What is the surface area of the solid created by revolving #f(x) = 2x^2-6x+18 , x in [2,3]# around the x axis?

Question

What is the surface area of the solid created by revolving #f(x) = 2x^2-6x+18  , x in [2,3]# around the x axis?

Evelyn Agard · Accepted Answer

#A=94pi/3#

Given:

#f(x)=2x^2-6x+18# The surface of solid is made up of number of rings with centers lying on x axis and having radius as ordinate of the function. The area of the ring when advanced through an increment of dx is given by the product of the perimeter with the advancement dx

#dA=2pif(x)dx=2pi(2x^2-6x+18)dx#

The area enclosed between the limits x=2, and x=3 can be obtained by integrating the above expression between the limits x=2 to x=3

#intdA=int2pi(2x^2-6x+18)dx#between x=2 and 3

#=2pi(2/3x^3-6/2x^2+18x)#

#2pi(2/3(3^3-2^3)-3(3^2-2^2)+18(3-2))#

#=pi(4/3(27-8)-6(9-4)+36(3-2))#

#=pi(4/3xx19-6xx5+36xx1)#

#=pi/3xx(4xx19-30xx3+36xx3)#

#=pi/3xx(76-90+108)#94 #=pi/3xx94# #A=94pi/3#

Samantha Adkison · Answer

undefined To find the surface area of the solid created by revolving the function $f(x) = 2x^2 - 6x + 18$ over the interval $x \in [2, 3]$ around the x-axis, we use the formula for surface area of revolution:

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

Where $f'(x)$ is the derivative of $f(x)$. Let's first find $f'(x)$:

$$ f'(x) = \frac{d}{dx}(2x^2 - 6x + 18) = 4x - 6 $$

Now, we substitute $f(x)$ and $f'(x)$ into the surface area formula:

$$ S = 2\pi \int_{2}^{3} (2x^2 - 6x + 18) \sqrt{1 + (4x - 6)^2} \, dx $$

Now, we compute this integral to find the surface area of the solid.