Find the length of the longest straight line that lies entirely within the surface: This will from A(r = 2,0 = 50°, ) = 20°) to B(r= 4,0 = 30°, ø = 60°) or A(x: = 2 sin 50° cos20°, y = 2 sin50° sin20°, z = 2 cos 50°) to B(x = 4sin 30° cos60°, y = 4sin 30° sin 60°, z = 4 cos 30°) or finally A(1.44, 0.52, 1.29) to B(1.00, 1.73, 3.46). Thus B – A = (-0.44, 1.21, 2.18) and Length = |B – A| = 2.53

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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please explain the solution of branch (d)

The surfaces r =
2 and 4, 0 = 30° and 50°, and = 20° and 60° identify a closed surface.
a) Find the enclosed volume: This will be
r60°
r50°
Vol =
p2 sin Odrdod = 2.91
20°
30°
where degrees have been converted to radians.
b) Find the total area of the enclosing surface:
r60°
r(sin 30° + sin 50°)drdo
20°
r60°
50°
-" |"
(4² +22)sin Od0dø +
Area =
20°
J30°
50°
+2
rdrde = 12.61
30°
c) Find the total length of the twelve edges of the surface:
r50°
Langth – 1 dr
= 4
dr + 2
(4+2)d0+
(4sin50° + 4sin 30° + 2 sin 50° + 2 sin 30°)do
2
30°
20°
= 17.49
d) Find the length of the longest straight line that lies entirely within the surface: This will be
from A(r = 2,0 = 50°, 6 = 20°) to B(r= 4,0 = 30°, o = 60°) or
A(x
= 2 sin 50° cos20°, y = 2 sin50° sin20°, z = 2 cos 50°)
to
B(x = 4sin 30° cos60°, y = 4sin 30° sin 60° , z = 4 cos 30°)
or finally A(1.44,0.52, 1.29) to B(1.00, 1.73, 3.46). Thus B- A = (-0.44, 1.21, 2.18) and
Length = |B – A| = 2.53
Transcribed Image Text:The surfaces r = 2 and 4, 0 = 30° and 50°, and = 20° and 60° identify a closed surface. a) Find the enclosed volume: This will be r60° r50° Vol = p2 sin Odrdod = 2.91 20° 30° where degrees have been converted to radians. b) Find the total area of the enclosing surface: r60° r(sin 30° + sin 50°)drdo 20° r60° 50° -" |" (4² +22)sin Od0dø + Area = 20° J30° 50° +2 rdrde = 12.61 30° c) Find the total length of the twelve edges of the surface: r50° Langth – 1 dr = 4 dr + 2 (4+2)d0+ (4sin50° + 4sin 30° + 2 sin 50° + 2 sin 30°)do 2 30° 20° = 17.49 d) Find the length of the longest straight line that lies entirely within the surface: This will be from A(r = 2,0 = 50°, 6 = 20°) to B(r= 4,0 = 30°, o = 60°) or A(x = 2 sin 50° cos20°, y = 2 sin50° sin20°, z = 2 cos 50°) to B(x = 4sin 30° cos60°, y = 4sin 30° sin 60° , z = 4 cos 30°) or finally A(1.44,0.52, 1.29) to B(1.00, 1.73, 3.46). Thus B- A = (-0.44, 1.21, 2.18) and Length = |B – A| = 2.53
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