How do you find the exact area of the surface obtained rotating the curve about the #x#-axis of #y=sqrt(8-x)#, #2<=x<=8#?

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Answer 1 · 2018-06-27T10:34:52

graph{y = sqrt(8-x) [-1.07, 10.97, -1.25, 5.76]}
Rotating about x-axis:

#dS = 2 pi y ds#

Arc length is:

#ds = sqrt(1 + x'^2) dy#

#y = sqrt (8-x)qquad x = 8- y^2 qquad x' = - 2y #

#x: 2 rarr 8 qquad harr qquad y: sqrt6 rarr 0 #

# S = 2 pi int_C y sqrt(1 + x'^2) dy#

# = 2 pi int_0^sqrt6 y sqrt(1 + 4y^2) dy#

# = 2 pi int_0^sqrt6 d/dy(1/12(1 + 4y^2)^(3/2)) dy#

# = pi/6 [(1 + 4y^2)^(3/2)]_0^sqrt6 #

# = pi/6( [(1 + 24)^(3/2)] - [(1 + 0)^(3/2)] )#

#=(62 pi)/3#