What is the surface area of the solid created by revolving #f(x)=xsqrt(x+1)# for #x in [0,1]# around the x-axis?

Question

Luke Alvarez · Accepted Answer

# (7 pi) /12#

Small element of width #Delta x# has volume #Delta V = pi f^2(x) Delta x# so

#V = pi \int_0^1 \ (x(sqrt(x+1))^2 \ dx#

# = pi \int_0^1 \ x^2(x+1) \ dx = pi \int_0^1 \ x^3+ x^2 \ dx#

# = pi [ (x^4)/4+ (x^3)/3]_0^1#

#= (7 pi) /12#

Samantha Adkison · Answer

undefined To find the surface area of the solid created by revolving the function $ f(x) = x\sqrt{x+1} $ for $ x $ in the interval $[0,1]$ around the x-axis, you can use the formula for surface area of revolution:

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

where $ a $ and $ b $ are the limits of integration, and $ y = f(x) $.

First, find $ \frac{dy}{dx} $ by differentiating $ f(x) $ with respect to $ x $, then substitute into the formula and integrate over the interval $[0,1]$.

$$ \frac{dy}{dx} = \frac{d}{dx}(x\sqrt{x+1}) $$

$$ = \sqrt{x+1} + \frac{x}{2\sqrt{x+1}} $$

Now, substitute into the formula:

$$ S = \int_{0}^{1} 2\pi x\sqrt{x+1} \sqrt{1 + \left(\sqrt{x+1} + \frac{x}{2\sqrt{x+1}}\right)^2} \, dx $$

Simplify the expression under the square root, then integrate over the interval $[0,1]$.

$$ S = \int_{0}^{1} 2\pi x\sqrt{x+1} \sqrt{1 + x+1 + \frac{x^2}{4(x+1)}} \, dx $$

$$ = \int_{0}^{1} 2\pi x\sqrt{x+1} \sqrt{\frac{5x+5}{4(x+1)}} \, dx $$

$$ = \int_{0}^{1} 2\pi x\sqrt{\frac{5x+5}{4}} \, dx $$

After simplifying, you can integrate this expression over the interval $[0,1]$ to find the surface area of the solid.