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 = 1 + x^2, y = 0, x = 0, x = 2 rotated about the x-axis?

1 Answer
Sep 26, 2015

See explanation below.

Explanation:

Here is the region to be revolved about the x axis:

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If we must use shells, we will need to take our representative slices horizontally and use thickness dy.

We need to rewrite the boundaries as functions of y so the curve becomes x=sqrt(y-1).
Note that, in the region y varies from 0 to 5. The radius of the shells will involve y values.
For the 'height' of the shell (which is lying on its side), we will use the x on the right minus the x on the left.

For y = 0 to 1, the 'height' is 2-0=2, The resulting solid is a cylinder with radius 1 and height 2. Its volume is 2pi.
(We could do an integral for this, but we learned the volume of a cylinder in geometry class.)

From y=1 to y=5, the 'height' is 2-sqrt(y-1).
The radius of the representative shell is y and the thickness is dy

We integrate:

int 2 pi rh dy = 2 pi int_1^5 y(2-sqrt(y-1))dy = (176pi)/15

Adding the two volumes, we get:

V = 2pi+176/15 pi = 206/15 pi

Note
If we had been allowed to use disks, we would integrate:

pi int_0^2 (1+x^2)^2 dx = pi int_0^2 (1+2x^2+x^4) dx

= pi [x+(2x^3)/3 + x^5/5]_0^2

= pi [2+16/3+32/5] = pi [(30+80+96)/15] = 206/15 pi

Of course, we get the same answer, but we only need to do one calculation and the integral is (I think) simpler.