\mathbf{b}=(1,-1)$ (image of the $v$-edge). Its area by the Cross Product and Area rule is $|a_x{b}_y-a_y{b}_x|=|1\cdot (-1)-1\cdot 1|=2$. It matches $|J|$ exactly, as it must.

Figure — read it like this: the left panel is the plain unit square in $uv$ (area $1$); its bottom yellow arrow is the $u$-edge, its left pink arrow the $v$-edge. The right panel is the image in $xy$: the yellow arrow $\mathbf{a}=(1,1)$ is where the $u$-edge landed, the pink arrow $\mathbf{b}=(1,-1)$ where the $v$-edge landed. The tilted blue parallelogram they span has area $2$ — exactly $|J|$, the number written inside it.

Let $J_T$ be the Jacobian matrix of $T:(u,v)\mapsto (x,y)$ and $J_{T^{-1}}$ that of the inverse map. Composing $T^{-1}\circ T$ gives the identity, so by the multivariable chain rule (which multiplies Jacobian matrices) $J_{T^{-1}}{J}_T=I$. Taking determinants and using $\det (AB)=\det A\det B$: $\det J_{T^{-1}}\cdot \det J_T=\det I=1.$ Since these determinants are exactly $\partial (u,v)/\partial (x,y)$ and $\partial (x,y)/\partial (u,v)$, their product is $1$. What it means geometrically: if $T$ triples areas, $T^{-1}$ must shrink them to a third — the stretches undo each other.

At $r=0$ the map is not injective: every angle $\theta$ maps to the same point $(0,0)$. The Jacobian vanishing signals the whole $\theta$-line collapsing to one point — the cell is squashed to zero area. The theorem's hypotheses (injective, nonzero Jacobian) fail on this single line/point. But the origin is a set of area zero, so it contributes nothing to the integral; we may simply exclude it. This is the standard "measure-zero exception" — see Double Integrals over General Regions.

Invert: $x=\frac{u+v}{2},y=\frac{u-v}{2}$. Jacobian $J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}1/2& 1/2\\ 1/2& -1/2\end{array}\right)=-\frac{1}{2},|J|=\frac{1}{2}.$ New region: the vertices give $u=x+y\in [0,2]$ and $v=x-y\in [0,2]$ — a clean square $0\le u\le 2,0\le v\le 2$. Integrand $x+y=u$. So ${\iint }_R(x+y)dA={\int }_0^2{\int }_0^{2}u\cdot \frac{1}{2}dvdu=\frac{1}{2}{\int }_0^{2}2udu=\frac{1}{2}\cdot 4=2.$ Answer: $\boxed{2}$.

Invert: adding/subtracting, $y=\frac{u+v}{2},x=\frac{v-u}{2}$. $J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}-1/2& 1/2\\ 1/2& 1/2\end{array}\right)=\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=-\frac{1}{2},|J|=\frac{1}{2}.$ New region: the triangle's edges $x=0,y=0,x+y=1$ become $v=u$ (from $x=0$: $v-u=0$), $v=-u$ (from $y=0$), and $v=1$ (from $x+y=1$). Together: $0\le v\le 1$ and $-v\le u\le v$. Integrand $=e^{u/v}$. ${\iint }_R{e}^{(y-x)/(y+x)}dA={\int }_0^1{\int }_{-v}^v{e}^{u/v}\frac{1}{2}dudv.$ Inner: ${\int }_{-v}^v{e}^{u/v}du=v(e^1-e^{-1})$. Then $\frac{1}{2}(e-e^{-1}){\int }_0^{1}vdv=\frac{1}{2}(e-e^{-1})\cdot \frac{1}{2}=\frac{e-e^{-1}}{4}.$ Numerically $\frac{e-e^{-1}}{4}\approx 0.5875$. Answer: $\boxed{\frac{e-e^{-1}}{4}}$.

Figure — read it like this: the left panel is the original triangle in $xy$ with its slanted hypotenuse $x+y=1$ marked in yellow. The right panel is the transformed region in $uv$-space: it is bounded by the two pink slanted lines $u=v$ and $u=-v$ (from the two legs of the triangle) and the horizontal yellow line $v=1$ (from the hypotenuse). We integrate $u$ from $-v$ to $v$, then $v$ from $0$ to $1$ — read those limits straight off the picture.

Cylindrical: $x=r\cos \theta ,y=r\sin \theta ,z=z$. The 3D Jacobian is $r$ (the $z$-row/column contribute a factor $1$; see Spherical and Cylindrical Coordinates). So $dV=rdrd\theta dz$ and ${\iiint }_{B}zdV={\int }_0^{2\pi }{\int }_0^1{\int }_0^{2}zrdzdrd\theta .$ Integrate in stages: ${\int }_0^{2}zdz=2$; then ${\int }_0^{1}2rdr=1$; then ${\int }_0^{2\pi }1d\theta =2\pi$. Product $=2\pi$. Answer: $\boxed{2\pi }$.

The map is linear and diagonal: $J=\frac{\partial (x,y,z)}{\partial (u,v,w)}=\det \left(\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right)=abc.$ Under this map the ellipsoid becomes the unit ball $u^2+v^2+w^2\le 1$ (volume $\frac{4}{3}\pi$). Therefore $V={\iiint }_{\text{ball}}|abc|dudvdw=abc\cdot \frac{4}{3}\pi =\frac{4}{3}\pi abc.$ Sanity check: setting $a=b=c=R$ gives $\frac{4}{3}\pi R^3$, the sphere. Answer: $\boxed{\frac{4}{3}\pi abc}$.

$J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}{x}_u& x_v\\ y_u& y_v\end{array}\right)=\det \left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)=1\cdot 0-0\cdot 1=0.$ The Jacobian is $0$ everywhere, not just at a point. What it looks like: the two image edge vectors ${\mathbf{r}}_u=(1,1)$ and ${\mathbf{r}}_v=(0,0)$ are — the second is the zero vector, so the "parallelogram" is squashed flat onto the line $y=x$, which has zero area. The map is not injective and the theorem does not apply; the image is a curve of area zero, and any double integral over it is $0$. This is the fully degenerate cousin of the origin-in-polar case (3.3): here it fails on all of space, so there is no rescue.

• L1: the Jacobian is the local area-scaling factor (a number or a function).
• L2: compute it, then $dxdy=|J|dudv$ — bars always.
• L3: forward and inverse Jacobians are reciprocals; zeros on measure-zero sets are harmless.
• L4: transform integrand and region before integrating.
• L5: linear maps give constant $|\det |$; identically-zero Jacobians mean genuine degeneracy — no shortcut survives.