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8 votes
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Area of loop in graph

First determine where the curve self-intersects. Remember that in polar coordinates, $r$ is an indication of distance from the origin. From the plot, it's evident that the origin is where the self-...
user170231's user avatar
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6 votes

Area of loop in graph

Another way of calculation. Changing to cartesian coordinates we have $x^3+y^3=3xy$ which gives a nodal cubic symmetric respect to the diagonal $y=x$. In order of calculate easier the integral we ...
Piquito's user avatar
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6 votes
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The length of the curve

Let $t:=\operatorname{arsinh}(r/\sqrt2)$ ($0\le t\le\operatorname{arsinh}\sqrt2$). Then, $$r=\sqrt2\sinh t,\quad r'_t=\sqrt2\cosh t,$$$$\theta=\sinh t\cosh t+t+\ln\sqrt2,\quad\theta'_t=2\cosh^2t,$$ ...
Anne Bauval's user avatar
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3 votes
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Epsilon-Delta proof for Multivariate Limit in Polar Coordinates

Your argument is wrong, because you made an $|r|$ appear out of thin air. Without it, your expression does not go to zero. Using polar coordinates here is overkill. What you should note is that the ...
Martin Argerami's user avatar
3 votes

Finding real and imaginary part with polar coordinates

The angle is in the second quadrant because real part is negative and imaginary part is positive, so can be represented by a complex number $$ r= e^ {i \frac{2 \pi}{3}}$$ $$ r^3 =(e^ {i \frac{2 \pi}{3}...
Narasimham's user avatar
3 votes

Area inside the circle $x^2+y^2=4$, under $y=x\sqrt{3}$ and above $y=1$

$y=\sqrt3 x\to\theta=\tfrac\pi 3$,$(\sqrt3,1)\to\theta=\tfrac\pi 6$. The given region is spanned by $\tfrac\pi 6\leq\theta\leq\tfrac\pi 3$. As you determined, the outer radius of the region is $r_1=2$,...
Bob Dobbs's user avatar
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3 votes

Area inside the circle $x^2+y^2=4$, under $y=x\sqrt{3}$ and above $y=1$

$$\begin{align}A&=\iint_{1\le y\le\sqrt3\atop y/\sqrt3\le x\le\sqrt{4-y^2}}dxdy\\&=\int_1^{\sqrt3}\left(\sqrt{4-y^2}-\frac y{\sqrt3}\right)dy\\&=\int_{\pi/6}^{\pi/3}\sqrt{4-(2\sin t)^2}\;2\...
Anne Bauval's user avatar
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3 votes

Need hint: Prove that $r = a (\sin t) + b (\cos t)$ is a circle, where $ab \neq 0$

It's been 11 years, I think it's safe to say it doesn't matter anymore if the entire solution is given. I've managed to get through the problem, so if it can help anyone looking for a fuller ...
Emmannuelle_Legolas's user avatar
3 votes
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Show that $y=\frac{1}{x}$ is a hyperbola

The hyperbola $xy=1$ is rotated at an angle of $45^∘=\pi/4$ relative to the $x$-axis. We can use a transformation to rotate our coordinate system by $45^∘$ to find the equation of this hyperbola ...
Sebastiano's user avatar
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3 votes
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Area between a circle and an spiral.

Back to basics: The area in polar coordinates is given by $$ A=\frac{1}{2}\int r^2d\theta $$ So, here we have $$ \begin{align} A&=\frac{1}{2}\int_0^4 (r^2-\theta^2)d\theta\\ &=\frac{1}{2}\...
Cye Waldman's user avatar
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3 votes
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Is knowing unit basis vector enough to specify a coordinate system?

A vector field is a coordinate basis (i.e. defines a coordinate system) if and only if their pairwise Lie brackets are zero. Intuitively, the vanishing of the Lie brackets means that the coordinate ...
Vincent Thacker's user avatar
3 votes

Find $x$ such that $\cosh(a + bx) + 1 = cx$

Substitute $ax+b=w$ to simplify the equation: $\cosh(ax+b)+1=cx\iff \cosh(w)=\frac cbw-\frac{ac}b-1=uw+v$ We use Lagrange reversion to get a real root: $$\cosh(w)-uw=v\implies w=-\frac vu+\sum_{n=1}^\...
Тyma Gaidash's user avatar
3 votes
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Solve Double Integral with Polar System

The area is bounded by a circle around $(1/2,1/2)$ with radius $1/\sqrt{2}\,:$ $$ 0=x^2-x+y^2-y=(x-\tfrac12)^2+(y-\tfrac12)^2-\tfrac12\,. $$ Expressing the integral in $(r,\theta)$ has to take into ...
Kurt G.'s user avatar
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3 votes
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Can't figure out multivariable limit of $\frac{x^3-x^2y}{x^2+y^6}$ with polar coordinate sub.

$$L =\lim_{(x, y)\to(0, 0)}\frac{x^3-x^2y}{x^2+y^6} = \lim_{(x, y)\to(0, 0)}\frac{x^2}{x^2+y^6}(x-y)$$ Notice $\displaystyle\lim_{(x, y)\to(0, 0)} x-y = 0$ and $\displaystyle\left|\frac{x^2}{x^2+y^6}\...
Alma Arjuna's user avatar
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3 votes
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Clarifications on the solution of a double integral: $\iint_X\frac{x^2y}{x^2+y^2}dxdy$

You can transform the conditions to polar coordinates as well. From the first condition, you get $r^2\cos^2\vartheta+r^2\sin^2\vartheta\geq1$, which reduces to $r\geq1$. From the second condition, you ...
Hume2's user avatar
  • 2,709
2 votes

Bézier approximation of archimedes spiral?

Bézier cubic has eight parameters, four for x function, four for y function. Impose on the spiral segment that: Endpoints go through correct places (four constraints); Tangent at endpoints have ...
jdaw1's user avatar
  • 193
2 votes
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Finding real and imaginary part with polar coordinates

There are two issues: Firstly, $\arctan (-\sqrt{3}) = -\frac{\pi}{3}$, not $-\frac{\pi}{6}$ (perhaps you computed $\arctan\left(-\frac{1}{\sqrt{3}}\right)$ instead). Secondly, $\arctan\frac{y}{x}$ ...
Travis Willse's user avatar
2 votes

Non homogeneous Heat equation in polar coordinates with non homogeneous BC's

You are doing nothing wrong. Write $u(r,t)=w(r,t)+v(r)$ and choose $v(r)$ as the solution to the ODE $$ a\frac{1}{r}\frac{d}{dr}\left(r\frac{dv}{dr}\right)+be^{-y(r-a)}=0 \tag{1} $$ with boundary ...
Gonçalo's user avatar
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2 votes
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Graphing $r^3\cos{3\theta}-1=0$ in the $xy$ plane.

First, let us consider searching for when asymptotes occur, that is, when $r \rightarrow \infty$. Rearranging, we see this is equivalent to when $\frac{1}{\cos(3\theta)} \rightarrow \infty$. This ...
invariant's user avatar
2 votes
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Which is the area of intersection, using polar coordinate

First of all, it is important to have an accurate drawing (using free software like Geogebra or Desmos) : Then, as you would do for the difference of two areas between curves having a cartesian ...
Jean Marie's user avatar
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2 votes
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How to get to $dr=\frac{x}{r}dx+\frac{y}{r}dy$ and $d\theta=\frac{-y}{r^2}dx+\frac{x}{r^2}dy$

One writes $r^2=x^2+y^2$ so that $2 r dr= 2x dx+2y dy$, $\theta = \arctan ({y\over x})+C$. As $d \arctan u= {du\over 1+u^2}$, $d \theta = ({dy\over x}-{y d x\over x^2}).{1\over 1+{y^2\over x^2}}$ =${x ...
Thomas's user avatar
  • 7,615
2 votes

Polar coordinates limits multivariable calculus

Maybe this counterexample can help you: Consider the limit $$\lim_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}$$ Then you might think that switching to polar coordinates it results in $$\tag{*}\lim_{r\to 0}\...
b00n heT's user avatar
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2 votes
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Average over sphere identity

Let us choose a coordinate system with the $z$-axis along the direction of $y\in\mathbb{R}^3$ such that $\omega = (\cos \phi \sin\theta, \sin\phi \sin\theta,\cos\theta)$ and $y= (0,0, |y|)$. Then we ...
Fabian's user avatar
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2 votes

Circle rolling between two functions

The fact that your curves are given in polar coordinates is less important than the fact that you need curves having a good representation in terms of curvilinear abscissa. There are very few of them, ...
Jean Marie's user avatar
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2 votes

double integral convert from cartesian to polar

You've done the transformation correctly: the resulting integral is $$I = \int_{\theta=-\pi/4}^{\pi/3} \cos \theta \, d\theta \int_{r=1}^2 r^2 \sin r^2 \, dr.$$ Although the integral with respect to $...
heropup's user avatar
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2 votes
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how do I find the intersection points of two curves?

COMMENT AS A HINT.-Calculation of some points in the figure of circle and lemniscata is easy and you know the polar coordinates, say $r=f_1(\theta)$ and $r=f_2(\theta)$ for each curve. A way you can ...
Piquito's user avatar
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2 votes
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Intersection of two hyperbolas in polar coordinates.

Indeed, there is an apparent paradox as can be seen on the following example (similar to yours) with two hyperbolas (featured with their asymptotes) : $$r=\frac{p_0}{1 - e_0 \cos \varphi} \ \text{and}...
Jean Marie's user avatar
  • 83.7k
2 votes

Error in graphing the polar equation $r=-1+\cos(\theta)$. I get a different answer than the book for cos(30degrees). Am I wrong?

It's wrong. Here is why. You can directly put the value and check. It's wrong for $90°$ part as well. For a more general approach:- $f(x)=-1+\cosθ$ has maximum value of $0$ at $θ=0°$ and minimum ...
Gwen's user avatar
  • 2,711
2 votes

Solving Laplace's equation on semi-annular domain

The solution form you have written is for a disk not an annulus or half-annulus Separation of variables can be used in your half-annulus too, and the form you will get is $$A + B\ln r + \sum_{n=1}^{\...
Zarrax's user avatar
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1 vote
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Can I see an example of transforming a tensor from polar to cartesian coordinates?

Is seems like you are familiar with the transformations of basis one forms $dr,d\theta$ to $dx,dy\,.$ The similar formulas for the basis vector fields $\partial_r,\partial_\theta,\partial_x,\partial_y$...
Kurt G.'s user avatar
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