Question

What is the surface area produced by rotating `f(x)=(1-x)/(x^2+6x+9), x in [0,3]` around the x-axis?

Answer 1

Surface Area `color(red)(S=-0.1637650055" ")`square unit

Explanation:

Given `y=(1-x)/(x^2+6x+9)" "` and `x=0` to `x=3`

The formula for finding the surface area or revolution

`S=2pi int_a^b y *ds`

`S=2pi int_a^b y *sqrt(1+(dy/dx)^2) dx`

`S=2pi int_a^b (1-x)/(x^2+6x+9) *sqrt(1+(dy/dx)^2) dx`

Also `dy/dx=(x-5)/(x+3)^3`

`S=2pi int_a^b (1-x)/(x^2+6x+9) *sqrt(1+((x-5)/(x+3)^3)^2) dx`

I suggest Simpson's Rule to calculate the integration and

`S=2pi int_a^b (1-x)/(x^2+6x+9) *sqrt(1+((x-5)/(x+3)^3)^2) dx`

`color(red)(S=-0.1637650055" ")`square unit

God bless...I hope the explanation is useful.