Skip to main content

New answers tagged

0 votes

Arc length of a polar curve $r = -8\cos(t)$

It is $\theta=0$ to $\pi$ because the circle $r=-8\cos\theta$ lies entirely in the left half-plane. Remember that $\theta$ measures the angle around the origin, not the circle's centre. In the polar ...
Pustam Raut's user avatar
  • 2,061
1 vote

how to write this region $D$ in relation to $r,\theta$ in this $\iint_Df(x,y)dxdy$ where $D=\{x^2+y^2 \le1,x+y\le 1\}$ and $D=\{x^2+y^2\le1,x+y\ge1\}$

In the case $D=\{x^2+y^2 \le 1,\, x+y\ge 1\}$, you have $0\le r\le1$ and $r(\cos \theta+\sin \theta)\ge 1$. The angle $\theta$ varies between $0$ and $\pi/2$, so the integral becomes: $$\iint_D f(x,y)...
Sine of the Time's user avatar
1 vote

Is there a way to derive the Polar Curve Area Formula using Parametrics?

The mapping of the Cartesian to Polar coordinates is $x=r\cos\theta$, $y=r\sin\theta$. The transformation of areas is $$ \int\int dx dy = \int\int ||\begin{array}{cc} \partial x/ \partial r & \...
R. J. Mathar's user avatar
  • 2,935
0 votes

Laplace equation in polar coordinates in $\mathbb{R}^n$

The Laplace(-Beltrami) operator on the unit sphere $S^{n-1}$ is \begin{align} \Delta_\varphi f=\frac 1{\sqrt{|g|}}\partial_i\left(\sqrt{|g|}\,g^{ij}\, \partial_j\,f\right) \end{align} where $|g|$ is ...
Kurt G.'s user avatar
  • 15.3k
1 vote

Area of a cardioid and a circle

It's not the first half of the circle but the full circle. The origin lies on the circumference, not at the centre. All of the circle lies between the polar arguments $0$ and $\pi.$
Lieven's user avatar
  • 1,953
0 votes

How to modify off-center circle in polar coordinates so that input angle has a linear relationship with angle on circle?

You need to express the polar angle $\theta$ as a function of central angle $\phi$ (see figure below). From the sine rule applied to triangle $OCP$ we get: $$ {r_0\over\sin\theta}={d\over\sin(\phi-\...
Intelligenti pauca's user avatar
0 votes

Volume of shifted and modified cylinder using polar coordinates

The latter volume is the correct one. Take a closer look at the $r$-integral: $$\begin{align*} \int_0^{4\cos\theta} r \sqrt{16 - r^2} \, dr &= 64 \int_0^{\cos\theta} r \sqrt{1 - r^2} \, dr \\ &...
user170231's user avatar
  • 20.2k

Top 50 recent answers are included