## Area Using Integration Solution3

In this page area using integration solution3 we are going to see solution of some practice questions.

(5) Find the area of the region bounded by x² = 36 y , y - axis , y = 2 and y = 4

Solution: Here we need to find the area of the bounded by the curve x² = 36 y , y - axis , y = 2 and y = 4. By graphing the given line we come to know that the required area lies right side of the y-axis.

x = √36 y

x = 6√y

b

Required area = ∫ x dy

a

4

= ∫ 6 √y dy

2

4

= ∫ 6 (y)^1/2 dy

2

4

= 6 [((y)^3/2)/(3/2)]

2

4
= (6x2)/3[(y)^3/2]

2

= 4 [4^3/2 - 2^3/2]

= 4 [4√4 - 2√2]

= 4 [4(2) - 2√2]

= 4 [8 - 2√2]

= 8 [4 - √2] square units

(6) Find the area included between the parabola y² = 4 a x and its latus rectum.

Solution:

Here we need to find the area of the bounded by the curve y² = 4 a x and its latus rectum. By graphing the given line we come to know that the required area lies right side of the y-axis.  area using integration solution3 From the above graph we come to know that we have to find the area of shaded portion. For that we are going to take the limits from x = 0 to x = a. By using these limits we can find the area above x axis. To find the total shaded area we have to multiply the area above the x-axis by 2.

b

Required area = ∫ y dx

a

a

= ∫2 √(ax) dx

0

We have to find the area above the x-axis and we have to multiply by 2.

a

= 2  ∫ 2 √a √x dx

0

a

= 4 √a [x^3/2/(3/2)]

0

= 4 x (2/3) √a [a√a - 0]

= (8/3) a² square units area using integration solution1 area using integration solution1 