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July 24, 2014

July 24, 2014

Posted by **Michael** on Thursday, April 19, 2012 at 1:12pm.

4x+y^2=9, x=2y

- Calculus -
**Reiny**, Thursday, April 19, 2012 at 3:51pmI assume you want the area of the region

you will need their intersection

use substitution ....

4(2y) + y^2 = 9

y^2 + 8y - 9 = 0

(y-1)(y+9) = 0

y = 1 or y -9

if y = 1 , x = 2 --> (2,1)

if y = -9, x = -18 --> (-18, -9)

use horizontal slices

x = -(1/4)y^2 + 9/2

the value of x for the region

= (-1//4)y^2 + 9/2 - 2y

area = ∫(-1/4 y^2 + 9/2 - 2y ) dy from -9 to 1

= [(-1/20)y^5 + (9/2)y - y^2 ] from -9 to 1

= (-1/20) + 9/2 - 1 - ( (-1/20)(-59049) + 81/2 - 18)

= ....

you do the button - pushing.

(you might also want to check my arithmetic, my weakness)

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