Question

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

Answer 1

Anyone

Answer 2

Lets do brainstorming everyone give all ideas coming to mind

Answer 3

integrate?

Answer 4

I got it
ans is
\[2\int\limits_{-1}^{0}x^2\sqrt{\frac{1+x}{1-x}}\]

Answer 5

\[2-\frac{\pi}{2}\]

Answer 6

\[y=2\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}dx\]isn't it?
why is yours still x^2 squared?

Answer 7

A=...
sorry

Answer 8

Sorry

Answer 9

is that a no? is should be x^2

Answer 10

or should not?

Answer 11

not x^2

Answer 12

turing yours was correct

Answer 13

Should be

Answer 14

\[A=\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}dx\]\[u=1-x\]groovy :)

Answer 15

but my sub wont' work

Answer 16

that won't work.. i don't think

Answer 17

Easier if you rationalise
miltiply sqrt(1-x) in numerator and denominator

Answer 18

oh yeah, ok

Answer 19

I thought it was arc....I assume everyone else is looking at area...lol

Answer 20

Sorry 1+x

Answer 21

compute the area of the loop of the curve

Answer 22

=area under curve

Answer 23

Simplifies to
\[\int\limits\limits_{-1}^{2}(\frac{x}{\sqrt{1-x^2}}-\sqrt{1-x^2}+\frac{1}{{\sqrt1-x^2}})dx\]

Answer 24

where did the 2 come from? I have something else

Answer 25

Typing error i menat it to be outside

Answer 26

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.

Answer 27

ok, I still have a different expression though
I'm trying to figure out if they're the same...

Answer 28

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?

Answer 29

2 times all that*

Answer 30

yeah
(1-x)(1+x)=1-x^2
you look tired

Answer 31

ok that's just a typo though

Answer 32

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

Answer 33

If that is the case add 1and subtract 1 from x^2 to arrive at mine

Answer 34

oh gotcha :P

Answer 35

Nice!

Answer 36

I feel double integrals no worth this problem it is not that difficult