Question: Find the area of the surface generated when the given curve is revolved about the given axis. \[…

Find the area of the surface generated when the given curve is revolved about the given axis.

\[ y = 7x + 2, \text{ for } 0 \leq x \leq 2; \text{ about the x-axis} \]

Solution

To find the area of the surface generated when the curve is revolved about the x-axis, we use the formula for the surface area of revolution: \[ A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] First, we identify \( y \) and the given bounds: \( y = 7x + 2 \) with \( 0 \leq x \leq 2 \). Next, find the derivative \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = 7 \] Substitute \(\frac{dy}{dx}\) into the surface area formula: \[ A = 2\pi \int_{0}^{2} (7x + 2) \sqrt{1 + (7)^2} \, dx \] Simplify \(\sqrt{1 + 49}\): \[ \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2} \] Now, substitute back: \[ A = 2\pi \int_{0}^{2} (7x + 2) \cdot 5\sqrt{2} \, dx \] \[ A = 10\pi\sqrt{2} \int_{0}^{2} (7x + 2) \, dx \] Evaluate the integral: \[ \int_{0}^{2} (7x + 2) \, dx = \left[ \frac{7}{2}x^2 + 2x \right]_{0}^{2} \] Calculate the definite integral: \[ = \left( \frac{7}{2}(2)^2 + 2(2) \right) - \left( \frac{7}{2}(0)^2 + 2(0) \right) \] \[ = \left( \frac{7}{2} \cdot 4 + 4 \right) \] \[ = (14 + 4) \] \[ = 18 \] Finally, multiply by \( 10\pi\sqrt{2} \): \[ A = 10\pi\sqrt{2} \cdot 18 \] \[ A = 180\pi\sqrt{2} \] The area of the surface generated is \( 180\pi\sqrt{2} \).