Exercises — Double integrals in polar coordinates — Jacobian r

$R$ is a disk of radius $3$. Every point in a full disk has $r$ from $0$ to $3$ and $\theta$ sweeping a full turn $0$ to $2\pi$. Attach the Jacobian $r$ to the constant integrand $1$: ${\int }_0^{2\pi }{\int }_0^{3}1\cdot rdrd\theta .$ That is the whole task — recognising that $dA$ becomes $rdrd\theta$.

The factor is ==$r$==. What "Jacobian" means here. When you change variables, the little area box changes size. The Jacobian matrix $J$ is the grid of all four partial derivatives that records how $x$ and $y$ respond to a nudge in $r$ and a nudge in $\theta$: $J=\left(\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }\end{array}\right)=\left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right).$ Its determinant $\det J$ measures the area-scaling factor of that box (this is the Jacobian Determinant, proved in the parent): $\det J=\cos \theta (r\cos \theta )-(-r\sin \theta )(\sin \theta )=r{\cos }^2\theta +r{\sin }^2\theta =r.$ The change-of-variables theorem uses its absolute value $|\det J|=r$. Since $r$ is a distance it satisfies $r\ge 0$, so $|r|=r$. The factor is never negative because distance is never negative.

Since $x^2+y^2=(r\cos \theta {)}^2+(r\sin \theta {)}^2=r^2({\cos }^2\theta +{\sin }^2\theta )=r^2$: $x^2+y^2=r^2,\sqrt{x^2+y^2}=r.$ This is the single most-used simplification in polar integrals: any expression built from $x^2+y^2$ collapses to a function of $r$ alone.

Inner integral in $r$: ${\int }_0^{3}rdr={\left[\frac{r^2}{2}\right]}_0^3=\frac{9}{2}.$ Outer integral in $\theta$ (the inner result is a constant, so just multiply by the sweep): ${\int }_0^{2\pi }\frac{9}{2}d\theta =\frac{9}{2}\cdot 2\pi =9\pi .$ Check: this is the area of a disk of radius $3$, and $\pi r^2=\pi \cdot 3^2=9\pi$. ✓

Integrand $x^2+y^2\to r^2$; disk radius $2$ means $r:0\to 2$, $\theta :0\to 2\pi$; attach $r$: ${\int }_0^{2\pi }{\int }_0^2{r}^2\cdot rdrd\theta ={\int }_0^{2\pi }{\int }_0^2{r}^{3}drd\theta .$ Inner: ${\int }_0^2{r}^{3}dr={\left[\frac{r^4}{4}\right]}_0^2=\frac{16}{4}=4.$ Outer: ${\int }_0^{2\pi }4d\theta =4\cdot 2\pi =8\pi .$

Convert the integrand fully (this is L1's second trap in reverse): $x\to r\cos \theta$. Upper half-disk means $\theta :0\to \pi$, radius $1$ so $r:0\to 1$. Attach $r$: ${\int }_0^{\pi }{\int }_0^1(r\cos \theta )rdrd\theta ={\int }_0^{\pi }\cos \theta d\theta {\int }_0^1{r}^{2}dr.$ Split because integrand factors into an $r$-part and a $\theta$-part. ${\int }_0^1{r}^{2}dr=\frac{1}{3},{\int }_0^{\pi }\cos \theta d\theta =[\sin \theta {]}_0^{\pi }=0-0=0.$ Answer $=\frac{1}{3}\cdot 0=0$. By symmetry this is expected: the upper half-disk is symmetric about the $y$-axis, so positive $x$ cancels negative $x$.

Understand the boundary first. Multiply $r=2\cos \theta$ by $r$: $r^2=2r\cos \theta$, i.e. $x^2+y^2=2x$, i.e. $(x-1{)}^2+y^2=1$ — a circle of radius $1$ centred at $(1,0)$. The figure below draws this circle together with several rays from the origin; each ray is one value of $\theta$, and $r$ runs from the origin out to the lavender boundary. Angle range: this circle sits in the right half-plane and touches the origin. As $\theta$ swings from $-\pi /2$ to $+\pi /2$, the ray from the origin sweeps across the whole disk (watch the dashed coral rays fan out in the figure); outside that range $2\cos \theta <0$ and there is no region. So $\theta :-\frac{\pi }{2}\to \frac{\pi }{2}$. Radial range: along each ray, $r$ runs from the origin $0$ out to the boundary $2\cos \theta$. This is the key analysis step — the outer $r$-limit is a function of $\theta$. ${\int }_{-\pi /2}^{\pi /2}{\int }_0^{2\cos \theta }rdrd\theta ={\int }_{-\pi /2}^{\pi /2}\frac{(2\cos \theta {)}^2}{2}d\theta ={\int }_{-\pi /2}^{\pi /2}2{\cos }^2\theta d\theta .$ Why the half-angle identity? We are stuck with ${\cos }^2\theta$, which has no elementary antiderivative in that raw squared form. The identity ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$ rewrites the square as a sum of things we can integrate term-by-term: the constant $\frac{1}{2}$ and the plain cosine $\frac{1}{2}\cos 2\theta$. (It comes straight from the double-angle formula $\cos 2\theta =2{\cos }^2\theta -1$; solving that for ${\cos }^2\theta$ gives exactly the line above.) So: ${\int }_{-\pi /2}^{\pi /2}2{\cos }^2\theta d\theta ={\int }_{-\pi /2}^{\pi /2}(1+\cos 2\theta )d\theta ={\left[\theta +\frac{1}{2}\sin 2\theta \right]}_{-\pi /2}^{\pi /2}=\left(\frac{\pi }{2}+0\right)-\left(-\frac{\pi }{2}+0\right)=\pi .$ Check: a circle of radius $1$ has area $\pi \cdot 1^2=\pi$. ✓ (See Area of Regions Bounded by Polar Curves.)

Integrand $y\to r\sin \theta$; limits are constant here (a genuine quarter ring). Attach $r$: ${\int }_0^{\pi /2}{\int }_1^2(r\sin \theta )rdrd\theta ={\int }_0^{\pi /2}\sin \theta d\theta {\int }_1^2{r}^{2}dr.$ ${\int }_1^2{r}^{2}dr={\left[\frac{r^3}{3}\right]}_1^2=\frac{8-1}{3}=\frac{7}{3},{\int }_0^{\pi /2}\sin \theta d\theta =[-\cos \theta {]}_0^{\pi /2}=0-(-1)=1.$ Answer $=\frac{7}{3}\cdot 1=\frac{7}{3}$.

Whole plane: $r:0\to \infty$, $\theta :0\to 2\pi$. Integrand $e^{-(x^2+y^2)}\to e^{-r^2}$; attach $r$: ${\int }_0^{2\pi }{\int }_0^{\infty }{e}^{-r^2}rdrd\theta .$ Why the $r$ is the hero: it is exactly the derivative-piece needed for $u=r^2,du=2rdr$: ${\int }_0^{\infty }{e}^{-r^2}rdr=\frac{1}{2}{\int }_0^{\infty }{e}^{-u}du=\frac{1}{2}{\left[-e^{-u}\right]}_0^{\infty }=\frac{1}{2}(0-(-1))=\frac{1}{2}.$ Then ${\int }_0^{2\pi }\frac{1}{2}d\theta =\frac{1}{2}\cdot 2\pi =\pi .$ Now connect to the 1-D integral. Let $I={\int }_{-\infty }^{\infty }{e}^{-x^2}dx$. Because $e^{-(x^2+y^2)}=e^{-x^2}{e}^{-y^2}$ and the double integral factors, $I^2={\int }_{-\infty }^{\infty }{e}^{-x^2}dx{\int }_{-\infty }^{\infty }{e}^{-y^2}dy=\pi \Rightarrow I=\sqrt{\pi }.$ This is the Gaussian Integral — impossible in closed form without the polar $r$.

Integrand $\frac{1}{\sqrt{x^2+y^2}}=\frac{1}{r}$. Watch what happens with the Jacobian: the $r$ from $dA$ cancels the $\frac{1}{r}$: ${\int }_0^{2\pi }{\int }_1^4\frac{1}{r}\cdot rdrd\theta ={\int }_0^{2\pi }{\int }_1^{4}1drd\theta .$ Inner: ${\int }_1^{4}dr=3$. Outer: ${\int }_0^{2\pi }3d\theta =6\pi$. Note on the annulus, not a disk: we started at $r=1$, avoiding $r=0$. Had the region been a full disk down to the origin, the integrand $\frac{1}{r}$ blows up there and we would need to check convergence of an improper integral. Here the inner radius $1$ keeps everything finite.

$z\ge 0$ means $4-x^2-y^2\ge 0$, i.e. $x^2+y^2\le 4$ — a disk of radius $2$. Volume $={\iint }_{R}zdA$. Integrand $z=4-x^2-y^2\to 4-r^2$; disk gives $r:0\to 2$, $\theta :0\to 2\pi$; attach $r$: $V={\int }_0^{2\pi }{\int }_0^2(4-r^2)rdrd\theta ={\int }_0^{2\pi }{\int }_0^2(4r-r^3)drd\theta .$ Inner: ${\int }_0^2(4r-r^3)dr={\left[2r^2-\frac{r^4}{4}\right]}_0^2=(8-4)-0=4.$ Outer: ${\int }_0^{2\pi }4d\theta =8\pi .$ So $V=8\pi$. (Sanity: it should be positive and less than the cylinder $\pi \cdot 2^2\cdot 4=16\pi$ — indeed $8\pi <16\pi$. ✓)

Region: the cardioid $r=1+\cos \theta$ is traced once as $\theta :0\to 2\pi$ (at $\theta =0$, $r=2$; at $\theta =\pi$, $r=0$ — the cusp at the origin). For each $\theta$, $r$ runs $0\to 1+\cos \theta$. Integrand $x^2+y^2\to r^2$; attach $r$ to get $r^3$: ${\int }_0^{2\pi }{\int }_0^{1+\cos \theta }{r}^{3}drd\theta ={\int }_0^{2\pi }\frac{(1+\cos \theta {)}^4}{4}d\theta .$ Expand $(1+\cos \theta {)}^4=1+4\cos \theta +6{\cos }^2\theta +4{\cos }^3\theta +{\cos }^4\theta$. Over $[0,2\pi ]$ the standard averages give: ${\int }_0^{2\pi }1d\theta =2\pi ,{\int }_0^{2\pi }\cos \theta d\theta =0,{\int }_0^{2\pi }{\cos }^2\theta d\theta =\pi ,$ ${\int }_0^{2\pi }{\cos }^3\theta d\theta =0,{\int }_0^{2\pi }{\cos }^4\theta d\theta =\frac{3\pi }{4}.$ Sum inside: $2\pi +0+6\pi +0+\frac{3\pi }{4}=8\pi +\frac{3\pi }{4}=\frac{35\pi }{4}$. Divide by $4$: $\frac{1}{4}\cdot \frac{35\pi }{4}=\frac{35\pi }{16}$.

First quadrant quarter disk: $r:0\to a$, $\theta :0\to \pi /2$. Convert $xy\to (r\cos \theta )(r\sin \theta )=r^2\cos \theta \sin \theta$; attach $r$: ${\int }_0^{\pi /2}{\int }_0^a{r}^2\cos \theta \sin \theta \cdot rdrd\theta ={\int }_0^{\pi /2}\cos \theta \sin \theta d\theta {\int }_0^a{r}^{3}dr.$ ${\int }_0^a{r}^{3}dr=\frac{a^4}{4},{\int }_0^{\pi /2}\cos \theta \sin \theta d\theta ={\left[\frac{{\sin }^2\theta }{2}\right]}_0^{\pi /2}=\frac{1}{2}.$ Answer $=\frac{a^4}{4}\cdot \frac{1}{2}=\frac{a^4}{8}.$ Interpretation: unlike L2.3, here nothing cancels by symmetry because the whole region has $x,y\ge 0$, so $xy\ge 0$ everywhere and the integral is genuinely positive.

Every problem here is the same three moves. Move 1 ::: Convert the integrand: replace $x^2+y^2\to r^2$, $x\to r\cos \theta$, $y\to r\sin \theta$ — no $x,y$ may survive. Move 2 ::: Replace $dxdy$ by $rdrd\theta$ (never drop the $r$). Move 3 ::: Read limits from the geometry — constant for full disks/annuli, $\theta$-dependent for curves like $r=2\cos \theta$ or the cardioid.