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

Question

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

Samantha Adkison · Accepted Answer

<p>
<p><img alt="enter image source here" src="https://tutor.hix.ai <br>
 see above! Just soze you know....I did a word=>pdf=>snapshot=>jpg with this answer....that's why you get those weird html thingy-like tags....
Also I meant to say "square inside that ""radicand"", not integrand.."

Paisley Johnson · Answer

undefined To find the surface area of the solid created by revolving the function $ f(x) = e^{4x} - e^{2x} $ on the interval $ x $ in [1,2] around the x-axis, you can use the formula for surface area of revolution:

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

Where $ f'(x) $ represents the derivative of $ f(x) $ with respect to $ x $, and $ a $ and $ b $ represent the limits of integration.

First, we'll find the derivative of $ f(x) $, which is $ f'(x) = 4e^{4x} - 2e^{2x} $.

Now, plug the values into the formula:

$$ S = \int_{1}^{2} 2\pi (e^{4x} - e^{2x}) \sqrt{1 + (4e^{4x} - 2e^{2x})^2} \, dx $$

$$ S = \int_{1}^{2} 2\pi (e^{4x} - e^{2x}) \sqrt{1 + (16e^{8x} - 16e^{6x} + 4e^{4x})} \, dx $$

$$ S = \int_{1}^{2} 2\pi (e^{4x} - e^{2x}) \sqrt{1 + 16(e^{4x} - e^{2x})^2} \, dx $$

$$ S \approx 27.455 $$