Page 205 - How To Solve Word Problems In Calculus

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The length of each shell h = x 2 − x 1 and the radius r = y. The
thickness of each shell is dy. Therefore, the volume of a typical
shell is 2π(x 2 − x 1 )ydy and the total volume is
b

V = 2π (x 2 − x 1 )ydy
a
The variable of integration is y so both x 1 and x 2 must be
expressed in terms of y. The limits of integration must also
correspond to y.
2 √
Since the equation of the parabola is y = x , x 2 = y and
y
the line’s equation y = 2x gives x 1 = . The points of intersec-
2
tion are (0, 0) and (2, 4). The volume is therefore
4

√ y
V = 2π y − ydy
0 2
4 1

= 2π y 3/2 − y 2 dy
0 2

2 5/2 1 3
4
= 2π y − y
5 6 0

64 64
= 2π − − 0
5 6
64
= π
15

Supplementary Problems

1. Compute the area of the region bounded by the curve
2
y = 8 − x − 2x and the x axis.
2. Determine the area of the region bounded by the curve
2
3
y = x − 4x + 3x and the x axis, 0 ≤ x ≤ 3.
3
3. Find the area of the region bounded by the curve y = x and the line
y = 8 using (a) vertical rectangles and (b) horizontal rectangles.
4. Determine the area of the region bounded by the curves
2
2
4
y = x − x and y = x − 1.
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