Problem

Find the surface area of the surface generated by revolving \( y=e^{2 x} \) with \( 0 \leq x \leq 1 \) about the \( x \)-axis.

Solution

Step 1 :\int_{0}^{1} 2\pi y \sqrt{1+{\left( \frac{dy}{dx} \right)}^2} dx

Step 2 :\frac{dy}{dx} = 2e^{2 x}

Step 3 :1 + {\left(\frac{dy}{dx}\right)}^2 = 1 + 4e^{4 x}

Step 4 :\int_{0}^{1} 2\pi e^{2 x} \sqrt{1+4e^{4 x}} dx

From Solvely APP
Source: https://solvelyapp.com/problems/8568/

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