What is the surface area of the solid created by revolving #f(x)=x-1# for #x in [1,2]# around the x-axis?

Question

Christopher Agresta · Accepted Answer

#S = sqrt(2)pi #

If you consider the interval #(x,x+dx)# the length of the arc element of the curve #y=f(x)# is:

#dl = sqrt(dx^2+dy^2) = sqrt(dx^2 + f'(x)^2dx^2) = dx sqrt(1+f'(x)^2)#

so the area of the surface element generated by the rotation around the #x#-axis is:

#dS = 2piydl = 2pi f(x) sqrt(1+f'(x)^2) dx#

Integrating over the interval, the surface is then calculated as:

#S = 2pi int_1^2 f(x) sqrt(1+f'(x)^2) dx #

As #f'(x) = 1#:

#S = 2sqrt(2) pi int_1^2 (x-1)dx = sqrt2pi [(x-1)^2]_1^2 = sqrt(2)pi #

Christopher Allers · Answer

undefined The surface area of the solid created by revolving $ f(x) = x - 1 $ for $ x $ in $[1,2]$ around the x-axis can be calculated using the formula for surface area of revolution:

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

where $ a $ and $ b $ are the limits of integration, and $ \frac{dy}{dx} $ is the derivative of $ f(x) $ with respect to $ x $.

In this case, $ a = 1 $, $ b = 2 $, and $ f(x) = x - 1 $. The derivative of $ f(x) $ is $ \frac{dy}{dx} = 1 $.

Substitute these values into the formula:

$$ S = 2\pi \int_{1}^{2} (x - 1) \sqrt{1 + (1)^2} \, dx $$

$$ S = 2\pi \int_{1}^{2} (x - 1) \sqrt{2} \, dx $$

$$ S = 2\pi \sqrt{2} \int_{1}^{2} (x - 1) \, dx $$

$$ S = 2\pi \sqrt{2} \left[ \frac{x^2}{2} - x \right]_{1}^{2} $$

$$ S = 2\pi \sqrt{2} \left[ \frac{2^2}{2} - 2 - \left( \frac{1^2}{2} - 1 \right) \right] $$

$$ S = 2\pi \sqrt{2} \left[ 2 - 2 - \frac{1}{2} + 1 \right] $$

$$ S = 2\pi \sqrt{2} \left[ - \frac{1}{2} + 1 \right] $$

$$ S = 2\pi \sqrt{2} \left[ \frac{1}{2} \right] $$

$$ S = \pi \sqrt{2} $$

Therefore, the surface area of the solid is $ \pi \sqrt{2} $ square units.