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

1 Answer

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.