Izoh:
0
2
lim
2
1
lim
1
ln
lim
ln
lim
2
0
3
0
2
0
2
0
x
x
x
x
x
x
x
x
x
x
x
endi 6-shaklda ifodalangan sohaning yuzasini
hisoblang.
1
0
1
0
2
1
0
.
3
1
2
2
b
kv
dx
x
x
dx
y
dydx
S
x
x
x
x
Natijani aniq integral yordamida tekshirib ko’rish mumkin.
b
kv
x
x
dx
x
dx
x
S
.
3
1
3
1
3
2
3
3
2
1
0
3
1
0
1
0
3
1
0
2
13.8
.
12
1
12
1
6
1
6
2
1
3
2
1
2
1
2
1
0
1
0
6
1
0
3
4
2
1
0
1
0
2
2
x
x
dx
x
x
x
dx
y
x
xydy
dx
x
x
x
x
.
13.9. 9 13.10. 20 13.11.
2
1
e
e
1312. 2
13.13
. 9 .
13.14
.
2
2
a
13.15.
5
6
20
13.16
.
5
3
1
13.17
.
21
19
7
13.18
.
4
0
4
2
3
4
2
2
0
3
0
4
0
0
2
2
3
2
2
3
4
2
3
14
12
4
1
y
dy
y
dy
y
x
arctg
y
y
dy
dx
y
x
y
dx
y
x
y
dx
J
y
y
y
13.19
. Bu yerda D soha
y
x
x
y
y
sin
1
,
0
,
2
,
0
chiziqlar bilan chegaralangan bo’lib, Ox
o’qiga nisbatan to’g’ri soha bo’ladi.
2
6
3
2
2
3
6
1
2
sin
4
1
cos
2
2
3
6
1
2
cos
2
1
sin
2
2
3
6
1
2
2
cos
1
sin
2
1
6
1
sin
sin
2
1
6
1
sin
1
6
1
2
3
1
3
2
0
2
0
2
0
2
0
2
2
0
2
0
2
sin
1
0
2
sin
1
0
2
0
y
y
y
dy
y
y
dy
y
y
dy
y
y
dy
y
dy
x
dx
x
dy
y
y
13.20
.
D soha
0
,
2
1
,
0
y
x
x
to’g’ri chiziqlar va
2
4
1
x
y
egri chiziq bilan chegralangan D
soha Oy o’qi bo’yicha to’g’ri sohadan iborat.
4
1
12
1
4
1
1
0
12
1
3
2
2
1
2
3
4
1
8
1
2
2
1
4
1
4
1
8
1
2
2
1
4
1
2
2
1
0
3
2
1
0
2
1
0
2
3
2
2
1
0
2
2
1
0
2
1
0
2
2
1
2
2
1
0
2
2
2
1
0
4
1
0
2
4
1
0
2
1
0
2
2
x
x
x
dx
x
dx
x
d
x
dx
x
x
x
dx
y
xy
dy
y
x
dx
x
x
13.21.
x
dy
y
x
f
dx
2
4
2
,
13.22.
2
2
1
2
/
1
1
2
/
1
x
x
zdy
dx
zdy
dx
13.23
.
d
f
d
arctg
cos
8
cos
4
2
2
/
sin
,
cos
13.24.
d
f
d
sec
0
4
/
0
sin
,
cos
13.25.
Doira koordinata o`qlariga nisbatan simmetrik bo`lgani uchun uning
4
1
bo`lagi yuzini topib, 4 ga
ko`paytirish mumkin. D soha
x
0
.
2
2
0
x
y
tengsizliklar bilan aniqlanadi.
birlik
kv
S
yoki
t
t
dt
t
tdt
tdt
t
t
x
tdt
dx
t
x
t
x
dx
x
dx
y
dy
dx
S
x
x
.
,
4
sin
2
1
2
2
2
2
sin
2
2
cos
1
2
cos
cos
sin
0
0
,
cos
2
,
sin
4
2
2
2
2
0
2
2
0
2
0
2
2
2
2
0
2
2
2
0
0
0
0
2
2
0
2
2
2
2
13.26
.
Berilgan jism yuqoridan
y
z
2
tekislik bilan chegralangan. Shuning uchun
D
dxdy
y
V
2
D soha parabolik segmentdan iborat bo’lib, xOy tekislikda
2
y
to’g’ri
chiziq va
2
x
y
parabola bilan
chegaralanganligi sababli berilgan jism yOz tekislik bo’yicha simmetrik bo’ladi. Shuning uchun hajmning
yarmini hisoblab, natijani 2 ga ko’paytiramiz.
birlik
kub
V
y
y
dy
y
dy
y
dy
y
y
y
dy
x
y
dx
y
dy
V
y
y
.
15
2
32
15
2
16
5
2
8
3
2
8
2
5
2
3
2
2
2
2
2
2
2
0
2
5
2
0
2
3
2
0
2
3
2
0
2
1
2
0
2
0
0
0
2
0
13.27.
4
1
1
13.28.
1
2
e
13.29.
.
3
ln
13.30.
4
2
1
13.31.
3
2
10
13.32.
2
13.33.
3
1
5
13.34
. D soha markazi koordinatalar boshida va radiusi 1 ga teng bo’lgan doiradan iborat. Shuning uchun
2
2
2
y
x
bo’lib,
1
u holda (13
/
) formulaga ko’ra
3
2
3
1
3
1
2
3
1
2
1
1
1
2
0
2
0
2
0
1
0
2
0
1
0
2
3
2
2
2
2
d
d
d
d
dxdy
y
x
D
13.35
.
sin
,
cos
b
y
a
x
almashtirish bajaramiz.
U vaqtda
ab
ab
ab
b
b
a
a
J
2
2
sin
cos
cos
sin
sin
cos
cos
sin
sin
cos
2
2
2
2
2
2
2
2
2
1
sin
cos
1
1
b
y
a
x
(13.11) formulaga asosan
D
D
d
d
ab
dxdy
b
y
a
x
2
2
2
2
2
1
1
.
Bu yerda
2
0
va
1
bo’lib,
1
0
bo’ladi, u
holda
3
2
3
3
2
3
1
2
1
1
2
0
2
0
2
0
1
0
2
3
2
2
0
1
0
2
2
ab
ab
d
ab
d
ab
d
d
ab
d
d
ab
D
13.36.
2
1
2
2
/
0
sin
cos
d
f
d
ab
J
b
y
a
x
13.37
.
2
/
0 0
3
2
2
6
c
a
d
d
c
13.38
.
2
/
1
0 0
4
3
8
c
a
d
d
13.39
. D sohani chegaralovchi aylana chizig’i qutb koordinatalarida quyidagicha bo’ladi:
cos
2
cos
2
2
,
cos
2
0
,
4
0
u vaqtda
birlik
kv
d
d
d
d
d
S
.
2
1
4
0
sin
2
sin
2
1
0
4
2
sin
2
1
2
cos
1
cos
2
2
4
0
4
0
4
0
4
0
2
cos
2
0
4
0
cos
2
0
2
4
0
13.40
.
3
0
3
0
2
2
3
0
3
0
2
2
1
2
4
1
2
4
dx
dy
y
x
dxdy
y
x
V
x
x
x
x
y
birlik
kub
x
x
x
x
dx
x
x
x
dx
x
x
x
dx
y
y
x
dx
dy
y
dy
x
x
x
x
x
45
2
189
162
2
171
63
6
7
6
2
19
21
3
2
4
18
19
21
3
3
2
3
1
4
3
2
1
4
2
1
4
3
0
4
3
2
3
0
3
0
3
2
3
2
3
0
3
0
3
3
0
2
3
0
3
0
2
3
0
2
13.41
.
x
musbat bo’lgan to’rtta oktantning birinchisidan jism hajmining
4
1
qismini topamiz.
dx
x
b
y
x
b
y
x
b
y
b
c
dydx
y
x
b
b
c
V
x
b
k
k
x
b
0
0
2
2
2
0 0
2
2
arcsin
2
2
4
1
Bu
yerda amallarni bajarsak,
2
0
8
1
4
4
1
bck
xdx
bc
V
k
yoki
2
2
1
bck
V
hosil bo’ladi.
13.42
. Ellipsoid koordinata tekisliklari bilan 8 ta teng qismga bo’linadi. Shuning uchun bir qismning hajmini
hisoblab 8 ga ko’paytiramiz.
6
3
1
4
4
1
1
8
1
0
3
2
2
0
2
2
2
2
0
2
0 0
2
2
2
2
2
2
0
0
2
2
2
2
2
2
abc
x
x
a
a
bc
dx
a
x
a
b
b
c
dx
u
b
c
dydx
y
u
b
c
b
u
a
x
dydx
b
y
a
x
c
V
a
a
a
a u
a
x
a
a
b
abc
abc
V
abc
V
3
4
6
8
;
6
8
1
Agar
c
b
a
bo’lsa, ellipsoid sharga aylanadi va
3
3
4
a
V
bo’ladi;
R
a
desak, u
holda shar
hajmining ma’lum formulasi kelib chiqadi, ya’ni:
3
R
3
4
V
.
13.43.
3
2
186
13.44.
6
abc
13.45.16 13.46.
2
a
13.47.
3
