## Converting unit square domain in (x,y) to polar coordinates

I have the following double integral \int_{0}^{1}\int_{0}^{1}\frac{x}{\sqrt{x^2+y^2}}dxdy The integrand is fairly simple: \frac{x}{\sqrt{x^2+y^2}}dxdy=\frac{r\cos(\theta )}{\sqrt{r^2}}rd\theta{}dr=r\cos{(\theta)}d\theta{}dr My trouble is with the limits of integration. I’ve tried: 0 \leq y \leq 1 means 0 \leq{} r\sin(\theta{}) \leq 1 so 0 \leq \theta \leq arcsin(1/r) But why isn’t it just 0<\theta < \pi{}/2 since the unit square is in … Read more

## Evaluating ∫∞−∞e−x2dx\int_{-\infty}^{\infty}e^{-x^2}dx using polar coordinates. [duplicate]

This question already has answers here: Proving ∫∞0e−x2dx=√π2 (19 answers) Closed 6 years ago. I need to express the following improper integral as a double integral of x and y and then, using polar coordinates, evaluate it. I=∫∞−∞e−x2dx Plotting it, we find a Gaussian centered at x=0 which tends to infinity to both sides. We … Read more

## How to reverse the integration order of the double integral ∫2πθ=0∫1+cosθr=0r2(sinθ+cosθ)drdθ\int_{\theta=0}^{2\pi}\int_{r=0}^{1+\cos\theta}r^2(\sin\theta+\cos\theta)drd\theta.

I am given the integral ∬H(x+y)dA where H is the area of the cardioid r=cos(\theta)+1. I have translated the double integral to polar coordinates in order to solve it, where \mathrm{d} A equals r\mathrm{d}r\mathrm{d}\theta, x=r\cos\theta and y=r \sin\theta: \int_{\theta=0}^{2\pi}\int_{r=0}^{1+\cos\theta}r^2(\sin\theta+\cos\theta)\mathrm{d}r\mathrm{d}\theta which equals: \frac{1}{3} \int_{\theta=0}^{2\pi} (\sin\theta + \cos\theta + 3 \sin\theta \cos\theta + 3\cos^2 \theta + 3\sin\theta … Read more

## Expressing double integrals in polar coordinates

I have the double integrals $\int_{x}^\infty \int_0^\infty e^{-\Lambda^a(x^a+y^a)}dxdy$ where $a$ and $\Lambda$ are real positive constants. I would like to express it in polar coordinates. What are my $r$ and $\theta$? Thanks. Answer The given integral equals $$\frac{1}{\Lambda^2}\iint_{0\leq x \leq y}\exp\left[-(x^a+y^a)\right]\,dx\,dy =\frac{4}{a^2\Lambda^2}\int_{0\leq u\leq v} (uv)^{2/a-1}e^{-(u^2+v^2)}\,du\,dv$$ or $$\frac{4}{a^2\Lambda^2}\int_{0}^{+\infty}\int_{\pi/4}^{\pi/2}\rho(\rho^2\sin\theta\cos\theta)^{2/a-1} e^{-\rho^2}\,d\theta\,d\rho$$ or  \frac{2}{a^2\Lambda^2}\,\Gamma\left(\frac{1}{2}+\frac{2}{a}\right)\,2^{-2/a}\int_{0}^{\pi/2}\left(\sin t\right)^{2/a-1}\,dt = … Read more

## Converting bounds of double integral to polar coordinates

I’m trying to convert the following to polar coordinates: ∫∞0∫−x−∞12πe−(x2+y2)/2dxdy After converting to polar coordinates, it should be: ∫∞0∫(7/4)π(3/2)π12πe−r2/2drdθ My question is how do we arrive at the bounds of (3/2)π and (7/4)π. Answer The first integral should be ∫∞0dx∫−x−∞12πe−(x2+y2)/2dy wich represent the integral over the half of the 4 quadrant between y axis and … Read more

## Calculate ∬\iint\frac{dxdy}{(1+x^2+y^2)^2} over a triangle

Calculate \iint\frac{dxdy}{(1+x^2+y^2)^2} over the triangle (0,0), (2,0), (1,\sqrt{3}). So I tried changing to polar coordinates and I know that the angle is between 0 and \frac{\pi}{3} but I couldn’t figure how to set the radius because it depends on the angle. Answer Yes, using polar coordinates is a good idea. We find \iint_T\frac{dxdy}{(1+x^2+y^2)^2}=\int_{\theta=0}^{\pi/3}d\theta\int_{\rho=0}^{f(\theta)}\frac{\rho d\rho}{(1+\rho^2)^2} =-\frac{1}{2}\int_{\theta=0}^{\pi/3}\left[\frac{1}{1+\rho^2 … Read more

## Finding self-intersections on a polar curve

I have a polar curve $r = \frac{2}{\theta}$ (which is a hyperbolic spiral) and I need to find out where it self-intersects. When $\theta$ is restricted to positive values it never intersects, but when it is allowed to be negative it intersects itself infinitely many times. My question is how can we prove that the … Read more

## Polar coordinates confusion

This seems to be very easy, however I cannot understand, where I am mistaking. Here’s the integral to be computed: ∬Dx2+y2dydx with D:=\left\{(x,y)\in \mathbb{R}^2:x \ge0, \; x^2+y^2-2y\leq0\right\} Clearly this integral can be written as \int_0^1\!\!\!\int_0^2x^2+y^2dydx={10\over 3} However when I try to switch to polar coordinates I get a different result. The integral in polar coordinates … Read more

## Write ODE in Polar Coordinates [closed]

Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it’s on-topic for Mathematics Stack Exchange. Closed 7 years ago. Improve this question I want to write this ODE system in polar coordinates (r,θ). ˙x=x−y−x3 ˙y=x+y−y3 Answer We use: x=rcosθ,y=rsinθ,r2=x2+y2,θ=tan−1(yx) From: x2+y2=r2⟹xx′+yy′=rr′ We have: xx′+yy′=x(x−y−x3)+y(x+y−y3)=rcosθ(rcosθ−rsinθ−(rcosθ)3)+rsinθ(rcosθ+rsinθ−(rsinθ)3)=rr′ … Read more

## How do I calculate the area of Bernoulli’s Lemniscate?

can anyone help me calculate this area? I have to use double integrals, and the question sounds like this: ” Calculate the area bounded by the curve (x2+y2)2=a2(x2−y2), where a is a real constant. I have searched online and found that this type of curve is a lemniscate, but I do not know how to … Read more