Challenge: Compute the are of the loop of curve \[y^2=x^2\frac{1+x}{1-x}\]

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Challenge: Compute the are of the loop of curve \[y^2=x^2\frac{1+x}{1-x}\]

Mathematics
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Anyone
Lets do brainstorming everyone give all ideas coming to mind
integrate?

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I got it ans is \[2\int\limits_{-1}^{0}x^2\sqrt{\frac{1+x}{1-x}}\]
\[2-\frac{\pi}{2}\]
\[y=2\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}dx\]isn't it? why is yours still x^2 squared?
A=... sorry
Sorry
is that a no? is should be x^2
or should not?
not x^2
turing yours was correct
Should be
\[A=\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}dx\]\[u=1-x\]groovy :)
but my sub wont' work
that won't work.. i don't think
Easier if you rationalise miltiply sqrt(1-x) in numerator and denominator
oh yeah, ok
I thought it was arc....I assume everyone else is looking at area...lol
Sorry 1+x
compute the area of the loop of the curve
=area under curve
Simplifies to \[\int\limits\limits_{-1}^{2}(\frac{x}{\sqrt{1-x^2}}-\sqrt{1-x^2}+\frac{1}{{\sqrt1-x^2}})dx\]
where did the 2 come from? I have something else
Typing error i menat it to be outside
which should give the anwer approx.....using winplot .4292 But of course the mathematics is more fun than letting the computer do all of the work.
ok, I still have a different expression though I'm trying to figure out if they're the same...
am I doing something wrong?\[A=\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}\cdot\sqrt{\frac{1+x}{1+x}}dx=\int_{-1}^{0}x\frac{1+x}{\sqrt{1-x}}dx\]\[=\int_{-1}^{0}\frac{x}{\sqrt{1-x}}+\frac{x^2}{\sqrt{1-x}}dx\]because my expression looks different ...or am I just tired?
2 times all that*
yeah (1-x)(1+x)=1-x^2 you look tired
ok that's just a typo though
I meant that the expression is different as far as I seeam I doing something wrong?\[A=2\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}\cdot\sqrt{\frac{1+x}{1+x}}dx=2\int_{-1}^{0}x\frac{1+x}{\sqrt{1-x^2}}dx\]\[=2\int_{-1}^{0}\frac{x}{\sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}}dx\]is what I meant
If that is the case add 1and subtract 1 from x^2 to arrive at mine
oh gotcha :P
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Nice!
I feel double integrals no worth this problem it is not that difficult

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