What is the surface area of the solid created by revolving #f(x) = 3x, x in [2,5]# around the x axis?

Question

Charles Anctil · Accepted Answer

#A=63sqrt10 pi# #A=2pi int_2^5 f(x)*sqrt(1+(d f(x))/(d x))^2*d x# #f(x)=3x# #d/(d x) f(x)=3# #A=2 pi int_2^5 3x*sqrt(1+3^2)* d x# #A=2 pi int_2^5 3xsqrt 10 *d x# #A=6sqrt10 pi int_2^5 x *d x# #A=6 sqrt10[1/2 x^2]_2^5# #A=6sqrt10 pi[(1/2*5^2-1/2*2^2)]# #A=6sqrt 10 pi[25/2-4/2]# #A=6sqrt 10 pi[21/2]# #A=63sqrt10 pi#

Christopher Agresta · Answer

undefined To find the surface area of the solid created by revolving the function $ f(x) = 3x $ on the interval $[2,5]$ around the x-axis, you can use the formula for the surface area of revolution:

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

First, calculate $ f'(x) $, then substitute it into the formula.

Given $ f(x) = 3x $, the derivative $ f'(x) $ is $ 3 $.

Substitute $ f(x) $ and $ f'(x) $ into the formula:

$$ A = 2\pi \int_{2}^{5} 3x \sqrt{1 + 3^2} \, dx $$

$$ A = 2\pi \int_{2}^{5} 3x \sqrt{10} \, dx $$

$$ A = 6\pi \sqrt{10} \int_{2}^{5} x \, dx $$

$$ A = 6\pi \sqrt{10} \left[\frac{x^2}{2}\right]_{2}^{5} $$

$$ A = 6\pi \sqrt{10} \left(\frac{5^2}{2} - \frac{2^2}{2}\right) $$

$$ A = 6\pi \sqrt{10} \left(\frac{25}{2} - \frac{4}{2}\right) $$

$$ A = 6\pi \sqrt{10} \left(\frac{21}{2}\right) $$

$$ A = 63\pi \sqrt{10} $$

Therefore, the surface area of the solid is $ 63\pi \sqrt{10} $.