Problem 17 - Calculus Early Transcendentals

Question

Find the area of the surface generated when the given curve is revolved about the \(y\) -axis. $$y=(3 x)^{1 / 3}, \text { for } 0 \leq x \leq \frac{8}{3}$$

Step-by-Step Solution

Answer

Answer: The area of the surface generated is \(2\pi\left[6 - \ln(8)\right]\).

Step 1: Find the derivative of f(x)

To find the derivative of \(f(x) = (3x)^{1/3}\), we will apply the power rule for differentiation, which gives us: $$f'(x)=\frac{d}{dx}(3x)^{1/3} = \frac{1}{3}(3x)^{-2/3}(3)$$

Step 2: Plug the derivative into the formula

Now we plug our original function \(f(x)\) and its derivative \(f'(x)\) into the surface area formula: $$S = 2 \pi \int_0^{\frac{8}{3}} (3x)^{1/3} \sqrt{1 + [(\frac{1}{3}(3x)^{-2/3}(3))^2]} dx$$

Step 3: Simplify the integrand

We now simplify the integrand by squaring the derivative and adding 1: $$1 + [(\frac{1}{3}(3x)^{-2/3}(3))^2] = 1 + (\frac{9}{9x^{4/3}})=\frac{9x^{4/3} + 9}{9x^{4/3}} = [1+(3x)^{-4/3}]$$ Thus, the integral becomes: $$S = 2 \pi\int_0^{\frac{8}{3}}(3x)^{1/3} [1+(3x)^{-4/3}] dx$$

Step 4: Evaluate the integral

Now, we evaluate the integral by breaking it into two parts: $$S = 2 \pi \left[\int_0^{\frac{8}{3}}(3x)^{1/3} dx + \int_0^{\frac{8}{3}}(3x)^{1 / 3}(3x)^{-4 / 3} dx\right]$$ For the first integral, we can use a simple substitution \(u=3x\) and proceed with the integration. For the second integral, notice how the powers combine to make a simpler function: $$(3x)^{1 / 3}(3x)^{-4 / 3} = (3x)^{1/3-4/3} = (3x)^{-1} = \frac{1}{3x}$$ Now we can evaluate both integrals: $$S = 2\pi\left[\int_0^{\frac{8}{3}}(3x)^{1/3} dx + \int_0^{\frac{8}{3}}\frac{1}{3x} dx\right] = 2\pi\left[\frac{3}{4}(3x)^{4/3}\Bigg|_0^{\frac{8}{3}} - \ln(3x)\Bigg|_0^{\frac{8}{3}}\right]$$

Step 5: Calculate the surface area

Now we plug in our limits of integration and calculate the surface area: $$S = 2\pi\left[\frac{3}{4}(3\left(\frac{8}{3}\right))^{4/3} - \ln (3(\frac{8}{3})) - \frac{3}{4}(3(0))^{4/3} + \ln(3(0))\right] = 2\pi\left[\frac{3}{4}(8) - \ln(8)\right]$$ $$S = 2\pi\left[6 - \ln(8)\right]$$ So the area of the surface generated when the given curve is revolved about the \(y\)-axis is equal to \(2\pi\left[6 - \ln(8)\right]\).