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Here $\Lambda$ is planar Lebesgue measure. For a Borel set $B\subseteq [0,\infty )$, polar coordinates giveThe pushforward is therefore locally finite and atomless. By the mapping theorem for Poisson point processes, $f(\Pi )$ is a homogeneous Poisson process on $[0,\infty )$ with constant intensity $\pi$. This is the two-dimensional radial volume transform of a homogeneous Poisson point process.

Put $S_k=R_k^2$. The variable $S_k$ is the $k$th arrival time of that rate-$\pi$ Poisson process, hence has the gamma distribution with shape $k$ and rate $\pi$:Since $s=r^2$ and $ds/dr=2r$, the change-of-variables formula for a probability density gives the Kth-nearest-neighbour distance in a homogeneous Poisson point process

In polar coordinates,With the change of radial coordinate $u=\log r$, the detection functional becomesIt is therefore $\lambda$ times ordinary path length in the $(u,\theta )$-plane. The shortest path is the line segment from $(\log A,0)$ to $(\log B,\alpha )$, soReturning to $r$ gives the logarithmic spiralIts minimum detection probability isThe functional depends only on the geometric path and not its parametrization, so tiptoeing and running give the same probability.

For the improved sensor, omit the irrelevant positive factor $\lambda$ and use the LagrangianThe coordinate $\theta$ is cyclic, so its conjugate momentum is conserved. Equivalently,is constant. Because $L$ has no explicit time dependence, the associated conservation of energy gives the constantMultiplying by $r$ and using $h=r\dot{\theta }$ gives the required equation

Uniform area density in the unit disk is $1/\pi$. The polar coordinates Jacobian gives the joint densityThus the coordinates are independent and the uniform random point in a disk has

For independent points $A,B$, the area of triangle $OAB$ isNowIndependence therefore givesConditioned on $A,B$, the probability that $C$ lies inside $OAB$ is its area divided by the disk area $\pi$. Hence

Four points in general position fail to form a convex quadrilateral exactly when one lies inside the triangle formed by the other three. Each of $A,B,C$ lies inside the triangle formed by $O$ and the other two with probability $4/(9{\pi }^2)$, and these three events are disjoint. The remaining possibility is that $O$ lies inside $ABC$. By origin in a triangle of three radial random points, this has probability $1/4$. Therefore

LetThenUsing polar coordinates on the first quadrant,Since $I>0$, the Gaussian integral is

In polar coordinates, the circle is $r=2$, the line $y=\sqrt{3}x$ is $\theta =\pi /3$, and $y=1$ is $r\sin \theta =1$. The inequalities selectThe surface height isThus the required volume is

The method of images replaces the earthed plane by an image charge $-q$ at $(-a,0,0)$. The two Coulomb potentials cancel at $x=0$, so uniqueness for the Dirichlet problem makes the resulting field the physical field in $x>0$. With $\mathbf{r}=(x,y,z)$ and ${\mathbf{e}}_x=(1,0,0)$,and

The electric multipole expansion from part (a) gives, for $r\gg a$,Thus the leading field is that of an electric dipole.

On the plane,Taking the plane normal to be $+{\mathbf{e}}_x$ and using polar coordinates,With the outward normal of the region $x>0$, the sign is reversed. This is consistent with Gauss's law: all electric flux from the real charge terminates on the grounded conductor.

The electrostatic boundary condition gives the induced surface charge densityand its integral isIn the plane $z=0$, the field lines leave the positive charge, meet the conductor normally, and are the right-half-plane portions of the field lines joining the real charge to its negative image.

In polar coordinates, setaway from the origin, assigning any value at the origin. This is unbounded as $r=\sqrt{x^2+y^2}\downarrow 0$. It belongs to $L^2(U)$ becauseMoreoverand henceThus $u\in H^1(U)$ but $u\notin L^{\infty }(U)$, exhibiting the failure of first-order Sobolev embedding into Linfinity in two dimensions.

Take the punctured complex planeandIn polar coordinates, $u(r,\theta )=\log r$, soThus $u$ is harmonic on $D$.

If $u=\mathrm{Re}f$ for a holomorphic function $f$, the calculation from part (a) would giveThe integral of a derivative around a closed curve is zero, whereas the residue theorem givesThis contradiction proves that no such $f$ exists. It is precisely the log modulus has no global harmonic conjugate on the punctured plane obstruction.

Since $z>0$, the cone has the parametrizationIts coordinate derivatives satisfyso its first fundamental form is

On a sufficiently narrow angular patch, defineand map the cone to the plane point with polar coordinates $(R,\Theta )$. The Euclidean metric pulls back asThe map is therefore a local isometry from a circular cone to the plane, proving that $\Sigma$ is locally isometric to the Euclidean plane.

Use the conventionThe Riemann-Lebesgue lemma states that if $f\in L^1({\mathbb{R}}^n)$, then $\hat{f}$ is continuous andContinuity follows from the dominated convergence theorem, since $e^{-ix\cdot {\xi }_j}\to e^{-ix\cdot \xi }$ and the integrands are dominated by $|f(x)|$.

For the decay, first take $g\in C_c^1({\mathbb{R}}^n)$. Choose a coordinate $j$ for which $|{\xi }_j|\ge |\xi |/\sqrt{n}$. Integration by parts givesand henceSince $C_c^1({\mathbb{R}}^n)$ is dense in $L^1({\mathbb{R}}^n)$, choose $g$ with $\Vert f-g{\Vert }_1<\epsilon$. The elementary Fourier bound givesso the decay for $g$ implies the decay for $f$.

The Parseval identity says that for $f,g\in L^2({\mathbb{R}}^n)$, with their Fourier transforms defined by the Plancherel theorem,In particular,

For the given radial function, as $|x|\to 0$,while as $|x|\to \infty$,Using polar coordinates, local integrability of $|f{|}^q$ is therefore determined byand integrability at infinity byThe assumptions $2a>-n$ and $b>n$ imply both $f\in L^2({\mathbb{R}}^n)$ and $f\in L^1({\mathbb{R}}^n)$. The Riemann--Lebesgue lemma and the direct estimate $|\hat{f}|\le \Vert f{\Vert }_1$ give $\hat{f}\in L^{\infty }$, while Parseval gives $\hat{f}\in L^2$. Finally, for every $2\le p<\infty$,Thus

The classical canonical partition function uses one phase-space cell $h^2$ per state. In polar coordinates,Equivalently, using $h=2\pi \hslash$,This is the particle in a finite two-dimensional harmonic trap partition function.

In the flat-spacetime limit $M=0$, the orbit equation is the harmonic oscillator equationIts general solution is $y=A\cos \varphi +B\sin \varphi$. The conditions $r\to \infty$, hence $y\to 0$, at $\varphi =0$ and $\varphi =\pi$ set $A=0$. Choosing $r=b$ at closest approach $\varphi =\pi /2$ fixes $B=1/b$, soThis is a straight line at perpendicular distance $b$ from the origin in polar coordinates.

Plane polar coordinates describe a point by its radial distance $r$ and polar angle $\theta$.

The distance from a focus to a conic section along a line perpendicular to its symmetry axis. In focus-centred polar coordinates, an ellipse may be written $r(1-e\cos \theta )=\ell$, where $e$ is its eccentricity of a conic.

The planar system $\dot{x}=y+(1-x^2-y^2)x$, $\dot{y}=-x+(1-x^2-y^2)y$ has polar coordinates $\dot{r}=r(1-r^2)$, $\dot{\theta }=-1$. Every nonzero orbit approaches the clockwise unit-circle limit cycle, of period $2\pi$. For $r(t_0)=r_0>0$,Interior orbits approach the origin backward in time, but exterior orbits have backward finite-time blow-up of an ordinary differential equation at $t=t_0+\frac{1}{2}\log (1-r_0^{-2})$. Thus forward convergence to a limit cycle does not imply existence for all negative times.