Verified answer

Using shell method:

y = √(6x) . . . and x = 6

height ===> √(6x)

radius ====> x

limits:

x = 0 ----> x = 6

6

∫ 2π * x * √(6x) dx

0

6

∫ 2π * √(6) * √(x) * x dx

0

6

∫ 2√(6)π * x^(1/2 + 1) dx

0

6

∫ 2√(6)π * x^(3/2) dx

0

. . . . . . . . . . . . . . . . . . . . 6

2√(6)π * x^(3/2 + 1)/(3/2 + 1) ]

. . . . . . . . . . . . . . . . . . . . 0

. . . . . . . . . . . . . . . .6

2√(6)π * x^(5/2)/(5/2) ]

. . . . . . . . . . . . . . . .0

. . . . . . . . . . . . . . . . . 6

2√(6)π * (2/5) * x^(5/2) ]

. . . . . . . . . . . . . . . . .0

2√(6)π * (2/5) * ( 6^(5/2) - 0^(5/2) )

2√(6)π * (2/5) * ( 36√(6) - 0 )

√(6)π * (144√(6)/5)

(144 * 6π/5)

(864π/5)

using the Washer/Disk Method:

about y-axis

y = √(6x) ---> y^2 = 6x ---> y^2/6 = x

when x = 6

y = √(6 * 6)

y = √(36)

y = 6

A (y) = π ( outer radius )^2 - π ( inner radius )^2

A (y) = π ( 6 - 0 )^2 - π ( y^2/6 - 0 )^2

A (y) = π ( 36 - y^4/36 )

6

∫ π ( 36 - y^4/36 ) dy = (864π/5)

0

========

free to e-mail if have a question