How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #y=6x+7# and #y=x^2# rotated about the line #y=49#?

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Answer 1 · 2015-09-23T00:26:57

Given a choice, I would use washers, but. . .

Here is a (non-interactive) graph using Desmos (desmos.com)
I have included the line #y=49# in red.

enter image source here

Here is the region using the Socratic grapher. It isn't quite accurate, but you can zoom in and out and drag the graph around

graph{(y-x^2)(y-6x-7) (sqrt(x+1))/(sqrt(x+1))<= 0 [-8.29, 20.21, -6, 8.26]}

#y=6x+7# if and only if #x=1/6y-7/6#

#y=x^2# if and only if #x= +-sqrty#

To use shells we take our representative slices horizontally. So the bounds become

For #y=0# to #y=1#, #x# goes from #-sqrty# to #sqrty#.

From #y-1# to #y=49#, #x# goes from #1/6y-7/6# to #sqrty#.

Throughout the problem, the radius of the cylindrical shell will be #49-y#

So we need to evaluate two integrals:

#2pi int_0^1 (49-y)(sqrty-(-sqrty))dy = 4pi int_0^1 (49y^(1/2)-y^(3/2))dy#

# = 1936/15pi#

And

#2pi int_1^49 (49-y)(sqrty-(1/6y-7/6))dy = 25296/5pi#

The volume is the sum of the two integrals.

Washers

#pi int_-1^7 [(49-x^2)^2-(49-(6x+7))^2]dx = 77824/15pi#