B.tech ii unit-2 material beta gamma function

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 1
Unit-II: GAMMA, BETA FUNCTION
Sr. No. Name of the Topic Page No.
1 Definition of Gamma function 2
2 Examples Based on Gamma Function 3
3 Beta function 5
4 Relation between Beta and Gamma Functions 5
5 Dirichlet’s Integral 9
6 Application to Area & Volume: Liouville’s
extension of dirichlet theorem
11
7 Reference Book 13

GAMMA, BETA FUNCTION
 The Gamma function and Beta functions belong to the category of the
special transcendental functions and are defined in terms of improper
definite integrals.
1.1 Definition of Gamma function :
The gamma function is denoted and defined by the integral
Γ𝑚 = ∫ 𝑒−𝑥
𝑥 𝑚−1
𝑑𝑥 (𝑚 > 0)
∞
0
1.2 Properties of Gamma function :
1) Γ( 𝑚 + 1) = 𝑚Γ𝑚
2) Γ( 𝑚 + 1) = 𝑚! When m is a positive integer.
3) Γ( 𝑚 + 𝑎) = ( 𝑚 + 𝑎 − 1)( 𝑚 + 𝑎 − 2)……… 𝑎Γ𝑎, when n is a
positive integer.
4) Γ𝑚 = 2 ∫ 𝑒−𝑥2
𝑥2𝑚−1
𝑑𝑥 ( 𝑚 > 0)
∞
0
5)
Γ𝑚
𝑡 𝑚
= ∫ 𝑒−𝑡𝑥
𝑥 𝑚−1
𝑑𝑥 ( 𝑚 > 0)
∞
0
6) Γ
1
2
= √ 𝜋
7) ∫ 𝑒−𝑥2
𝑑𝑥 =
√𝜋
2
∞
0
8) ∫ 𝑥 𝑛
(𝑙𝑜𝑔 𝑥) 𝑚
𝑑𝑥 =
(−1) 𝑚
( 𝑛+1) 𝑚+1
Γ(𝑚 + 1)
1
0
2.1 Examples Based on Gamma Function:
Example 1: Evaluate 𝚪(−
𝟏
𝟐
).
Solution: We know that Γ( 𝑚 + 1) = 𝑚Γ𝑚

 Γ (−
1
2
+ 1) = −
1
2
Γ (−
1
2
)
 Γ (
1
2
) = −
1
2
Γ(−
1
2
)
 √ 𝜋 = −
1
2
Γ (−
1
2
)
∴ 𝚪(−
𝟏
𝟐
) = −𝟐√ 𝝅. __________Ans.
Example 2: Evaluate ∫ √ 𝒙𝟒
𝒆−√𝒙
𝒅𝒙
∞
𝟎
Solution: Let 𝐼 = ∫ 𝑥
1
4 𝑒−√𝑥
𝑑𝑥
∞
0
__________(i)
Putting √ 𝑥 = 𝑡 ⟹ 𝑥 = 𝑡2
so that 𝑑𝑥 = 2𝑡 in (i), we get
𝐼 = ∫ 𝑡1 2⁄
𝑒−𝑡
2𝑡 𝑑𝑡
∞
0
= 2∫ 𝑡3 2⁄
𝑒−𝑡
𝑑𝑡
∞
0
= 2∫ 𝑡
5
2
−1
𝑒−𝑡
𝑑𝑡
∞
0
= 2Γ(
5
2
)
= (2 ×
3
2
)Γ (
3
2
)
= (2 ×
3
2
×
1
2
)Γ (
1
2
)
=
3
2
√ 𝜋
∴ ∫ √ 𝒙𝟒
𝒆−√𝒙
𝒅𝒙
∞
𝟎
=
𝟑
𝟐
√ 𝝅 ________Ans.
Example 3: Evaluate ∫
𝒙 𝒂
𝒂 𝒙
𝒅𝒙
∞
𝟎
.
Solution: Let 𝐼 = ∫
𝑥 𝑎
𝑎 𝑥
𝑑𝑥
∞
0
_______ (i)
Putting 𝑎 𝑥
= 𝑒 𝑡

⟹ 𝑥 log 𝑎 = 𝑡
⟹ 𝑥 =
1
log 𝑎
⟹ 𝑑𝑥 =
𝑑𝑡
log 𝑎
in (i), we have
𝐼 = ∫ (
𝑡
log 𝑎
)
𝑎
𝑒−𝑡∞
0
𝑑𝑡
log 𝑎
=
1
(log 𝑎) 𝑎+1 ∫ 𝑒−𝑡
𝑡 𝑎
𝑑𝑡
∞
0
=
1
(log 𝑎) 𝑎+1 ∫ 𝑡( 𝑎+1)−1
𝑒−𝑡
𝑑𝑡
∞
0
=
1
(log 𝑎) 𝑎+1
Γ(𝑎 + 1)
∴ ∫
𝒙 𝒂
𝒂 𝒙
𝒅𝒙
∞
𝟎
=
𝟏
( 𝐥𝐨𝐠 𝒂) 𝒂+𝟏
𝚪(𝒂 + 𝟏) ________ Ans.
Example 4: Prove that ∫ ( 𝒙 𝒍𝒐𝒈𝒙) 𝟒
𝒅𝒙 =
𝟒!
𝟓 𝟓
𝟏
𝟎
Solution: We know that
∫ 𝑥 𝑛
(𝑙𝑜𝑔 𝑥) 𝑚
𝑑𝑥 =
(−1) 𝑚
( 𝑛+1) 𝑚+1
Γ(𝑚 + 1)
1
0
_______(i)
Now, ∫ ( 𝑥 𝑙𝑜𝑔𝑥)4
𝑑𝑥 =
1
0
∫ 𝑥41
0
( 𝑙𝑜𝑔𝑥)4
𝑑𝑥
Putting 𝑛 = 𝑚 = 4 in (i), we get
∫ 𝑥4
1
0
( 𝑙𝑜𝑔𝑥)4
𝑑𝑥 =
(−1)4
(4 + 1)4+1
Γ(4 + 1)
=
Γ5
55
=
4!
55
__________ proved.
2.2 EXERCISE:

1) Evaluate: (a) Γ(−
3
2
) (b) Γ (
7
2
) (c)Γ(0)
2) ∫ 𝑒−ℎ2 𝑥2
𝑑𝑥
∞
0
3) ∫
𝑑𝑥
√−𝑙𝑜𝑔𝑥
1
0
4) ∫ ( 𝑥 𝑙𝑜𝑔𝑥)3
𝑑𝑥
1
0
3.1 BETA FUNCTION:
Definition: The Beta function denoted by 𝛽( 𝑚, 𝑛) or 𝐵(𝑚, 𝑛) is defined as
𝐵( 𝑚, 𝑛) = ∫ 𝑥 𝑚−1
(1 − 𝑥) 𝑛−1
𝑑𝑥, (𝑚 > 0, 𝑛 > 0)
1
0
3.2 Properties of Beta function:
1) B(m,n)= B(n,m)
2) 𝐵( 𝑚, 𝑛) = 2 ∫ 𝑠𝑖𝑛2𝑚−1
𝜃 𝑐𝑜𝑠2𝑛−1
𝜃 𝑑𝜃
𝜋
2⁄
0
3) 𝐵( 𝑚, 𝑛) = ∫
𝑥 𝑚−1
(1+𝑥) 𝑚+𝑛
𝑑𝑥
∞
0
4) 𝐵( 𝑚, 𝑛) = ∫
𝑥 𝑚−1+𝑥 𝑛−1
(1+𝑥) 𝑚+𝑛
𝑑𝑥
1
0
4.1 RelationbetweenBeta and Gamma Functions:
Relation between Beta and gamma functions is
𝛽( 𝑚, 𝑛) =
Γm .Γn
Γ(m+n)
 Using above relation we can derive following results:
 ∫ 𝑠𝑖𝑛 𝑝
𝜃 𝑐𝑜𝑠 𝑝
𝜃 𝑑𝜃 =
1
2
𝛽 (
𝑝+1
2
,
𝑞+1
2
) =
Γ(
𝑝+1
2
).(
𝑞+1
2
)
2Γ(
𝑝+𝑞+2
2
)
𝜋
2⁄
0
 Γ (
1
2
) = √ 𝜋
 Euler’s formula:
Γ𝑛 . Γ(1 − 𝑛) =
𝜋
sin 𝑛𝜋
 Duplication formula:

Γ𝑛 . Γ(𝑛 +
1
2
) =
√𝜋 Γ(2𝑛)
22𝑛−1
4.2 EXAMPLES:
Example 1: Evaluate ∫ 𝒙 𝟒
(𝟏− √ 𝒙)
𝟓
𝒅𝒙
𝟏
𝟎
Solution: Let √ 𝑥 = 𝑡 ⟹ 𝑥 = 𝑡2
so that 𝑑𝑥 = 2𝑡 𝑑𝑡
∫ 𝑥4
(1− √ 𝑥)
5
𝑑𝑥 =
1
0
∫( 𝑡2)4 (1 − 𝑡)5
(2𝑡 𝑑𝑡)
1
0
= 2 ∫ 𝑡9
(1 − 𝑡)5
𝑑𝑡
1
0
= 2 𝐵(10,6)
= 2
Γ10 Γ6
Γ16
= 2 ×
9!5!
15!
=
2×1×2×3×4×5
15×14×13×12×11×10
=
1
11×13×7×15
=
1
15015
∴ ∫ 𝑥4
(1 − √ 𝑥)
5
𝑑𝑥 =
1
0
1
15015
_________ Ans.
Example 2: Find the value of 𝚪 (
𝟏
𝟐
).
Solution: We know that,
∫ 𝑠𝑖𝑛 𝑝
𝜃 𝑑𝜃 =
Γ(
𝑝+1
2
).(
𝑞+1
2
)
2Γ(
𝑝+𝑞+2
2
)
𝜋
2⁄
0
Putting 𝑝 = 𝑞 = 0, we get ∫ 𝑑𝜃 =
𝚪(
𝟏
𝟐
) 𝚪(
𝟏
𝟐
)
2 𝚪𝟏
𝜋
2
0
 [ 𝜃]0
𝜋 2⁄
=
1
2
(Γ
1
2
)
2


𝜋
2
=
1
2
(Γ
1
2
)
2
 (Γ
1
2
)
2
= 𝜋
 Γ (
1
2
) = √ 𝜋 _______Ans.
Example 3: show that ∫ √ 𝒄𝒐𝒕𝜽 𝒅𝜽 =
𝟏
𝟐
𝚪(
𝟏
𝟒
) 𝚪 (
𝟑
𝟒
)
𝝅
𝟐
𝟎
Solution: We know that,
∫ 𝑠𝑖𝑛 𝑝
𝜃 𝑑𝜃 =
Γ(
𝑝+1
2
).(
𝑞+1
2
)
2Γ(
𝑝+𝑞+2
2
)
𝜋
2⁄
0
∫ √ 𝑐𝑜𝑡𝜃𝑑𝜃 = ∫
𝑐𝑜𝑠1 2⁄
𝜃
𝑠𝑖𝑛1 2⁄ 𝜃
𝑑𝜃
𝜋
2
0
𝜋
2
0
= ∫ 𝑠𝑖𝑛−1 2⁄
𝜃
𝜋
2
0
𝑐𝑜𝑠1 2⁄
𝜃 𝑑𝜃
On applying formula (1), we have
∫ √ 𝑐𝑜𝑡𝜃𝑑𝜃 =
Γ(
−
1
2
+1
2
) Γ(
1
2
+1
2
)
2Γ(
−
1
2
+
1
2
+2
2
)
𝜋
2
0
=
Γ(
1
4
) Γ(
3
4
)
2 Γ(1)
=
1
2
Γ (
1
4
)Γ (
3
4
)
∴ ∫ √ 𝑐𝑜𝑡𝜃 𝑑𝜃 =
1
2
Γ (
1
4
)Γ (
3
4
)
𝜋
2
0
__________Ans.
Example 4: Evaluate ∫ ( 𝟏 + 𝒙) 𝒑−𝟏 ( 𝟏 − 𝒙) 𝒒−𝟏
𝒅𝒙
+𝟏
−𝟏
Solution: Put 𝑥 = 2cos 2𝜃, then 𝑑𝑥 = −2sin 2𝜃 𝑑𝜃 in
∫ (1 + 𝑥) 𝑝−1 (1 − 𝑥) 𝑞−1
𝑑𝑥
+1
−1
= ∫ (1 + 𝑐𝑜𝑠2𝜃) 𝑝−1(1 − 𝑐𝑜𝑠2𝜃) 𝑞−1(−2𝑠𝑖𝑛2𝜃)
0
𝜋
2
= ∫ (1 + 2𝑐𝑜𝑠2
𝜃 − 1) 𝑝−1(1 − 1 + 2𝑠𝑖𝑛2
𝜃) 𝑞−1(−4𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 𝑑𝜃)
0
𝜋
2
= 4∫ 2 𝑝−1
𝑐𝑜𝑠2𝑝−2
𝜃 . 2 𝑞−1
𝑠𝑖𝑛2𝑞−2
𝜃 . 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 𝑑𝜃
𝜋
2
0

= 2 𝑝+𝑞
∫ 𝑠𝑖𝑛2𝑞−1
𝜃 𝑐𝑜𝑠2𝑝−1
𝜃
∞
0
𝑑𝜃
= 2 𝑝+𝑞
Γ(
2𝑞
2
) Γ(
2𝑝
2
)
2Γ(
2𝑝+2𝑞
2
)
= 2 𝑝+𝑞−1 Γ( 𝑝) Γ( 𝑞)
Γ( 𝑝+𝑞)
__________Ans.
Example 5: Show that 𝚪( 𝒏) 𝚪( 𝟏− 𝒏) =
𝝅
𝒔𝒊𝒏 𝒏𝝅
(𝟎 < 𝑛 < 1)
Solution: We know that
𝛽( 𝑚, 𝑛) = ∫
𝑥 𝑛−1
(1 + 𝑥) 𝑚+𝑛
𝑑𝑥
∞
0
Γ𝑚 Γ𝑛
Γ(𝑚+𝑛)
= ∫
𝑥 𝑛−1
(1+𝑥) 𝑚+𝑛
𝑑𝑥
∞
0
Putting 𝑚 + 𝑛 = 1 𝑜𝑟 𝑚 = 1 − 𝑛, we get
Γ(1−𝑛) Γ𝑛
Γ1
= ∫
𝑥 𝑛−1
(1+𝑥)1
𝑑𝑥
∞
0
Γ(1 − 𝑛)Γ𝑛 = ∫
𝑥 𝑛−1
1+𝑥
𝑑𝑥
∞
0
[∵ ∫
𝑥 𝑛−1
1+𝑥
𝑑𝑥
∞
0
=
𝜋
𝑠𝑖𝑛 𝑛𝜋
]
∴ Γ( 𝑛)Γ(1− 𝑛) =
𝜋
𝑠𝑖𝑛 𝑛𝜋
______proved.
4.3 EXERCISE:
1) Evaluate ∫ (1 − 𝑥3)−1 2⁄
𝑑𝑥
1
0
2) Evaluate ∫
𝑥 𝑚−1+𝑥 𝑛−1
(1+𝑥) 𝑚+𝑛
𝑑𝑥
1
0
3) Evaluate ∫ (
𝑥3
1−𝑥3
)
1
2
𝑑𝑥
1
0
4) Prove that Γ (
1
4
) Γ (
3
4
) = 𝜋√2
5) Show that 𝛽( 𝑝, 𝑞) = 𝛽( 𝑝 + 1, 𝑞) + (𝑝, 𝑞 + 1)
5.1 DIRICHLET’S INTEGRAL:
If 𝑙, 𝑚, 𝑛 are all positive, then the triple integral
∭ 𝑥 𝑙−1
𝑦 𝑚−1
𝑧 𝑛−1
𝑑𝑥 𝑑𝑦 𝑑𝑧
𝑉
=
Γ(l)Γ(m)Γ(n)
Γ(𝑙 + 𝑚 + 𝑛 + 1)

Where V is the region 𝑥 ≥ 0, 𝑦 ≥ 0, 𝑧 ≥ 0 and 𝑥 + 𝑦 + 𝑧 ≤ 1.
Note:
∭ 𝑥 𝑙−1
𝑦 𝑚−1
𝑧 𝑛−1
𝑑𝑥 𝑑𝑦 𝑑𝑧𝑉
=
Γ(l)Γ(m)Γ(n)
Γ(𝑙+𝑚+𝑛+1)
ℎ𝑙+𝑚+𝑛
Where V is the domain, 𝑥 ≥ 0, 𝑦 ≥ 0, 𝑧 ≥ 0 and 𝑥 + 𝑦 + 𝑧 ≤ ℎ
5.2 Corollary: Dirichlet’s theorem for n variables, the theorem status that
∭…∫ 𝑥1
𝑙1−1
𝑥2
𝑙2−1
… 𝑥 𝑛
𝑙 𝑛−1
𝑑𝑥1 𝑑𝑥2 𝑑𝑥3 … 𝑑𝑥 𝑛
=
Γ𝑙1Γ𝑙2Γ𝑙3 … Γ𝑙 𝑛
Γ(1 + 𝑙1 + 𝑙2 + ⋯+ 𝑙 𝑛)
ℎ𝑙1+𝑙2+⋯+𝑙 𝑛
Example 1: Prove that ∫
𝒙 𝟒(𝟏+𝒙 𝟓)
(𝟏+𝒙 𝟏𝟓)
𝒅𝒙 =
𝟏
𝟓𝟎𝟎𝟓
∞
𝟎
Solution: Let 𝐼 = ∫
𝒙 𝟒(𝟏+𝒙 𝟓)
(𝟏+𝒙) 𝟏𝟓
𝒅𝒙
∞
𝟎
 𝐼 = ∫
𝑥4
(1+𝑥)15
𝑑𝑥
∞
0
+ ∫
𝑥9
(1+𝑥)15
𝑑𝑥
∞
0
 𝐼 = 𝐼1 + 𝐼2 __________ (i)
Now, put 𝑥 =
𝑡
1+𝑡
, when 𝑥 = 0, 𝑡 = 0; when 𝑥 = ∞, 𝑡 = 1
1 + 𝑥 = 1 +
𝑡
1−𝑡
=
1
1−𝑡
 𝑑𝑥 =
𝑑𝑡
(1−𝑡)2
∴ 𝐼1 = ∫ (
𝑡
1−𝑡
)
4
. (1 − 𝑡)15
.
1
(1−𝑡)2
𝑑𝑡
1
0
= ∫ 𝑡4
(1− 𝑡)9
𝑑𝑡
1
0
= 𝛽(5,10) _______(2)
And 𝐼2 = ∫ (
𝑡
1−𝑡
)
9
. (1 − 𝑡)15
.
1
(1−𝑡)2
𝑑𝑡
1
0

= ∫ 𝑡9
(1− 𝑡)4
𝑑𝑡
1
0
= 𝛽(10,5) ________(3)
∴ 𝐼 = 𝐼1 + 𝐼2
= 𝛽(5,10) + 𝛽(10,5) [Using(2) and (3)]
= 𝛽(5,10) + 𝛽(5,10) [𝛽( 𝑚, 𝑛) = 𝛽(𝑛, 𝑚)]
= 2𝛽(5,10)
=
2Γ5Γ10
Γ15
=
2×4!×9!
14!
=
2×4×3×2×1×9!
14×13×12×11×10×9!
=
1
5005
_______ Proved.
5.3 EXERCISE:
1) Find the value of ∫
𝑥3−2𝑥4+𝑥5
(1+𝑥)7
𝑑𝑥
1
0
2) Show that ∫
𝑥 𝑚−1(1−𝑥) 𝑛−1
(𝑎+𝑥) 𝑚+𝑛
𝑑𝑥 =
𝛽(𝑚,𝑛)
𝑎 𝑛(1+𝑎) 𝑚
1
0
3) 𝛽( 𝑚 + 1, 𝑛) =
𝑚
𝑚+𝑛
𝛽(𝑚, 𝑛)
6.1 Application to Area & Volume:
 Liouville’s extension of dirichlet theorem:

∭ 𝑓(𝑥 + 𝑦 + 𝑧)𝑥 𝑙−1
𝑦 𝑚−1
𝑧 𝑛−1
=
Γ(l)Γ(m)Γ(n)
Γ(l + m + n)
∫ 𝑓( 𝑢) 𝑢𝑙+𝑚+𝑛−1
𝑑𝑢
ℎ2
ℎ1
Example1: Show that ∭
𝒅𝒙 𝒅𝒚 𝒅𝒛
(𝒙+𝒚+𝒛+𝟏) 𝟑
=
𝟏
𝟐
𝒍𝒐𝒈𝟐−
𝟓
𝟏𝟔
, the integral being
takenthroughout the volume bounded by
𝒙 = 𝟎, 𝒚 = 𝟎, 𝒛 = 𝟎, 𝒙 + 𝒚 + 𝒛 = 𝟏.
Solution: By Liouville’s theorem when 0 < 𝑥 + 𝑦 + 𝑧 < 1
∭
(𝑥+𝑦+𝑧+1)3
= ∭
𝑥 𝑙−1 𝑦 𝑚−1 𝑧 𝑛−1 𝑑𝑥 𝑑𝑦 𝑑𝑧
(𝑥+𝑦+𝑧+1)3
(0 ≤ 𝑥 + 𝑦 + 𝑧 ≤ 1)
=
Γ1Γ1Γ1
Γ(l+m+n)
∫
1
(u+1)3
u3−1
du
1
0
=
1
2
∫
𝑢2
(𝑢+1)3
𝑑𝑢
1
0
= ∫ [
1
𝑢+1
−
2
(𝑢+1)2
+
1
(𝑢+1)3
] 𝑑𝑢
1
0
(Partial fractions)
=
1
2
[log( 𝑢 + 1) +
2
𝑢+1
−
1
2(𝑢+1)2
]
0
1
=
1
2
[𝑙𝑜𝑔2 + 2(
1
2
− 1) − (
1
8
−
1
2
)]
=
1
2
𝑙𝑜𝑔2 −
5
16
∴ ∭
𝒅𝒙 𝒅𝒚 𝒅𝒛
(𝒙+𝒚+𝒛+𝟏) 𝟑
=
𝟏
𝟐
𝒍𝒐𝒈𝟐−
𝟓
𝟏𝟔
_______Proved.
Example 2: Find the mass of an octantof the ellipsoid
𝒙 𝟐
𝒂 𝟐
+
𝒚 𝟐
𝒃 𝟐
+
𝒛 𝟐
𝒄 𝟐
= 𝟏,
the density at any point being 𝝆 = 𝒌 𝒙 𝒚 𝒛.
Solution: Mass = ∭ 𝜌 𝑑𝑣
= ∭( 𝑘 𝑥 𝑦 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧

= 𝑘 ∭( 𝑥 𝑑𝑥)( 𝑦 𝑑𝑥)(𝑧 𝑑𝑧) _______(1)
Putting
𝑥2
𝑎2
= 𝑢,
𝑦2
𝑏2
= 𝑣,
𝑧2
𝑐2
= 𝑤 and 𝑢 + 𝑣 + 𝑤 = 1
So that
2𝑥 𝑑𝑥
𝑎2
= 𝑑𝑢,
2𝑦 𝑑𝑦
𝑏2
= 𝑑𝑣,
2𝑧 𝑑𝑧
𝑐2
= 𝑑𝑤
Mass= 𝑘∭ (
𝑎2 𝑑𝑢
2
)(
𝑏2 𝑑𝑣
2
)(
𝑐2 𝑑𝑤
2
)
=
𝑘 𝑎2 𝑏2 𝑐2
8
∭ 𝑑𝑢 𝑑𝑣 𝑑𝑤, Where 𝑢 + 𝑣 + 𝑤 ≤ 1
=
8
∭ 𝑢𝑙−1
𝑣 𝑙−1
𝑤 𝑙−1
𝑑𝑢 𝑑𝑣 𝑑𝑤
=
8
Γ1Γ1Γ1
Γ3+1
=
8×6
=
48
∴ 𝑴𝒂𝒔𝒔 =
𝒌 𝒂 𝟐 𝒃 𝟐 𝒄 𝟐
𝟒𝟖
Ans.

6.2 EXERCISE:
1) Find the value of ∭ 𝑙𝑜𝑔( 𝑥 + 𝑦 + 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧 the integral extending
over all positive and zero values of 𝑥, 𝑦, 𝑧 subject to the condition 𝑥 +
𝑦 + 𝑧 < 1.
2) Evaluate ∭
√1−𝑥2−𝑦2−𝑧2
1+𝑥2+𝑦2+𝑧2
𝑑𝑥 𝑑𝑦 𝑑𝑧, integral being taken over all
positive values of 𝑥, 𝑦, 𝑧 such that 𝑥2
+ 𝑦2
+ 𝑧2
≤ 1.
3) Find the area and the mass contained m the first quadrant enclosed by
the curve (
𝑥
𝑎
)
𝛼
+ (
𝑦
𝑏
)
𝛽
= 1 𝑤ℎ𝑒𝑟𝑒 𝛼 > 0, 𝛽 > 0 given that density at
any point 𝑝(𝑥𝑦) is 𝑘 √ 𝑥𝑦.
7.1 REFERENCEBOOK:
1) Introduction to Engineering Mathematics
By H. K. DASS. & Dr. RAMA VERMA
2) Higher Engineering Mathematics
By B.V.RAMANA
3) A text bookof Engineering Mathematics
By N.P.BALI
4) www1.gantep.edu.tr/~olgar/C6.SP.pdf

B.tech ii unit-2 material beta gamma function

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