Question

the total area bounded by the curve
y^2(1-x) = x^2(1+x) and the line x = 1 ???

Answer 1

Can you give us a picture?: )

Answer 2

there is no picture

Answer 3

cant you make one by youself?

Answer 4

actually i don't know how 2 solve it

Answer 5

Have you learnt anything about integrals?

Answer 6

yes

Answer 7

so, firstly you need to solve eq y^2(1-x) = x^2(1+x) to see where this graphs intersect. Can you do it? Also, I add the picture how the scheme should look like.

Answer 8

But sadly, I dont know hot to solve it, maybe someone can help you : ) @Directrix

Answer 9

k

Answer 10

@amistre64

Answer 11

nice symmetric curves you got there; can we isolate one symmetry of it?

Answer 12

and maybe define the wonky loopy part in an equivalent manner?

Answer 13

ok, looks like the symetry is located at the origin

Answer 14

\[ y^2(1-x) = x^2(1+x)\]
\[ y^2 = x^2\frac{1+x}{1-x}\]
\[ y = x\sqrt{\frac{1+x}{1-x}}\]
which might be able to be integrated by parts; form x=0 to x=1

Answer 15

k

Answer 16

trying to solve for x is a pain :) so lets run that thru a grap and see if we get something useful

Answer 17

k

Answer 18

http://www.wolframalpha.com/input/?i=+y%5E2%281-x%2B1%29+%3D+%28x-1%29%5E2%281%2Bx-1%29%2C+y%3Dxsqrt%28%281%2Bx%29%5C%281-x%29%29

can you see that? i shifted the original over by 1 to compare with, so I think we got a match

Answer 19

k

Answer 20

therefore:\[2\int_{0}^{1}x\sqrt{\frac{1+x}{1-x}}dx\]

might have to do a limiting thing since at x=1 its a bad math nono

\[2~\lim_{b\to 1}\int_{0}^{b}x\sqrt{\frac{1+x}{1-x}}dx\]

Answer 21

if i could see a way to get an equivalent side view it might turn simpler, or even a way to go parametric

Answer 22

how

Answer 23

dunno yet, my brain has a mind of its own :)

Answer 24

.... and of course i have to deal with bad users in the biology chat :/

Answer 25

im back .... for the moment.

any progress?

Answer 26

i don't know how to solve further

Answer 27

well, the solution is (4+pi)/2

so im sure this thing is spose to convert to polars

Answer 28

k

Answer 29

y^2(1-x) = x^2(1+x)

r^2 = x^2 + y^2

x = rcos(t), y=rsin(t)

r^2sin^2(t)(1-rcos(t)) = r^2 cos^2(t)(1+rcos(t))

sin^2(t)(1-rcos(t)) = cos^2(t)(1+rcos(t))

sin^2(t)-rcos(t)sin^2(t) = cos^2(t)+rcos^3(t)

sin^2(t)-cos^2(t) = rcos^3(t) + rcos(t)sin^2(t)

\[r=\frac{sin^2(t)-cos^2(t)}{cos(t)(cos^2(t)+sin^2(t))}\]
\[r=\frac{sin^2(t)-cos^2(t)}{cos(t)}\]

Answer 30

http://www.wolframalpha.com/input/?i=polar+r%3D%28sin%5E2%28t%29-cos%5E2%28t%29%29%2F%28cos%28t%29%29

well, at least its the right shape still :)

Answer 31

can we by any chance simplify that r= any more?

Answer 32

sin^2 = 1 - cos^2

(1 - 2cos^2)/cos

sec - 2cos

Answer 33

i think thats as simple as we can get it

Answer 34

be back later, have a class to get to

Answer 35

k

Answer 36

ok, heres what of come up with in my deranged state :)