Problem 5 - Calculus Early Transcendentals

Question

Find the area of the surface generated when the given curve is revolved about the \(x\) -axis. $$y=3 x+4 \text { on }[0,6]$$

Step-by-Step Solution

Answer

Answer: The area of the surface generated is $$108\pi\sqrt{10}$$ square units.

Step 1: Find the derivative of the curve y = 3x + 4

In this problem, we are given the equation $$y = 3x + 4$$, so we'll find the derivative, $$\frac{dy}{dx}$$. Since this is a simple linear function, the derivative is just the slope: $$\frac{dy}{dx} = 3$$.

Step 2: Plug in the values into the surface area formula

Using the formula for the surface area of revolution: $$A = 2\pi \int_a^b y\sqrt{1+(\frac{dy}{dx})^2}dx$$ We will plug in the values we found for $$y$$, $$\frac{dy}{dx}$$, and the given interval $$[0, 6]$$: $$A = 2\pi \int_0^6 (3x + 4)\sqrt{1 + (3)^2} dx$$

Step 3: Simplify the integral expression

The integral expression is as follows: $$A = 2\pi \int_0^6 (3x + 4) \sqrt{1 + 9} dx = 2 \pi \int_0^6 (3x + 4) \sqrt{10} dx $$ Since $$\sqrt{10}$$ is a constant, we can pull it out of the integral: $$A = 2\pi\sqrt{10}\int_0^6 (3x + 4) dx$$

Step 4: Evaluate the integral

Now, we can evaluate the integral: $$A = 2\pi\sqrt{10} \left[\frac{3}{2}x^2 + 4x\right]_0^6$$ Plug in the upper limit, 6: $$ A = 2\pi\sqrt{10} \left[\frac{3}{2}(6)^2 + 4(6)\right] $$ Now plug in the lower limit, 0: $$ A = 2\pi\sqrt{10}\left(\frac{3}{2}(36) + 24\right) $$

Step 5: Solve for the area A

Now, we just need to solve for A: $$ A = 2\pi\sqrt{10} (54) = 108\pi\sqrt{10} $$ The area of the surface generated by revolving the curve $$y=3x+4$$ on the interval $$[0, 6]$$ about the $$x$$-axis is $$108\pi\sqrt{10}$$ square units.