What is the surface area of the solid created by revolving #f(x) = 2x^2+3 , x in [1,4]# around the x axis?

1 Answer

Hint is given below

Explanation:

HINT: The given function: #y=2x^2+3#

#\frac{dy}{dx}=\frac{d}{dx}(2x^2+3)=4x#

Now, the surface area of solid generated by revolving curve: #y=2x^2+3# around the x-axis is given by using integration with proper limits

#=\int 2\pi y\ ds#

#=\int_0^4 2\pi(2x^2+3)\sqrt{1+(\frac{dy}{dx})^2}\ dx#

#=2\pi\int_0^4 (2x^2+3)\sqrt{1+(4x)^2}\ dx#

#=2\pi\int_0^4 (2x^2+3)\sqrt{1+16x^2}\ dx#