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=x+2 and y=x^2 rotated about the x axis?

1 Answer
Oct 23, 2015

See the explanation section, below.

Explanation:

Graph the region, including the points of intersection, (-1,1) and (2,4).

In order to use shells, we must take our representative slices parallel to the axis of rotation. So the thickness of each shell will be dy.

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The thickness of each shell will be dy and we will be integrating with respect to y. So, we need to express the boundaries as function of y:

x=y-2 and x= +- sqrty

And the limits of integration are 0 to 4

The volume of the representative shell is 2pi*"radius"*"height"*dy

The radius of each shell will be y

The "height" of each cylindrical shell will be the distance between the x values. This will be the x on the right (x_"right") minus the x on the left (x_"left").

Notice in the picture, that as y goes from 0 to 4, the values of x_"right" and x_"left" change.

From y=0 to y=1 , we have:
"height" = x_"right" - x_"left" = sqrty - (-sqrty) = 2sqrty = (8pi)/5

So we need

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

From y=1 to y=4 , we have:
"height" = x_"right" - x_"left" = sqrty - (y-2)

So we need

int_1^4 2 pi y (sqrty-y+2) dy = 2pi int_1^4 (y^(3/2)-y^2+2y) dy = (64pi)/5

The total volume is (8pi)/5 + (84pi)/5 = 72/5 pi