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

Question

Ezekiel Alexander · Accepted Answer

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

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.

Scarlett Allison · Answer

undefined To find the surface area produced by rotating the function $ f(x) = \frac{1-x}{x^2 + 6x + 9} $ around the x-axis over the interval $[0, 3]$, we can use the formula for surface area of a curve rotated about the x-axis:

$$ A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left(f'(x)\right)^2} \, dx $$

Where $ f'(x) $ is the derivative of $ f(x) $ with respect to $ x $, and $ a $ and $ b $ represent the limits of integration, which are 0 and 3 in this case.

We first need to find $ f'(x) $ and then substitute into the formula to calculate the surface area.