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

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?

Cameron Alexander · Accepted Answer

Hint is given below

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#

Luke Alvarez · Answer

undefined To find the surface area of the solid created by revolving the function $ f(x) = 2x^2 + 3 $ around the x-axis over the interval $[1, 4]$, we'll use the formula for surface area of a solid of revolution:

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

where $ f(x) $ is the function being rotated, $ f'(x) $ is its derivative, and $ [a, b] $ is the interval of rotation.

First, we need to find $ f'(x) $, which is $ f'(x) = 4x $.

Now we can plug the function and its derivative into the formula and integrate over the interval $[1, 4]$:

$$ S = 2\pi \int_1^4 (2x^2 + 3) \sqrt{1 + (4x)^2} \, dx $$

After integration, the result will be the surface area of the solid of revolution.