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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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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
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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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Average distance from a point on a circle to the y-axis.

On the circle $x^2+y^2=9$ we have identically $r\equiv 3$, so in the integral we should consider the length of the circle (which is $2\pi r=6\pi$) and integrate only on $\theta$. Thus, the integral is ...
Julio Puerta's user avatar
  • 7,832

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