Question
Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis.
y=3(2-x),y=0,x=0
use shell method
y=3(2-x),y=0,x=0
use shell method
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GPT 3.5
To find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis using the shell method, we need to first determine the limits of integration.
The region is bounded by y=3(2-x), y=0, and x=0. To find the limits of integration, we need to find the x-values where the two curves intersect. Setting 3(2-x) = 0, we get x = 2. Therefore, the limits of integration are from x=0 to x=2.
The volume V of the solid generated by revolving the region about the y-axis can be calculated using the formula:
V = ∫[0,2] 2πx (3(2-x)) dx
V = 2π∫[0,2] (6x - 3x²) dx
V = 2π[3x² - x³] from 0 to 2
Now, plug in the limits of integration:
V = 2π[(3(2)^2 - (2)^3) - (3(0)^2 - (0)^3)]
V = 2π[(3(4) - 8) - 0]
V = 2π[12 - 8]
V = 2π(4)
V = 8π cubic units
Therefore, the volume of the solid generated by revolving the region about the y-axis is 8π cubic units.
The region is bounded by y=3(2-x), y=0, and x=0. To find the limits of integration, we need to find the x-values where the two curves intersect. Setting 3(2-x) = 0, we get x = 2. Therefore, the limits of integration are from x=0 to x=2.
The volume V of the solid generated by revolving the region about the y-axis can be calculated using the formula:
V = ∫[0,2] 2πx (3(2-x)) dx
V = 2π∫[0,2] (6x - 3x²) dx
V = 2π[3x² - x³] from 0 to 2
Now, plug in the limits of integration:
V = 2π[(3(2)^2 - (2)^3) - (3(0)^2 - (0)^3)]
V = 2π[(3(4) - 8) - 0]
V = 2π[12 - 8]
V = 2π(4)
V = 8π cubic units
Therefore, the volume of the solid generated by revolving the region about the y-axis is 8π cubic units.
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