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the region enclosed by the graphs of y=x^2 and y=4x-x^2 is rotated about the line y=6 what is the volume of the solid

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Answer: To find the volume of the solid obtained by rotating the region enclosed by the graphs of y = x^2 and y = 4x - x^2 about the line y = 6, we can use the method of cylindrical shells.

The region enclosed by the graphs of the two equations looks like this:

/|

4x-x^2 / |

/ |

/ |

/ |

y=x^2 ------

To rotate this region about the line y = 6, we need to shift the entire region 6 units upward, like this:

/|

/ |

/ |

/ |

/ |

y=10 ------

| /|

| / |

|/ |

y=6 ------

Now we can see that the solid we want to find is the result of rotating the region bounded by the two curves around the line y = 10 (which is the same as rotating the region bounded by the shifted curves around the line y = 6).

To use the method of cylindrical shells, we need to integrate over the range of x values that define the region. The limits of integration are where the two curves intersect, which can be found by setting them equal to each other:

x^2 = 4x - x^2

2x^2 - 4x = 0

2x(x - 2) = 0

x = 0 or x = 2

So the limits of integration are x = 0 and x = 2. The height of each cylindrical shell is the difference between the two curves at the given x value, so it is:

y = 4x - x^2 - x^2 = 4x - 2x^2

The radius of each cylindrical shell is the distance from the x-axis to the line y = 6, which is:

r = 6 - y = 6 - (4x - 2x^2)

Now we can set up the integral to find the volume:

V = ∫[0,2] 2πr y dx

V = ∫[0,2] 2π(6 - 4x + 2x^2)(4x - 2x^2) dx

V = ∫[0,2] 16πx - 24πx^2 + 8πx^3 dx

V = [8πx^2 - 8πx^3 + 2πx^4]₀²

V = 32π/3

Therefore, the volume of the solid obtained by rotating the region enclosed by the graphs of y = x^2 and y = 4x - x^2 about the line y = 6 is 32π/3 cubic units.

Explanation:

User Amir Keshavarz
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