Question

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

Answer 1

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`