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DrunkEngineer
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Homework Statement
The planar region bounded by [itex]y = x, y = \sqrt{x}[/itex] is rotated about the line x = -2.
Find the Volume.
Homework Equations
[itex]V = 2\pi\int_{0}^{4} R dA[/itex]
The Attempt at a Solution
Solution:
y = -2
(-2)^2 = x
x = 4
y^2 = +- 2
the point of intersection should be (0,0) and (4,2)
now I am using the cylindrical method in obtaining the volume
[itex]V = 2\pi\int_{0}^{4} R dA[/itex]
where dA = (y2-y1)dx
for
y2 = sqrt(x)
y1 = x
and
radius be r = x+2 since the revolved line is at the 2nd quadrant and x in the 1st quadrant and it needs to be added
[tex]V = 2\pi\int_{0}^{4} (x+2)(y_{2}-y_{1}) dx [/tex]
[tex]V = 2\pi\int_{0}^{4} (x+2)(\sqrt{x}-{x}) dx [/tex]V = -(416/15)pi ?
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