Posted by **Amy** on Thursday, November 21, 2013 at 10:22pm.

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

3y+x=3 , y^2-x=1

- Chris - Amy -Calculus Area between curves -
**Reiny**, Thursday, November 21, 2013 at 10:56pm
Why are you switching names ??

find the intersection;

from the first:

x = 3 - 3y

into the 2nd:

y^2 - (3-3y) = 1

y^2 + 3y - 4 = 0

(y+4)(y-1) = 0

y = -4 or y = 1

in x = 3-3y ....

if y = 1, x = 0 ---->(0,1)

if y = -4 , x = 15 , ---> (15, -4)

Your sketch should look like this and my answers are confirmed

http://www.wolframalpha.com/input/?i=plot+3y%2Bx%3D3+%2C+y%5E2-x%3D1

judging from the graph I would integrate with respect to y

so the effective width of my horizontal slice

= 3-3y - (y^2 - 1)

= 4 - 3y - y^2

area = ∫(4 - 3y - y^2) dy from -4 to 1

= (4y - (3/2)y^2 - (1/3)y^3) | from -4 to 1

= (4 - 3/2 - 1/3) - (-16 -24 + 64/3)

= 13/6 - (-56/3)

= 125/6

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