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

Question

Isabella Ackerman · Accepted Answer

#pi/6[27-5sqrt5]# units^#2#

The formula for surface area of revolution is, #SA=2piint_a^bf[x]sqrt[1+[d/dxf[x]]^2dx#.

From the question, #SA#=#2piint_1^2sqrtxsqrt[1+[d/dxsqrtx]^2# [where #f[x]#=#sqrtx]#.

Differentiating# d/dxsqrtx#= #1/[2sqrtx# and so#[d/dx[sqrtx]]^2#=#1/[4x#

Therefore, #SA#=#2piint_1^2sqrtxsqrt[1+1/[4x]dx# = #2piint_1^2sqrtxsqrt[[4x+1]/[4x]dx#.

That is, #2piint _1^2sqrtx/[2sqrtx]##sqrt[4x+1#dx = #piint_1^2sqrt[4x+1#dx..........#[1]#

Integrating ....#[1]# w.r.t. gives us #SA# =, #pi/6sqrt[[4x+1]^3]# and when evaluted for #x= 1, x=2# will result inthe above answer when simplified.

Hope this helpful.

Samantha Adkison · Answer

undefined To find the surface area of the solid created by revolving the function $ f(x) = \sqrt{x} $ for $ x $ in the interval $[1,2]$ around the x-axis, you can use the formula for surface area of a solid of revolution:

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

Where $ f'(x) $ represents the derivative of $ f(x) $.

In this case, $ f(x) = \sqrt{x} $. The derivative $ f'(x) = \frac{1}{2\sqrt{x}} $.

So,

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