Question

find the volume of the solid formed by revolving the region bounded by the graphs of y = 4x - x^2 and f(x) = x^2 from [0,2] about the... a) x-axis.

Answer 1

`v = (32\pi)/(3)`

or

`v=33.52`

Explanation:

Given

`f(x) = 4x - x^2`

`g(x) = x^2`

`[a,b] = [0,2]`

Required

The volume of the solid formed

Rotating about the x-axis.

Using the washer method to calculate the volume, we have:

`\int dv = \int\limit^b_a \pi(f(x)^2 - g(x)^2) dx`

Integrate

`v = \int\limit^b_a \pi(f(x)^2 - g(x)^2)\ dx`

`v = \pi \int\limit^b_a (f(x)^2 - g(x)^2)\ dx`

Substitute values for a, b, f(x) and g(x)

`v = \pi \int\limit^2_0 ((4x - x^2)^2 - (x^2)^2)\ dx`

Evaluate the exponents

`v = \pi \int\limit^2_0 (16x^2 - 4x^3 - 4x^3 + x^4 - x^4)\ dx`

Simplify like terms

`v = \pi \int\limit^2_0 (16x^2 - 8x^3 )\ dx`

Factor out 8

`v = 8\pi \int\limit^2_0 (2x^2 - x^3 )\ dx`

Integrate

`v = 8\pi [ (2x^(2+1))/(2+1) - (x^(3+1))/(3+1) ]|\limit^2_0`

`v = 8\pi [ (2x^(3))/(3) - (x^(4))/(4) ]|\limit^2_0`

Substitute 2 and 0 for x, respectively

`v = 8\pi ([ (2*2^(3))/(3) - (2^(4))/(4) ] - [ (2*0^(3))/(3) - (0^(4))/(4) ])`

`v = 8\pi ([ (2*2^(3))/(3) - (2^(4))/(4) ] - [ 0 - 0])`

`v = 8\pi [ (2*2^(3))/(3) - (2^(4))/(4) ]`

`v = 8\pi [ (16)/(3) - (16)/(4) ]`

Take LCM

`v = 8\pi [ (16*4- 16 * 3)/(12)]`

`v = 8\pi [ (64- 48)/(12)]`

`v = 8\pi * (16)/(12)`

Simplify

`v = 8\pi * (4)/(3)`

`v = (32\pi)/(3)`

or

`v=(32)/(3) * (22)/(7)`

`v=(32*22)/(3*7)`

`v=(704)/(21)`

`v=33.52`