# Integration-Area & Volume

(a) Determine the area of the bounded region enclosed between the curves: y = (x + 1)^2 and y = 1 − x.

(b) Find the volume of the solid formed by rotating y = 2x^2 + 1 about the line y = 3.

### 1 Answer

- 自由自在Lv 71 decade agoFavorite Answer
(a) y = (x + 1)^2 & y = 1- x

1 - x = (x + 1)^2

x^2 + 2x + 1 + x - 1 = 0

x^2 + 3x = 0

x(x + 3) = 0

x = 0 or x = -3

Area = ∫[(1 - x) - (x + 1)^2]dx from -3 to 0

= x - x^2/2 - [(x + 1)^3] / 3 from -3 to 0

= 0 - 0 - 1/3 + 3 + 9/2 - 8/3

= 9/2

(b) y = 2x^2 + 1 & y = 3

3 = 2x^2 + 1

2x^2 = 2

x = +/-1

Points of intersection are (-1,3) and (1, 3)

Volume = ∫π(y - 3)^2 dx from -1 to 1

= π∫(2x^2 + 1 - 3)^2 dx from -1 to 1

= π∫(2x^2 - 2)^2 dx from -1 to 1

= 4π∫(x^2 - 1)^2 dx from -1 to 1

= 4π∫ (x^4 - 2x^2 + 1) dx from -1 to 1

= 4π[x^5/5 - 2x^3/3 + x] from -1 to 1

= 4π[1/5 - 2/3 + 1 + 1/5 - 2/3 + 1]

= 64π/15

- Login to reply the answers