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Let D be the region in the x,y-plane enclosed by the curves?
Let D be the region in the x,y-plane enclosed by the curves x = y^2-8 and x = 2y-5. Find the volume of the solid with base D that lies under the surface z = 2y^2.
Not sure on how to set up this problem.
2 Answers
- PaulR2Lv 77 years ago
First you need to find the limits.
x = y² - 8
x = 2y - 5
2y - 5 = y² - 8
y² - 2y - 3 = 0
( y - 3 )( y + 1 ) = 0
y = -1 and y = 3
So the points of intersection are ( -7 , -1 ) and ( 1 , 3 ). This means our x limits are from x = y² - 8 to x = 2y - 5 and our y limits are from -1 to 3.
∫ [-1,3] ∫[y²-8,2y-5] ( 2y² ) dxdy
∫ [-1,3] ( 2xy² from [y²-8,2y-5] ) dy
∫ [-1,3] ( 2(2y - 5 )y² - ( 2(y² - 8 )y² ) ) dy
∫ [-1,3] ( 4y³ - 10y² - 2y&8308; - 16y² ) dy
∫ [-1,3] ( -2y⁴ + 4y³ - 26y² ) dy
( -0.4y⁵ + y⁴ - ( 26y³ / 3 ) ) from [-1,3]
192 / 5 = 38.4
So the volume is 38.4 units.