Find the area of the surface generated when the given curve is revolved about the given axis. $y=6 x+2$, for $0 \leq x \leq 2$; about the $x$-axis
The surface area is square units.
(Type an exact answer in terms of $\pi$.)
Answer
Final Answer: The area of the surface generated when the curve \(y=6x+2\) for \(0 \leq x \leq 2\) is revolved about the x-axis is \(\boxed{32\sqrt{37}\pi}\) square units.
Step 1 ：We are given the curve \(y=6x+2\) for \(0 \leq x \leq 2\) and asked to find the area of the surface generated when this curve is revolved about the x-axis.
Step 2 ：The formula for the surface area of a solid of revolution, when rotating a curve \(y=f(x)\) from \(x=a\) to \(x=b\) around the x-axis, is given by: \(A = 2\pi \int_{a}^{b} y \sqrt{1 + (y')^2} dx\), where \(y'\) is the derivative of \(y\) with respect to \(x\).
Step 3 ：In this case, \(f(x) = 6x + 2\), \(a = 0\), \(b = 2\), and the derivative \(f'(x) = 6\).
Step 4 ：So, we need to calculate the integral: \(A = 2\pi \int_{0}^{2} (6x + 2) \sqrt{1 + 6^2} dx\).
Step 5 ：By calculating the integral, we find that the area \(A = 32\sqrt{37}\pi\).
Step 6 ：Final Answer: The area of the surface generated when the curve \(y=6x+2\) for \(0 \leq x \leq 2\) is revolved about the x-axis is \(\boxed{32\sqrt{37}\pi}\) square units.
