Question

Sketch the region bounded by the curves and find its area.

x+y-y^3=0 , x-y+y^2=0

how can you tell which to use? in terms of x or y? (please explain in village idiot terminology b/c my algebra is weak!!! I am trying to improve it lol. calculus is easy, algebra is not :/ )

Answer 1

either or. you can solve this for x or y

Answer 2

since x is by itself i think that'd be the easiest however you'd have to switch your graphing calculator to graph the curves

Answer 3

without a calculator :o and how do you get those to be in terms of x ? y=.....Myt algebra deficient brain is just not seeing it.

Answer 4

to be honest i really think you shouldn't since it much easier to find when x=

Answer 5

\[x=y^3-y\]
\[x=y-y^2 \]

Answer 6

when y = 0 both = 0 so that can be one of your your limits

Answer 7

haha wait thats all you do ! -_-

Answer 8

yes then you integrate in respects to y (dy

Answer 9

now you have to find when these two intersect

Answer 10

\[y^3-y=y-y^2 \]
\[y^3 +y^2 -2y=0\]
\[y(y^2+y-2)=0\]

Answer 11

there is three intersectiosn so you're going to need two limits =]

Answer 12

i mean two integrations

Answer 13

yeah. adding them.


so i use quadratic equation two find other two intersections right

Answer 14

they factor (x+2)(x-1)

Answer 15

(y+2)(y-1)*

Answer 16

oh derp. -.- i see now. that was simpler than i imagine it. i got it from here. thank you !

Answer 17

also since the area of one of the regions is below the x axis... it's going to come uot negative... which will create a net, you might want to htink about that