Polar coordinates

Polar coordinates use $x=r\cos \theta$, $y=r\sin \theta$ and area element $rdrd\theta$.

In planar polar coordinates $(r,\theta )$, the polar radius $r\ge 0$ is the Euclidean distance to the origin. It satisfies $x=r\cos \theta$ and $y=r\sin \theta$. For a differentiable plane curve with parameters $r(t),\theta (t)$, the orthogonality of radial and angular directions gives squared speed ${\dot{r}}^2+r^2{\dot{\theta }}^2$.

Suppose $0\le r_1(\theta )\le r_2(\theta )$ and the region is described without overlap by $\alpha \le \theta \le \beta$ and $r_1(\theta )\le r\le r_2(\theta )$. Integrating the polar area element gives its area:For a region bounded by a single radial graph and two rays, take $r_1=0$. If the two bounding graphs cross, split the angular interval at their intersections so that the larger radius is used consistently.

This change of variables replaces radius by its negative logarithm: $r=\sqrt{x^2+y^2}=e^{-s}$. The angular convention in the title measures $t$ from the positive $y$ axis toward the positive $x$ axis. Its Jacobian determinant is $e^{-2s}>0$, so it supplies local coordinates on the punctured plane. For a single-valued function $u$, its expression $v(s,t)=u(x,y)$ is periodic in $t$ with period $2\pi$.

The multivariable chain rule givesand, for a twice continuously differentiable function,The first-order correction comes from differentiating the coefficients $x,y$ in the angular operator. Consequently the partial differential equation with the right-hand expression equal to zero becomes the heat equation $v_s=v_{tt}$, with the logarithmic radial coordinate acting as the evolution variable.

For polar coordinates $x=r\cos \theta$, $y=r\sin \theta$ with $r\ge 0$, the Jacobian determinant is $r$. The change of variables formula therefore gives the area element $dA=rdrd\theta$. Integrating over the radial interval $r_1(\theta )\le r\le r_2(\theta )$ gives $[r_2(\theta {)}^2-r_1(\theta {)}^2]/2$ per unit angle. Bounds must describe the intended region without counting it more than once.

The cylindrical coordinate system extends planar polar coordinates by an axial coordinate: $(x,y,z)=(R\cos \varphi ,R\sin \varphi ,z)$.

For $x=r\cos \varphi$, $y=r\sin \varphi$ and unchanged $z$, the Jacobian determinant of Cartesian coordinates with respect to cylindrical coordinates is $r$. Hence a volume integral acquires the displayed factor for $r\ge 0$. The inverse determinant is $1/r$ where $r>0$; the axis is a coordinate degeneracy.

The azimuthal angle $\varphi$ in cylindrical coordinates is measured in the horizontal plane from a chosen reference axis. In the usual mathematical convention $x=s\cos \varphi$, $y=s\sin \varphi$, so positive $\varphi$ runs counterclockwise from positive $x$ when viewed from positive $z$. This same angular coordinate occurs in spherical coordinates. It is defined modulo $2\pi$ away from the polar axis and is nonunique on the axis.

The local orthonormal basis associated with cylindrical coordinates: ${\mathbf{e}}_r=(\cos \theta ,\sin \theta ,0)$, ${\mathbf{e}}_{\theta }=(-\sin \theta ,\cos \theta ,0)$, and ${\mathbf{e}}_z=(0,0,1)$. The first two basis vectors depend on the angular coordinate and are undefined on the axis without a choice of angle.

The azimuthal unit vector is tangent to increasing azimuthal angle at fixed cylindrical radius and height. For $r>0$ it equals $(-y/r,x/r,0)$. A counterclockwise circle has tangent displacement $d\mathbf{r}=r\hat{\varphi }d\varphi$. Unlike a fixed Cartesian basis vector, its direction depends on position.

For a differentiable scalar field and cylindrical coordinates with $r>0$, the gradient has the displayed components in the local orthonormal basis. The chain rule applied to $r=\sqrt{x^2+y^2}$ and a local angle $\theta$ gives $r_x=\cos \theta$, $r_y=\sin \theta$, ${\theta }_x=-\sin \theta /r$, and ${\theta }_y=\cos \theta /r$. Substitution in $\nabla \varphi ={\varphi }_x{\mathbf{e}}_x+{\varphi }_y{\mathbf{e}}_y+{\varphi }_z{\mathbf{e}}_z$ proves the formula. The angular factor $1/r$ appears because an angular change $d\theta$ sweeps arc length $rd\theta$.

Differentiate the position vector $\mathbf{r}=r{\mathbf{e}}_r+z{\mathbf{e}}_z$ using ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$, ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$, and fixed ${\mathbf{e}}_z$. The first derivative is the velocity $\dot{r}{\mathbf{e}}_r+r\dot{\theta }{\mathbf{e}}_{\theta }+\dot{z}{\mathbf{e}}_z$; differentiating once more gives the displayed acceleration. In particular, $r$ times the azimuthal acceleration is $d(r^2\dot{\theta })/dt$.

If the cylindrical components of a vector field have dependence $e^{im\varphi }$, differentiation of the rotating basis adds the displayed cross product term. Scalar-component differentiation alone misses this contribution. For $\mathbf{B}=(J/2)s\hat{\mathbf{\varphi }}$, one also has $(\mathbf{v}\cdot \nabla )\mathbf{B}=(J/2)\hat{\mathbf{z}}\times \mathbf{v}$. These identities are useful in cylindrical magnetohydrodynamic waves.

The cylindrical radius $R=\sqrt{x^2+y^2}$ is the perpendicular distance from the symmetry axis.

A cardioid is an epicycloid with one cusp; a standard polar equation is $r=a(1+\cos \theta )$.

For $a>0$, the cardioid $r=a(1+\cos \theta )$ traces its boundary once over $0\le \theta \le 2\pi$. The polar area element givesThe polar arc length integrand is $2a|\cos (\theta /2)|$. Splitting at the cusp parameter $\theta =\pi$ gives perimeter $8a$. The powers of $a$ distinguish the scaling of area and length.

For $a>0$ and $0<c<2a$, consider the region $c\le r\le a(1+\cos \varphi )$ in polar coordinates. It exists precisely for $|\varphi |\le \beta$, where $\beta =\arccos (c/a-1)\in (0,\pi )$. The polar area element $rdrd\varphi$ givesExpanding the square and integrating its even terms gives the title formula. This cap lies inside the cardioid and outside the concentric circle, so reversing the radial bounds would select the wrong region or a signed area.