8.2, #6. Slice the solid by planes perpendicular to the axis of rotation. The solid is gotten by rotating the finite region enclosed by the graphs of y = x^{3}, y = −1, and x = 1. This region is illustrated below.

.^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} (−1, −1)  (1, 1) (0, 0)  • •  •_{.}_{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}

The graph of x^{3} meets the line y = −1 at (−1, −1) and the line x = 1 at (1, 1).) A slice meeting the axis at [x, x + Δx] has the approximate shape of a disk of radius x^{3} + 1 and thickness Δx. Since the disk has volume π(x^{3} + 1)^{2} Δx, the total volume of all slices is ∑ π(x^{3} + 1)^{2} Δx and the solid has volume

∫_{−1}^{1} π(x^{6} + 2x^{3} + 1) dx = π(\frac{x^{7}}{7} + \frac{x^{4}}{2} + x)|_{−1}^{1} = \frac{16π}{7} = 7.1808.

8.2, #7. Slice the solid by planes perpendicular to the axis of rotation. A slice meeting the axis at [y, y + Δy] has the approximate shape of a disk of radius 1 − y^{2} (since y = \sqrt{x}) and thickness Δy. The volume of the disk is π(1 − y^{2})^{2}Δy, so the total volume of all slices is ∑ π(1 − y^{2})^{2}Δy. Thus the solid has volume

∫_{0}^{1} π(1 − 2y^{2} + y^{4}) dy = π(y − \frac{2}{3}y^{3} + \frac{y^{5}}{5} )|_{0}^{1} = \frac{8π}{15} = 1.6755.

8.2, #11. Slice the solid by planes perpendicular to the axis of rotation. A slice meeting the x-axis at x has the approximate shape of an annulus with outer radius 7, inner radius 7−e^{x}, and thickness Δx. The annulus has volume π(7^{2} −(7−e^{x})^{2})Δx = π(14e^{x} −e^{2x})Δx. So the solid has volume

∫_{0}^{1} π(14e^{x} − e^{2x}) dx = π(14e^{x} − (e^{2x}/2))|_{0}^{1} = π(14e − 14 − (e^{2} − 1)/2) = 65.5381

8.2, #12. Slice the solid by planes perpendicular to the x-axis. A slice meeting the axis at x has the approximate shape of a rectangular box of side lengths e^{x}, e^{x}, and Δx (since the line segment connecting the boundary points (x, 0) and (x, e^{x}) of the region has length e^{x}). Since the box has volume e^{2x}Δx, the solid has volume

∫_{0}^{1} e^{2x} dx = (e^{2x}/2))|_{0}^{1} = (e^{2} − 1)/2) = 3.1945.

8.2, #15. Slice the tree into 6 circular cylinders of length 20 feet. The area of a circle of circumference C is C^{2}/4π. The corresponding approximations for the volume in cubic feet of the tree are

LEFT = \frac{20}{4π} (26^{2} + 22^{2} + 19^{2} + 14^{2} + 6^{2} + 3^{2}) = 2804.31
RIGHT = \frac{20}{4π} (22^{2} + 19^{2} + 14^{2} + 6^{2} + 3^{2} + 1^{2}) = 1730.01
TRAP = \frac{1}{2}(LEFT + RIGHT) = 2267.16

8.2, #18. This graph is a quarter of the circle of radius 2, so the arclength will be π. The function f(x) = \sqrt{4 − x^{2}} for 0 ≤ x ≤ 2. Its derivative is −x/\sqrt{4 − x^{2}} so

1 + (f′(x))^{2} = 1 + \frac{x}{4 − x^{2}} = \frac{4}{4 − x^{2}}

and the integral is

∫_{0}^{2} \frac{2}{\sqrt{4 − x^{2}}} dx

From tables or using the substitution x = 2 sin θ we find the arc lenght to be π.

8.2, #19. We have f(x) = x^{3/2}, so f′(x) = (3/2)x^{1/2} and the integral is

∫_{0}^{2} \sqrt{1 + (9/4)x} dx = (8/27)((11/2)^{3/2} − 1) = 3.5255.

This value can be checked by Simpson’s rule.

8.2, #21. y = \frac{e^{x} + e^{−x}}{2} , y′ = \frac{e^{x} − e^{−x}}{2} . So

\sqrt{1 + (y′)^{2}} = \sqrt{1 + \frac{e^{2x} − 2 + e^{−2x}}{4}} = \sqrt{\frac{e^{2x} + 2 + e^{−2x}}{4}} = \frac{e^{x} + e^{−x}}{2} .

The length of the catenary on [−1, 1] is

\frac{1}{2} ∫_{−1}^{1}(e^{x} + e^{−x}) dx = \frac{1}{2}(e^{x} − e^{−x})|_{−1}^{1} = e − e^{−1} = 2.35.