300
CHAPTER 35
35.82
Find the arc length of r = cos
2 (6/2) (Problem 35.45).
Then
35.83
Find the arc length of r = sin (0/3) from 0=0 to 0 = 3ir/2.
Then
35.84
Find the area of the surface generated by revolving the upper half of the cardioid /• = 1 — cos 8 about the polar
axis.
The general formula for revolution about the polar axis is
the disk formula). In this case, the calculation in Problem 35.78 shows that
Hence,
35.85
Find the surface area generated by revolving the lemniscate r
2 = cos 20 (Fig. 35-5) about the polar axis.
The required area is twice that generated by revolving the first-quadrant arc.
So, the surface area
35.86
Find the surface area generated by revolving the lemniscate r
2 = cos20 about the 90°-line.
The general form for revolutions about the 90°-line is
As
in Problem 35.85, the required area is twice that generated by revolving the first-quadrant arc. Thus,
35.87
Describe the graph of r = 2 sin 0 + 4 cos 0.
Multiply both sides by r: r
2 = 2r sin 0 + 4r cos 9, .v
2 + / = 2y + 4x, x
2 - 4x + y
2 - 2y =0, (x - 2)
2 +
(_y _ i)
2 = 5. Thus, the graph is a circle with center (2.1) and radius V5. Because (~2)
2 + (-1)
2 = 5, the
circle passes through the pole.
35.88
Find the centroid of the arc of the circle r = 2 sin 6 + 4 cos 6 from 6 = 0 to 6 = irl2.
The general formulas for the centroid of an arc are
where L is the arc length. In this case, the arc happens to be half of a circle of
radius
and, thus, its arc length
Since
and
we have:
and
35.89
Derive an expression for tan <£, where <£ is the angle made by the tangent line to the curve r = f(0) with the
positive jc-axis.
35.90
Find the slope of the tangent line to the spiral r = 6 at 0 = ir/3.
r' = l. Hence, by Problem 35.89, with tan 6 = V5,
_ Let a prime denote differentiation with respect to 6. Then tan <£ = dy/dx = y'/x'. Since y = rsinO,
y' = r cos 6 + r' sin ft Since x = r cos 0, x' = —r sin 0 + r' cos 6. Thus,
and
and