2. Double Integrals
Let f(x, y) be defined at each point in the given region R, then its double integral is
𝑅
𝑓 𝑥, 𝑦 𝑑𝑥 𝑑𝑦, where continuity of f(x, y) in the region R ensures the existence of the
integral.
Case 1 : x = c1 to x = c2, y = 𝜑1(x) to y = 𝜑2(x)
Case 2 : x = 𝜑1(y) to x = 𝜑2(y), y = d1 to y = d2
Case 3 : x = c1 to x = c2, y = d1 to y = d2
3. Double Integrals
Example 1 : The value of 𝑅
𝑥𝑦 𝑑𝑥𝑑𝑦, where R is a region in the +ve quadrant at the ellipse
𝑥2
𝑎2 +
𝑦2
𝑏2 = 1 is __________.
Solution : Consider the horizontal strip,
x = 0 to x =
𝑎
𝑏
𝑏2 − 𝑦2, y = 0 to y = b
4. Double Integrals
Example 2 : The value of 𝑅
𝑦 𝑑𝑥𝑑𝑦, where R is a region enclosed by y = x2, x + y = 2, x = 0
is __________.
Solution : x + x2 = 2 x (x + 1) = 2 x = 1, -2
Consider the vertical strip,
x = 0 to x = 1, y = x2 to y = 2 - x
5. Change of Order
Example :
Solution :- Given limits,
𝑦 =
𝑥
4
to y = 2, 𝑥 = 0 to 𝑥 = 8
By changing the limits,
𝑦 = 0 𝑡𝑜 𝑦 = 2, 𝑥 = 0 𝑡𝑜 𝑥 = 4𝑦
𝑞 = 4𝑦
6. Triple Integrals
Let f(x, y, z) be defined at each point in the region ‘R’ of space then its triple integral is
𝑅
𝑓 𝑥, 𝑦, 𝑧 𝑑𝑥𝑑𝑦𝑑𝑧.
𝑧 = 𝜑1 𝑥, 𝑦 to z = φ2(𝑥, 𝑦)
𝑦 = ψ1 𝑥 to y = ψ2 𝑥
𝑥 = 𝑐1 to 𝑥 = 𝑐2
𝑅
𝑓 𝑥, 𝑦, 𝑧 𝑑𝑥𝑑𝑦𝑑𝑧 =
7. Length Area and Volume Integrals
Length of an arc of a curve y = f(x) between the lines x = x1 & x = x2 is
𝑙 =
𝑥1
𝑥2
1 + (
𝑑𝑦
𝑑𝑥
)2 𝑑𝑥
Length of an arc of a curve x = f(t) and y = g(t) between t = t1 and t = t2
𝑙 =
𝑡1
𝑡2
(
𝑑𝑥
𝑑𝑡
)2+(
𝑑𝑦
𝑑𝑡
)2 𝑑𝑥
Area of the region bounded by the curve y = f(x) and y = g(x) between x = x1 & x = x2 is
A = 𝑥1
𝑥2
[𝑔 𝑥 − 𝑓(𝑥)] 𝑑𝑥
8. Length Area and Volume Integrals
The volume of solid generated by revolving y = f(x) between x = x1 & x = x2 about x-axis is
𝑉 = 𝑥1
𝑥2
𝜋𝑦2 𝑑𝑥
Similarly, about y-axis 𝑉 = 𝑦1
𝑦2
𝜋𝑥2 𝑑𝑦
9. Length Area and Volume Integrals
Example 1 : Length of the curve 𝑦 =
2
3
𝑥3/2 between x = 0 & 1 is _________.
Solution:- 𝑙 = 𝑥1
𝑥2
1 + (
𝑑𝑦
𝑑𝑥
)2 𝑑𝑥
𝑑𝑦
𝑑𝑥
=
2
3
3
2
𝑥(
3
2−1)
= 𝑥
1
2
𝑙 =
0
1
1 + 𝑥 𝑑𝑥 =
1 + 𝑥
3
2
3
2
0 → 1 = 1.22
10. Length Area and Volume Integrals
Example 2 : Area bounded by parabola 𝑦 = 2 𝑥2 & straight line x = y - 4 is _________.
Solution:- Point of intersection,
𝑦 = 2(𝑦 − 4)2
𝑦 = 2(𝑦2
− 8𝑦 + 16)
2𝑦2 − 17𝑦 + 32 = 0
𝑦 =
17 ± 289 − 256
4
=
17 ± √33
4
𝐴 =
17−√33
4
17+√33
4
𝑦 − 4 −
𝑦
2
𝑑𝑦 −−−− −Ans