# What is the surface area of the solid created by revolving #f(x) = x^2-3x+2 , x in [3,4]# around the x axis?

##### 1 Answer

Feb 25, 2017

#### Explanation:

The surface area of a solid created by revolving

#SA = 2pi \ int_(x=alpha)^(x=beta) \ f(x) \ sqrt(1+(f'(x))^2) \ dx#

So we have;

# \ \ \ \ \ \ f(x) = x^2 - 3x + 2 #

# :. f'(x) = 2x-3 #

And so the Surface Area is;

# SA = 2pi \ int_(3)^(4) \ (x^2-3x+2) \ sqrt(1+(2x-3)^2) \ dx #

#\ \ \ \ \ = 2pi \ int_(3)^(4) \ (x^2-3x+2) \ sqrt(1+(4x^2-12x+9)) \ dx#

#\ \ \ \ \ = 2pi \ int_(3)^(4) \ (x^2-3x+2) \ sqrt( 4x^2-12x+10 ) \ dx#

Although the integral can be established it is quite complex,so I will just quote the result

# SA = pi/32 (5sinh^-1(3)-45sqrt(10) -5sinh^-1(5) +235sqrt(26))#

#\ \ \ \ \ = 103.426839 ... # #