Question

Find the area enclosed by parabolas x=y^2-4y, x=2y-y^2

Answer 1

start by drawing a rough sketch and find the intersecting points

Answer 2

to find the intersection points do i equal them to each other?

Answer 3

yes solve them simultaneously,

but before even solving this, its a good to have some idea of how the area u wanted looks like..

Answer 4

clearly the given equations represent parabolas, facing east and west
also if u think a bit, u wil notice that both will go through the point (0, 0)
so, (0, 0) is one intersecting point.

u still need to find other intersecting point by equating them...

Answer 5

how can i know how the graph looks like?

Answer 6

two ways :-
1) if u knw a bit about parabolas, :
change the given equations to vertex form, then it becomes easy to visualize/sketch
2) simply enter the equations @ :
https://www.desmos.com/calculator

Answer 7

pick one..

Answer 8

thanks i got the graph

Answer 9

so now i equal them to eachother?

Answer 10

yes

Answer 11

would the intervals be from 2 to 4?

Answer 12

nope

Answer 13

thats the area between parabolas u wanto find right ?

Answer 14

yea

Answer 15

intersecting points :
(0, 0) and (-3, 3)

so the area wud be :
\(\large \mathbb{\int_0^3 (2y-y^2) - (y^2-4y) dy}\)

Answer 16

see if that makes more or less sense..

Answer 17

is the answer 27?

Answer 18

u should get 9 :

http://www.wolframalpha.com/input/?i=area+between++x%3Dy%5E2-4y%2C+x%3D2y-y%5E2

Answer 19

in wolframalpha do you how they simplify it to 6y-2y^2?

Answer 20

we have below :
\(\large \mathbb{\int_0^3 (2y-y^2) - (y^2-4y) dy}\)

group like terms :

\(\large \mathbb{\int_0^3 (2y-y^2 - y^2+4y) dy}\)

\(\large \mathbb{\int_0^3 (2y-2y^2+4y) dy}\)

\(\large \mathbb{\int_0^3 (6y-2y^2) dy}\)

Answer 21

ok thank you for helping me