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

Hint is given below

#### Explanation:

HINT: The given function: $y = 2 {x}^{2} + 3$

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{d}{\mathrm{dx}} \left(2 {x}^{2} + 3\right) = 4 x$

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

$= \setminus \int 2 \setminus \pi y \setminus \mathrm{ds}$

$= \setminus {\int}_{0}^{4} 2 \setminus \pi \left(2 {x}^{2} + 3\right) \setminus \sqrt{1 + {\left(\setminus \frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \setminus \mathrm{dx}$

$= 2 \setminus \pi \setminus {\int}_{0}^{4} \left(2 {x}^{2} + 3\right) \setminus \sqrt{1 + {\left(4 x\right)}^{2}} \setminus \mathrm{dx}$

$= 2 \setminus \pi \setminus {\int}_{0}^{4} \left(2 {x}^{2} + 3\right) \setminus \sqrt{1 + 16 {x}^{2}} \setminus \mathrm{dx}$