Calculus: Analytic Geometry and Calculus, with Vectors

(^246) Integrals
perpendicular to the y axis intersect the solid in plane sets the areas ()f
which are not easily found.
Finally, we illustrate the "cylindrical shell method" for finding volumes
of solids by finding the volume of a solid cone in another way. We
consider the solid cone to be the solid obtained by rotating, about thex
axis, the triangular region T in which the solid cone intersects the first
h
quadrant of the xy plane. This re-
gion appears in Figure 4.56. This
time we make a partition P of the

interval 0 5 y s b of the y axis.

When y7,_1 < yk 5 yk, the lines in

z the xy plane having the equations

y = Yk-1 and y = yk cut from T a

strip approximating a rectangular

Figure 4.56 region of length [h - (h/b)yk] and

width Dyk. When this rectangular

region is rotated about the x axis, it generates a cylindrical shell resem-

bling a tin tomato can from which both top and bottom have been

removed. Different points in this shell have different distances from the

x axis, but when JPJ, the norm of P, is small, these distances are all

nearly y*. Taking 2iryk* to be the circumference of the shell, we use the

number

(4.57) 21ryk*(ii - b ya

/

to approximate the area of the shell. Multiplying this by Ayk, the

thickness of the shell, gives an approximation to the volume of the shell.

This leads us to the formulas

(4.571) V = lim I 2lryk (h - b yk LIyk

and

f

(4.58) V=2irh Ibyt1-_y1dy=*rb2h.

For finding volumes of cones, the slab/methodproduces answers much

more easily and quickly than the cylindrical shell method. Most of the

problems at the end of this section should be solved by the slab method,

but the cylindrical shell method sometimes works better than the slab

method.

Problems 4.59

(^1) Find the volume of the spherical solid (or ball) of radiusa which has its
center at the origin. Find out whether it is easier to partition the whole inter-