At the point N xN = 0 , thereby cos t = 0 and t N
intersection of the line and ellipse, thereby 2 3 = 4 sin t The area of the rectangle OMCD is equal to:
SOMCD = OM MC = 2 3 6 cos |
π |
= |
Evaluating the area of the region ONCD : |
3 |
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= π2 . The point C , sin t = 23 , tC =
6 3 .
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SONCD = ∫3 |
4 sin td(6 cos t) = − 24∫3 sin 2 tdt = |
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Hence, S = 2(2π + 3 3 − 6 3) = 4π − 6 3 .
5. Find the area of the region bounded by the Bernoulli’s lemniscate (x2 + y 2 )2 = x2 − y 2 (fig. 2.18).
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y N |
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ϕ = |
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4 |
B |
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у |
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S1 |
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A |
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D x |
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Fig. 2.17 |
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Fig. 2.18 |
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Solution. Transform the given equation into a corresponding polar equation. In this case the substitutions x = ρcos ϕ , y = ρsin ϕ are appropriated
ρ 4 = ρ 2 (cos2 ϕ − sin 2 ϕ) , thereby ρ = cos 2ϕ .
Notice that there is symmetry with respect to the x axis and y axis. Consequently,
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4 |
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S = 4S1 |
= 4 |
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cos 2ϕ d ϕ = sin 2ϕ |
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= 1 . |
2 |
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6. Find the area of the region bounded by the curves ρ = 6 sin 3ϕ and ρ = 3 |
( ρ ≥ 3 ). |
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171 |
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π |
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3 |
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ρ=3 |
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6 |
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ρ = 6 |
Fig. 2.19
Solution. The curve ρ = 6 sin 3ϕ is called a three-leaved rose. As ϕ
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increases from 0 up to |
π |
, 3ϕ increases from 0 up to π. Thus ρ, which is sin 3ϕ, |
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goes from 0 up to 6, then back to 0, for ϕ in |
0, |
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3 |
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three loops making up the graph of ρ = 6 sin 3ϕ . For 0 in |
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, ρ = 6 sin 3ϕ |
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3 |
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is negative (or 0). This yields the lower loop in fig. 2.19. For ϕ in |
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again positive, and we obtain the upper left loop. Further choices of |
ϕ lead only |
to repetition of the loops already shown.
The second function ρ = 3 is the circle. For ρ ≥ 3 the given area S is shaded
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in fig. 2.21 and is equal to S = 6S ABCA . |
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Finding the polar coordinates of |
А |
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and В: |
6 sin 3ϕ = 3 , sin 3ϕ = |
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ϕ A = |
. There is В on the ray with ϕ B |
= |
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6 |
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18 |
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As is shown the area of S ABCA = SOABO − SOACO , where ОАСО is a sector that is subtended by a central angle π6 . Evaluating the area of it:
SOACO = 12 R 2 α = 12 9 π6 −18π = π2 .
172
http://vk.com/studentu_tk, http://studentu.tk/
Thereafter
SOABO = |
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1 |
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∫6 |
36 sin 2 3ϕdϕ = 9 ∫6 |
(1− cos 6ϕ)dϕ = 9(ϕ − |
sin 6ϕ |
) |
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= π + |
3 3 . |
2 |
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9 |
3 |
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S = 6 |
π |
− |
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= 3π + |
. |
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6 |
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Arc length |
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7. Find the arc length of the curve y =x2 for x in [0,1]. |
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Solution. By the given formula l = ∫b |
1+ (y′)2 dx. |
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Since y = x2, |
y′ = 2x. Thus l = ∫1 |
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1 + 4x2 dx = |
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2x = tdt dx = |
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dt |
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, |
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1 arctg 2 |
dt |
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= |
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2cos2 t |
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= |
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∫0 |
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∫0 |
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sec3 tdt = |
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2 |
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cos3 t |
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x = 0 t = 0; x = 1 t = arctg 2 |
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1 |
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arctg 2 |
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arctg 2 |
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tg t 1+ tg |
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t |
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arctg 2 |
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= |
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sect tg t |
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+ |
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∫ |
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sectdt |
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1 arctg 2 sect (sect + tg t ) |
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1 arctg 2 sec2 t + sect tg t |
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+ |
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dt = |
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sect + tg t |
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sect + tg t |
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arctg 2 |
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arctg 2 |
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sect + tg t |
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1+ tg2 t |
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ln |
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= |
2 5 |
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1 |
ln( 5 +2) . |
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8. Find the length of one arch of the cycloid
x = a(t − sin t) , y = a(1− cos t) Solution. Here the parameter is t and we compute xt/
xt/ = a(1− cos t);yt/ = a sin t.
http://vk.com/studentu_tk, http://studentu.tk/
(xt/ )2 +(yt/ )2 dϕ .
To complete, one arch varies from 0 to 2π (see fig. 2.20). By given formula
β
the length of one arch is: l = ∫
α
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a |
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О• |
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2πa |
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x |
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Fig. 2.20 |
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Hence |
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2π |
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2π |
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2π |
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t dt = |
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l = ∫ |
a 2 (1− cos t)2 + a 2 sin 2 tdt = a ∫ |
2 − 2 cos tdt = a ∫ |
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4 sin 2 |
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2π |
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2π |
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t |
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2π |
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= 2a ∫ |
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2a ∫sin |
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4a cos |
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= 8a . |
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sin |
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dt = |
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9. Find the arc length of the curve ρ = sin ϕ (fig. 2.21). |
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y |
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Solution. Since ρ ≥ 0, then 0 ≤ ϕ ≤ π. Hence |
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1 |
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π |
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π |
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l = ∫ |
sin2 ϕ + cos2 ϕdϕ = ∫dϕ = π . |
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0,5 |
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Remark. It is no surprise that the graph appears to be a |
О• |
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circle. |
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The |
equation |
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in |
rectangular |
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coordinates |
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that |
x corresponds |
to |
the |
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polar |
equation |
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ρ = sin ϕ |
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is |
Fig. 2.21 |
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x 2 + ( y − 1/ 2)2 = 1/ 4 . |
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Volume of a solid |
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z |
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10. Find the volume |
of |
the |
solid |
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bounded by |
the |
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1 |
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paraboloid |
z = x |
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+ |
y 2 |
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y 2 |
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4 |
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(fig. 2.22). |
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Solution. There is a region between the paraboloid and |
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cone. The volume V of it is equal to |
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О |
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V = Vп − Vк , |
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x |
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where Vn is the volume of the paraboloid between the point |
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z = 0 and the plane z = 1 (they are the roots of the equation |
Fig. 2.22 |
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z = z 2 ) and Vk is the volume of the cone between them. |
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174 |
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z |
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x 2 |
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The perpendicular |
section |
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axis |
are |
the |
ellipses |
( z )2 + (2 |
z )2 |
= 1 |
(with the paraboloid) |
and |
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(with the cone). As is |
generally |
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(2z)2 |
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known the area of ellipse |
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+ |
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= 1 is πab. |
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a 2 |
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b2 |
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Hence |
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1 |
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1 |
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1 |
1 |
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V = Vп − Vк = π∫ |
z 2 |
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zdz − π∫ z 2zdz = π∫ 2zdz − π∫ 2z 2 dz = |
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= πz |
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πz |
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11. The region between |
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x 2 + y 2 |
= 1, |
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y = |
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2x 2 |
and x = 0 ( x ≥ 0 ) |
is |
revolved around the a) x axis (fig. 2.23); b) y axis (fig. 2.24). Find the volume of the solids of revolution produced.
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у |
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Fig. 2.23 |
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Fig. 2.24 |
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Solution. Evaluating the system |
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2 |
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3 |
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= x |
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2 = 16x |
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> 0, |
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we get x = |
and |
y = |
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11 2 |
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а) V = π ∫[(1− x |
2 |
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4 |
]dx = |
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5 2 |
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π x |
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= |
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π ; |
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175 |
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16 − 7 |
2 |
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∫ |
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− y2 )dy = π |
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b) V = π |
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dy + π |
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2 + π y − |
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= |
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π . |
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12. The asteroid x = a cos3 t , y = a sin 3 t
(fig. 2.25) is revolved around the x axis. Find the volume of the solids of revolution produced.
Solution. Notice that there is symmetry with respect to the x axis. We can find only the volume of
ООАВ, which is revolved around the x axis and multiple
xthe result by 2:
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a |
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Fig. 2.25 |
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V = 2π∫ y 2 (x)dx . |
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This prompts the following maneouver: |
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x = a cos3 t , |
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dx = −3a cos2 t sin tdt , |
y = a sin 3 t . |
If |
x = 0 , then tB = |
π |
and if |
x = a , then t A = 0 . |
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V = 2π∫ (a sin 3 t)2 (−3a) cos2 t sin tdt = 6πa3 ∫2 sin 7 t cos2 tdt = |
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π |
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2 π |
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π |
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We used the reduction formula: ∫2 sin n xdx = |
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Exercises for class and homework |
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Т.8 |
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Find the area of the region bounded by the given curves (or lines).
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xy = 4 , y = 1, y = x + 3 . |
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y = x 2 − 3x + 3 , y = − x 2 + x + 9 . |
3. |
y 2 − 2 y − 2x − 3 = 0 , y − x + 1 = 0 . |
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4. y = tg x , y = sin x − 2 , x = − π4 , x = π4 .
5.y = 16 , y = 17 − x 2 .
x2
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xy = 20 , |
x 2 + y 2 = 41 (Quadrant І). |
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y = |
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, 2y = x 2 . |
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8.y = 0 , y = arcsin x , y = arccos x .
9.ρ = 2 sin 2ϕ .
10.ρ = 4 cos 2ϕ , ρ = 2 ( ρ ≥ 2 ).
11.ρ = 2 + cos ϕ .
12.ρ = sin 2 ϕ2 (righter of ray ϕ = π2 ).
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13. ρ = |
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, ϕ = |
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14.(x 2 + y 2 )3 = 4xy(x 2 − y 2 ) .
15.x 4 + y 4 = x 2 + y 2 .
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x = a cos3 t , |
y = b sin 3 t . |
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x = 1 ( x ≥ 1). |
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18.x = 2(t − sin t) , y = 2(1− cos t) , y = 1( y ≥ 1 ). Find the arc length.
19.y = ln x from x = 3 to x = 8 .
20. y = ln(1− x 2 ) |
from x = 0 to |
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y = ln sin x from x = |
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22. |
y = |
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x = 1 . |
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23.x = 9(t − sin t) , y = 9(1− cos t) (only one arch of the cycloid) .
24.x = 8 sin t + 6 cos t , y = 6 sin t − 8 cos t , 0 ≤ t ≤ π2 .
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25. |
x = |
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26. |
y = x − x 2 |
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x = a cos5 t , y = a sin 5 t . |
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ρ = sin |
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29.ρ = 1+ cos ϕ .
30.ρ = ϕ 2 , 0 ≤ ϕ ≤ π .
31.ρ = ϕ1 , 34 ≤ ϕ ≤ 43 .
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Find the volume of the solid bounded by the paraboloid |
z = |
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and the plane z = 1 . |
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33. |
Find the volume of the solid bounded by the hyperboloid of one sheet |
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− z 2 = 1 and the planes z = −1 and z = 2 . |
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The region bounded by the given curves is revolved around the x axis. Find volume of the solid of revolution produced.
34. y = |
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, 8y = x 2 . |
35. y 2 = x , y = x 2 . |
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Answers
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1. 4ln 4 + 3/ 2 . 2. 22 |
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8. |
2 − 1 . 9. π . 10. |
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16. |
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16π(3π + 10) /5. 35. |
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Т.8 Individual test problems
8.1. Find the area of the region bounded by the given curves (or lines).
8.1.1. y = x2 − 2x − 1 , 2y = 3x − 2 . |
8.1.2. 4y = x2 , 2y = 6 − x2 . |
8.1.3. x = y 2 − 2 y , x = − y 2 + 2y + 6 . |
8.1.4. y = x 4 − x , y = 0 . |
8.1.5. y = x2 −6x +6, |
y =−x2 +2x . |
8.1.6. x = y 2 − 2 , y = − x . |
8.1.7. y = x2 + 4x + 2 , |
y = 2 + x . |
8.1.8. x = y 2 − 2y − 2 , y = − x . |
8.1.9. x = y 2 + 2y − 2 , y = −2 − x . |
8.1.10. y = x arctg x , 0 ≤ x ≤ 1 . |
8.1.11. y = |
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8.1.12. y = x2 +5x, y = 7 −x . |
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8.1.13. x = y 2 − 2y − 1, |
y = 1 − x . |
8.1.14. y = 3x −4, y =−x2 . |
8.1.15. x = y 2 + 2y − 1, |
y = −1 − x . |
8.1.16. y2 = 4 − x, x = y2 − 2 y . |
8.1.17. y = x tg2 x , 0 ≤ x ≤ π / 4 . |
8.1.18. y = x2 + 6x, y = − x2 . |
8.1.19. y = cos3 x sin 2x , 0 ≤ x ≤ π / 4 . |
8.1.20. y = xcos2 x , 0 ≤ x ≤ π / 2 . |
8.1.21. y = x |
4 − x2 , y = 0 , 0 ≤ x ≤ 2 . |
8.1.22. y = xsin2 x, 0 ≤ x ≤ π / 4 . |
8.1.23. y = sin 4 x sin 2x , 0 ≤ x ≤ π / 3 . |
8.1.24. y = x2− 4x + 2, y = 2 − x. |
8.1.25. y = |
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0 ≤ x ≤ 1 . |
8.1.26. x =y2+2 y +3, x=8−2 y. |
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8.1.27. y = 2x 2 − 12x + 16 , |
y = x 2 − 5x + 4 . |
8.1.28. y = x2 +8x +7, |
y =−x2 −2x −5 . |
8.1.29. y = x |
9−x2 , |
y = 0 , 0 ≤ x ≤ 3 . |
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8.1.30. x = 2y 2 − 8y + 6 , |
x = y 2 − 3y . |
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8.2. Find the area of the region bounded by the indicated curve.
8.2.1.ρ =1+cos ϕ , ρ =1 ( ρ ≥1 ).
8.2.2.ρ = 2 +cos ϕ .
8.2.3.ρ =1+cosϕ , ρ =3 / 2 ( ρ ≤3/ 2 ).
8.2.4.ρ = 2−sin ϕ .
8.2.5.ρ =1+sin ϕ , ρ =1/ 2 ( ρ ≥1/ 2 ).
8.2.6.ρ =3−cos ϕ .
8.2.7.ρ =1−sin ϕ , ρ =1 ( ρ ≤1 ).
8.2.8.ρ = 2 +cos 2ϕ .
8.2.9.ρ =1−cos ϕ , ρ =1 ( ρ ≥1 ).
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8.2.10.ρ =3+sin 2ϕ .
8.2.11.ρ =1−sin ϕ , ρ =3 / 2 ( ρ ≥3/ 2 ).
8.2.12.ρ =1+2cos ϕ .
8.2.13.ρ = 2 cos 2ϕ , ρ = 1 (ρ ≥ 1).
8.2.14.ρ =1+2sin ϕ .
8.2.15.ρ = 4 sin 2ϕ, ρ = 2 (ρ ≥ 2).
8.2.16.ρ = cosϕ + sinϕ.
8.2.17. ρ = 6 cos 3ϕ, ρ = 3 |
3 ( ρ ≥3 3 ). |
8.2.18. ρ =cosϕ – inϕ. |
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8.2.19. ρ = 2 sin 3ϕ, ρ = |
3 ( ρ ≥ 3 ). |
8.2.20.ρ = cos2 ϕ .
8.2.21.ρ = cos 2ϕ + sin 2ϕ.
8.2.22.ρ =sin2 ϕ .
8.2.23.ρ = 3 cos ϕ , ρ =sin ϕ .
8.2.24.ρ = 3 + cos 2ϕ.
8.2.25.ρ = tg ϕ , ϕ = π / 3 .
8.2.26.ρ = cos2 2ϕ .
8.2.27.ρ = 1+ tg ϕ , ϕ = π / 4 .
8.2.28.ρ = cos ϕ2 .
8.2.29.ρ = 4sin2ϕ , ρ = 2 3 ( ρ ≥ 2 3 ).
8.2.30.ρ = 2 − cos 2ϕ .
8.3.Find the area of the region bounded by the indicated curve.
8.3.1.x = 4 2 cos3 t, y = 2 2 sin3 t, x = 2 ( x ≥ 2 ).
8.3.2.x = 16 cos3 t, y = 2 sin3 t, x = 2 ( x ≥ 2 ).
8.3.3. x = 2 cos t, y = 6 sin t, y = 3 ( y ≥ 3 ).
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8.3.4. x = 2(t − sin t), y = 2(1− cos t), |
y = 3 ( y ≥ 3 , 0 ≤ x ≤ 4π ) . |
8.3.5. x = 16 cos3 t, y = sin3 t, x = 2 , |
x = 6 |
3 ( 2 ≤ x ≤ 6 3 ). |
8.3.6. x = 6 cos t, y = 2 sin t, y = 1 , |
y = 3 (1 ≤ y ≤ 3 ). |
8.3.7. x = 3(t − sin t), y = 3(1− cos t), |
y = 3 |
( y ≥ 3 , 0 ≤ x ≤ 6π ) . |
8.3.8. x = 8 |
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x = 4 ( x ≤ 4 ). |
8.3.9. x = 2 |
2 cos t, y = 3 |
2 sin t, |
y = 3 ( y ≥ 3 ). |
8.3.10. x = 6(t − sin t), y = 6(1 − cost), |
y = 3 , |
y = 9 ( 3 ≤ y ≤ 9 , 0 ≤ x ≤ 2π ). |
8.3.11. x = 32 cos3 t, y = sin3 t, x = 4 ( x ≥ |
4 ). |
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