Let $R$ a region defined by the interior of the circle $x^2+y^2=1$ and the exterior of the circle $x^2+y^2=2y$ and $x\geq 0$, $y\geq 0$ Using polar coordinates $x=r\cos t$, $y=r\sin t$ to determine the region $D$ in $rt$ plane that corresponds to $R$ under this change of coordinate system (polar coordinates) i.e. since $T(r,t)=(r\cos t, […]

If by definition $r=\sqrt{x^2 + y^2}$, then why do we allow $r$ to be negative? Relatedly, I do not understand the last section of this conversation discussing points being represented by multiple $\theta$: Student: So a single point could have many different values? Mentor: Correct! The values for $r$ can be given as positive and […]

My textbook asks to find the orthogonal trajectory of the family of curves $r = 2Ccos(\theta)$ where C is a parameter. The original equation of the family of curves was $x^2 + y^2 = 2Cx$, but it led to an equation that was as yet unsolvable using the methods taught by the textbook up to […]

I’m getting different answers for this. Many websites say that when you get a negative value of r, you flip the coordinate 180 degrees across the pole. However my teacher says that you cannot have a negative value of r because the function should pick up this point when a value of theta that is […]

I’m trying to find the area between $r=4\sin(\theta)$ and $r=2$. I found the points of intersections to be $\pi/6,5\pi/6$. Which implies the area is $$A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\sin(\theta))^2-2^2d\theta.$$ Is this correct? Or did I find the area for the following region

I’m trying to derive the gradient in polar coordinates using the chain rule. So the idea is that when we have a function $f(x,y)$ and we switch to polar coordinates, we’re really composing $f$ with $P(r,\theta) = (r\cos(\theta),r\sin(\theta))$. So then the gradient of $f$ in polar coordinates should just be $\nabla (f\circ P)(r,\theta)$. From my […]

I understand how it is symmetric about the $y$-axis. because $r(-\theta) = \sin \left(-\frac\theta2\right)=-\sin \left(\frac\theta2\right)=-r(\theta)$ But how is it symmetric about $x$-axis?

I’m having a problem converting $\int\limits_1^2 \int\limits_0^ \sqrt{2x-x^2} \frac{1}{(x^2+y^2)^2} dy dx $ to polar coordinates. I drew the graph using my calculator, which looked like half a circle on the x axis. I know that $\frac{1}{(x^2+y^2)^2} dydx$ turns to $ \frac{r}{(r^2)^2}drd\theta$, which would be $ \frac{1}{r^3}$ The region of integration in $\theta $ is I […]

I need to show that the del operator in 2D polar coordinates is $\nabla=e_r\partial_r+\frac{1}{r}e_r+\frac{1}{r}e_{\phi}\partial_{\phi}$. I try the following approach: $\nabla=\partial_xe_x+\partial_ye_y=\bigg(\frac{\partial r}{\partial x}\partial_r+\frac{\partial \phi}{\partial x}\partial_{\phi}\bigg)\bigg(\cos(\phi)e_r-\sin(\phi)e_{\phi}\bigg)+\bigg(\frac{\partial r}{\partial y}\partial_r+\frac{\partial \phi}{\partial y}\partial_{\phi}\bigg)\bigg(\sin(\phi)e_r+\cos(\phi)e_{\phi}\bigg) $ and then use $r=\sqrt{x^2+y^2}$, $y=\arctan(y/x)$, $\partial_{\phi}e_r=e_{\phi}$ and $\partial_{\phi}e_{\phi}=-e_r$. The problem is, I end up with $\nabla=e_r\partial_r+\frac{1}{r}e_{\phi}\partial_{\phi}$, missing a term compared to the expression above. Any ideas? […]

How do you set up a double integral for an ellipse in polar coordinates without using Jacobian or Greens Theorem? I can’t seem to figure out what (or if) the limits of r can possible be. $x = a\cos(t), y = b\cos(t), ( z = 0)$ $x^2/a + y^2/b = 1$ Thank you.

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