Asked by Quay
Find the area of a shaded region whose bounded by equations x = y^2 - 4y, x = 2y - y^2, over the point (-3,3). Based on the subtraction/integration I did the answer should be 36. I'll show what I did below.
∫ (-3,3) 2y - y^2 - y^2 - 4y
But this is incorrect. Assistance would be appreciated.
∫ (-3,3) 2y - y^2 - y^2 - 4y
But this is incorrect. Assistance would be appreciated.
Answers
Answered by
oobleck
why integrate from -3 to 3?
The curves intersect at (0,0) and (-3,3)
The area is
∫[0,3] (2y-y^2)-(y^2-4y) dy
= ∫[0,3] -2y^2+6y dy
= -2/3 y^3 + 3y^2 [0,3]
= -2/3 * 27 + 3*9
= -18+27
= 9
The curves intersect at (0,0) and (-3,3)
The area is
∫[0,3] (2y-y^2)-(y^2-4y) dy
= ∫[0,3] -2y^2+6y dy
= -2/3 y^3 + 3y^2 [0,3]
= -2/3 * 27 + 3*9
= -18+27
= 9
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