Tagged Questions

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Two ways to evaluate $\int (\Delta u) v d\Omega$, two different results

I would like to evaluate the integral $\int (\Delta u) v d\Omega$, where the domain $\Omega$ is a cylinder. On the boundaries, either the normal derivative $\partial_n u$ is zero or $v$ is zero. An ...
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Evans 's PDE proof

Again, I got stuck. Please help me to understand the following: What is the meaning when you change from integration over the Ball B(x,r) to the surface integration dB(x,s), with another integral ...
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Evaluating a polar double integral on the semi disc

The integral: $$\iint_D (x^2-y^2)\,dx\,dy$$ where $D$ is defined as: $$\{(x,y)\in \mathbb R^2 \mid x^2+y^2\le 1, x\ge 0\}$$ Context I have solved double integrals on quarter discs but this semi ...
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Finding a mistake in the computation of a double integral in polar coordinates

I have to find $P\left(4\left(x-45\right)^2+100\left(y-20\right)^2\leq 2 \right)$ $f(x)$ and $f(y)$ are given, which I will use in my solution below . ...
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Evaluation of the integral of $e^{-(x^2+y^2)}$ over a disk

Show that $$\renewcommand{\intd}{\,\mathrm{d}} \iint_{D(R)} e^{-(x^2+y^2)} \intd x \intd y = \pi \left(1 - e^{-R^2}\right)$$ where $D(R)$ is the disc of radius $R$ with center $(0,0).$ I ...
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Use a double integral in polar coordinates to find the area

So the area is just an intersection of two circles Converting the two circles to polar coordinates, I get: $r(r-2\sin\theta)=0$, and $r(r-2\cos\theta)=0$ Ummm so $r =0$ and r = $2\sin\theta$ ...
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Compute double integral on polar coordinates, find $r(\phi)$

I have the function $f(x,y)=y^2-2x^2y+6x^3-3xy+2y-6x$ and the region $\{y\geq 2x^2-2, y\leq 3x\}$. The region is: To compute the integral in cartesian coordinates: ...
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Integrating $\int_0^2 \int_0^ \sqrt{1-(x-1)^2} \frac{x+y}{x^2+y^2} dy\,dx$ in polar coordinates

I'm having a problem integrating $\displaystyle\int_0^2 \int_0^ \sqrt{1-(x-1)^2} \frac{x+y}{x^2+y^2} \,dy\,dx$. I drew the graph, and it looks like half a circle on top of the $x$ axis. I tried ...
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Integrating $\int_1^2 \int_0^ \sqrt{2x-x^2} \frac{1}{((x^2+y^2)^2} dydx$ in polar coordinates

I'm having a problem converting $\int_1^2 \int_0^ \sqrt{2x-x^2} \frac{1}{(x^2+y^2)^2} dy dx$ to polar coordinates. I drew the graph using my calculator, which looked like half a circle on the x ...
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Tranforming to polar co-ordinates

$$I = \int_0^1\int_0^{\sqrt{1-x^2}} xy \, dy\, dx$$ By transforming to circular polar co-ordinates, evaluate I. How do I do this? Is there a formula/strategy for doing this that works with ...
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Limit is found using polar coordinates but it is not supposed to exist.

Consider the following 2-variable function: $$f(x,y) = \frac{x^2y}{x^4+y^2}$$ I would like to find the limit of this function as $(x,y) \rightarrow (0,0)$. I used polar coordinates instead of ...
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Polar equation — find area under graph using double integral

What is the area of the region in the plane bounded by the curve given in polar coordinates $r = 4 + 2\cos(2\theta)$? Could someone walk me through the conversion of polar coordinates to rectangular ...
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Find the image of a ring

I'm working on the following problem: Find the image of the ring defined by $4 \lt x^2 + y^2 \lt 16$ under the mapping $$F(x,y) = \left(\frac{x}{x^2+y^2} , \frac{y}{x^2+y^2}\right)$$ It looks to ...
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How do I define the limits of a double integral in polar coordinates over an annulus?

Evaluate the double integral by re-writing them in polar coordinates: $\displaystyle\iint\limits_{R}\frac{y^2}{x^2}\ dA$, where $R$ is part of the annulus (ring) $9\leq x^2+y^2\leq 25$ lying ...
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How to integrate over polar coordinates

Evaluate the following double integral by rewriting it in polar coordinates: $\displaystyle\iint\limits_Dxy\,dA$, where $D$ is the disc with center at the origin and radius 5 I have very little ...
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Polar Integration over intersection of two circles

Let $C_0$ denote a circle centered at $(0,0)$ with a radius of $r_0$ and let $C_1$ denote a circle of radius $r_1$ centered at a point $(x_1,0)$. Assume that we are given some function, $\phi(r)$ ...
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Area of $\left( \frac{x^2}{9}+\frac{y^2}{25} \right)^2 \le x^2 + y^2$

I've used the modified polar coordinates: $x = 3r \cos \theta$, $y =5r \sin \theta$, which got me to $$r^2 \le 9 \cos^2 \theta + 25 \sin^2 \theta$$ What now?
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Why is the formula for the area of a cardioid $\int_a^b \frac{1}{2} r^2 d \theta$

I've seen this expression in many places :$\int_a^b \frac{1}{2} r^2 d \theta$ and was wondering if someone can explain where this came from? I've noticed that it's sometimes explained in conjunction ...
$f(x,y)=\langle y- \cos y, x \sin y\rangle$
$f(x,y)=\langle y-\cos y,x\sin y\rangle$ $C$ is the circle $(x-3)^2 + (y+4)^2 = 4$ orientated clockwise. Relevant theorems: Green's theorem (this is under the Green's theorem section of our book). ...