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Find te Volumes of the solids generated by revolving the regions bounded by the graphs of the equations

0 votes
y = sqrt x

y = 0

x = 6

 

a. The x-axis

b. The y-axis

c. The line x = 6

d. the line x = 9
asked Apr 3, 2018 in CALCULUS by Zack909 Rookie
reshown Apr 3, 2018 by bradely

1 Answer

0 votes

c)

Using the method of cylindrical shells is preferable when rotating around a vertical axis. This method integrates the lateral surface area of one of those shells. In this case, the height of a cylindrical shell is √x and the radius to the line x=6 is 6-x.

2π ∫rh dx

= 2π ∫(6-x)√x dx, from x = 0 to 6

= 2π ∫6x^½ - x^(3/2) dx, from x = 0 to 6

= 2π [4x^(3/2) - (2/5)x^(5/2)], from x = 0 to 6

= 2π [4·6^(3/2) - (2/5)·6^(5/2)]

= 2π [(4/6)6^(5/2) - (2/5)·6^(5/2)]

= 2π [(2/3)6^(5/2) - (2/5)·6^(5/2)]

= 2π [(4/15)·6^(5/2)]

= 2π [(4/15)·36√6]

= 96π√6 / 5

d)

Using the method of cylindrical shells is preferable when rotating around a vertical axis. This method integrates the lateral surface area of one of those shells. In this case, the height of a cylindrical shell is √x and the radius to the line x=9 is 9-x.

2π ∫rh dx

= 2π ∫(9-x)√x dx, from x = 0 to 6

= 2π ∫9x^½ - x^(3/2) dx, from x = 0 to 6

= 2π [6x^(3/2) - (2/5)x^(5/2)], from x = 0 to 6

= 2π [6·6^(3/2) - (2/5)·6^(5/2)]

= 2π [6^(5/2) - (2/5)·6^(5/2)]

= 2π [(3/5)·6^(5/2)]

= 2π [(3/5)·36√6]

= 216π√6 / 5

answered Apr 4, 2018 by johnkelly Apprentice

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