296
CHAPTER 35
Find the intersection points of the curves r~ = 4 cos 20 and r
2 = 4 sin 20.
35.53
35.54
35.55
35.56
35.57
35.58
35.59
From 4cos29 = 4sin20, we obtain tan20 = l, 20 = ir/4 or S-ir/4. The latter is impossible, since
4cos(57r/4)<0 and cannot equal r
2
. Hence, 0 = -rr/8, r
2 = 2V2, r = ±2
3 '
4 . Putting (-r, 6 + IT)
in the second equation, we obtain the same equation as before and no new points are found. However, both
curves pass through the pole r = 0, which is a third point of intersection.
Find the points of intersection of the curves r = 1 and r = -1,
As r = l and r=-l both represent the circle *
2 + y
2 = l, there are an infinity of intersection points.
Find the area enclosed by the cardioid r = 1 + cos 6.
As was shown in Problem 35.30, the curve is traced out for 0=0 to 0 = 2-n. The general formula for the
area is
In this case, we have
Find the area inside the inner loop of the limacon r = 1 + 2 cos 0.
Problem 35.31 shows that the inner loop is traced out from 0 = 5ir/6 to 0 = 7ir/6. So, the area
Find the area inside one loop of the lemniscate r
2 = cos 20.
From Problem 35.32, we see that the area of one loop is double the area in the first quadrant. The latter area
is swept out from 6 = 0 to 0 = ir/4. Hence, the required area is
i(l-0)=|.
Find the area inside one petal of the four-leaved rose r = sin 26
From Problem 35.33, we see that the area of one petal is swept out from 0=0 to 9 = ir/2. Hence, the
area is
Find the area inside the limacon r = 4 + 2 cos 6.
35.60
Find the area inside the limacon r = l + 2 cos 0, but outside its inner loop.
From Problem 35.43, we see that the area is swept out from 0 = 0 to 6 = 2ir. Hence, the area is
\ J
2 " [4+ 4 cos 20+cos
2 20] d0 = i J
2 " [4 + 4cos0 + k(l + cos40)] d6 = $(1* + 2sin20 + i sin 40) I
2
," =
£(97r) = 97i72.
From Problem 35.38, we see that the area is swept out from 0=0 to 0 = 2ir. So, the area is
| J 0
2 "(16+16cos0 + 4cos20)d0 = | J2" [16 +16 cos 0+2(1 + cos 20)] dO = ^(180 + 16sin 0 + sin20) ]2" =
U367r> = 187r.
From Problem 35.31, we see that it suffices to double the area above the *-axis. The latter can be obtained
by subtracting the area above the je-axis and inside the loop from the area outside the loop. Since the desired
area outside the loop is swept out from 0 = 0 to 0 = 2?r/3 and the desired area inside the loop is swept
out from 0=7r to 0 = 4ir/3, we obtain 2(i J 0
2 '
/3 r
2 dO - i J^
17
" r
2 dO) = J" 0
2
"'
3
(l + 4cos 0 + cos
2 0) dO -
J*"
3 (1 + 4 cos 0 + 4 cos
2 0) dO = J"'3 [1 + 4 cos 0 + 2(1 + cos 20)] dO - J*"'3[l + 4 cos 0 + 2(1 + cos 20)] dO =
0
2
(30 + 4 sin 0 + sin 20) ]
2 "'
3 - (30 + 4 sin 0 + sin 20) ]*J
n = (2-rr + 2V5 - ^V^) - [(4ir - 2V3 + ^Vl) - (3ir)] =
7T+3V5.
35.61
Find the area inside r = 2 + cos 20.
CHAPTER 35
Find the intersection points of the curves r~ = 4 cos 20 and r
2 = 4 sin 20.
35.53
35.54
35.55
35.56
35.57
35.58
35.59
From 4cos29 = 4sin20, we obtain tan20 = l, 20 = ir/4 or S-ir/4. The latter is impossible, since
4cos(57r/4)<0 and cannot equal r
2
. Hence, 0 = -rr/8, r
2 = 2V2, r = ±2
3 '
4 . Putting (-r, 6 + IT)
in the second equation, we obtain the same equation as before and no new points are found. However, both
curves pass through the pole r = 0, which is a third point of intersection.
Find the points of intersection of the curves r = 1 and r = -1,
As r = l and r=-l both represent the circle *
2 + y
2 = l, there are an infinity of intersection points.
Find the area enclosed by the cardioid r = 1 + cos 6.
As was shown in Problem 35.30, the curve is traced out for 0=0 to 0 = 2-n. The general formula for the
area is
In this case, we have
Find the area inside the inner loop of the limacon r = 1 + 2 cos 0.
Problem 35.31 shows that the inner loop is traced out from 0 = 5ir/6 to 0 = 7ir/6. So, the area
Find the area inside one loop of the lemniscate r
2 = cos 20.
From Problem 35.32, we see that the area of one loop is double the area in the first quadrant. The latter area
is swept out from 6 = 0 to 0 = ir/4. Hence, the required area is
i(l-0)=|.
Find the area inside one petal of the four-leaved rose r = sin 26
From Problem 35.33, we see that the area of one petal is swept out from 0=0 to 9 = ir/2. Hence, the
area is
Find the area inside the limacon r = 4 + 2 cos 6.
35.60
Find the area inside the limacon r = l + 2 cos 0, but outside its inner loop.
From Problem 35.43, we see that the area is swept out from 0 = 0 to 6 = 2ir. Hence, the area is
\ J
2 " [4+ 4 cos 20+cos
2 20] d0 = i J
2 " [4 + 4cos0 + k(l + cos40)] d6 = $(1* + 2sin20 + i sin 40) I
2
," =
£(97r) = 97i72.
From Problem 35.38, we see that the area is swept out from 0=0 to 0 = 2ir. So, the area is
| J 0
2 "(16+16cos0 + 4cos20)d0 = | J2" [16 +16 cos 0+2(1 + cos 20)] dO = ^(180 + 16sin 0 + sin20) ]2" =
U367r> = 187r.
From Problem 35.31, we see that it suffices to double the area above the *-axis. The latter can be obtained
by subtracting the area above the je-axis and inside the loop from the area outside the loop. Since the desired
area outside the loop is swept out from 0 = 0 to 0 = 2?r/3 and the desired area inside the loop is swept
out from 0=7r to 0 = 4ir/3, we obtain 2(i J 0
2 '
/3 r
2 dO - i J^
17
" r
2 dO) = J" 0
2
"'
3
(l + 4cos 0 + cos
2 0) dO -
J*"
3 (1 + 4 cos 0 + 4 cos
2 0) dO = J"'3 [1 + 4 cos 0 + 2(1 + cos 20)] dO - J*"'3[l + 4 cos 0 + 2(1 + cos 20)] dO =
0
2
(30 + 4 sin 0 + sin 20) ]
2 "'
3 - (30 + 4 sin 0 + sin 20) ]*J
n = (2-rr + 2V5 - ^V^) - [(4ir - 2V3 + ^Vl) - (3ir)] =
7T+3V5.
35.61
Find the area inside r = 2 + cos 20.