Find the area enclosed by the curves: y^2=2x, and y=2x-2.?

Question

This is the graph of the functions: http://img2.freeimagehosting.net/image.php?021c196fae.jpg

Which method is better to use the integral, from top - bottom dx, or right - left dy? Any hint is appreciated, even to find the upper and lower limits, thanks!

Answer 1

Express x in terms of y in each equation:
x=(y^2)/2
x=y/2 + 1
Find points of intersection by setting
(y^2)/2=y/2 + 1
<=> y^2 - y - 2 =0
<=> y = -1, y = 2
Integrate the w.r.t. y the difference (y/2 + 1 - (y^2)/2) over the interval [-1,2]:
2
∫(y/2 + 1 - (y^2)/2)dy
-1

= [y^2/4 + y - y^3/6]_-1^2
= [4/4+2-8/6]-[1/4-1+1/6]
= 9/4

Answer 2

right to left integral is better i think.
consider a small rectangular strip parallel to y axis.
area of this strip is dx dy.
y changes from 2x-2 to sqrt(2x).
pt of intersection of the curves is
x=(y^2)/2
substituting it ,y=2(y^2)/2 -2
2y=2y^2-4
y^2-y-2=0
(y-2)(y+1)=0
y=2 ,x=2
y= -1 this pt is not needed since we need only the pt in the first quadrant .this can be seen from the graph.
x varies from 0 to 2.
integral of dy from 2x-2 to sqrt(2x) gives sqrt(2x)-2x+2
now integrate this from x=0 to x=2.

Answer 3

The answer is integral
y = (1/2) * ∫ (-x^2 +x+2)* dx
from -1 to 2.

See:
http://s210.photobucket.com/albums/bb64/oregfiu/?action=view¤t=integral.jpg

Note: I interchanged x and y, but the result is the same.
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