I derived the following equality using a few methods that aren't all that complicated. I'm wondering if there is any way to show this is true other than the way I used to derive it.

\[\int _{ -1 }^{ 1 }{ \frac { 2{ x }^{ 3 }+2bx+a+1 }{ \sqrt { 2{ x }^{ 2 }+2\left[ a+b \right] x+{ \left[ a+1 \right] }^{ 2 }+{ \left[ b-1 \right] }^{ 2 } } } dx } =\int _{ 0 }^{ 2 }{ \frac { a+2-y }{ \sqrt { { y }^{ 2 }-2\left[ a+2 \right] y+{ \left[ a+2 \right] }^{ 2 }+{ b }^{ 2 } } } dy } \]

where \(a\) and \(b\) are real numbers.

Something about it makes me think there's a trick that will make them look the same.

If there isn't any other generally simple way to get this, then I might mess around with what I did to see what can come of it.

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TopNewestShow your proof :) – Aman Rajput · 1 year ago

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\(F\left( x,y \right) =\left< \frac { -x }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } ,\frac { -y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } \right> \)

This field is conservative, so any closed path integral should result in zero. I decided to create an arbitrary path starting at \(\left( a,b \right) \) that follows a parabola to \(\left( a+2,b \right) \) then proceeds linearly back to \(\left( a,b \right) \). I set up the following parametric equations for these paths:

\({ r }_{ 1 }\left( t \right) =\left< t+a+1,{ t }^{ 2 }+b-1 \right> \) for \(-1\le t\le 1\)

\({ r }_{ 2 }\left( t \right) =\left< a+2-t,b \right> \) for \(0\le t\le 2\)

The line integrals give us

\(\int _{ { r }_{ 1 } }^{ }{ F\left( x,y \right) dr } \quad +\quad \int _{ { r }_{ 2 } }^{ }{ F\left( x,y \right) dr } \quad =\quad 0\)

\(\int _{ { r }_{ 1 } }^{ }{ F\left( x,y \right) ds } \quad =\quad -\int _{ { r }_{ 2 } }^{ }{ F\left( x,y \right) dr } \)

From here, if I have copied everything down correctly, it should give the claim above. I might elaborate on the last steps later. – Austin Antonacci · 12 months ago

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