Problem 18 - Calculus Early Transcendentals

Question

Find the area of the surface generated when the given curve is revolved about the \(y\) -axis. $$y=\frac{x^{2}}{4}, \text { for } 2 \leq x \leq 4$$

Step-by-Step Solution

Answer

Answer: The area of the surface generated is \(42\pi\) square units.

Step 1: Rewrite the function in terms of y

We need to rewrite the given curve \(y=\frac{x^2}{4}\) as a function of \(y\). We can do this by solving for \(x\) in terms of \(y\).$$x = \sqrt{4y}$$Now we have the function \(x = f(y)\).

Step 2: Compute the derivative dx/dy

We will find the derivative \(\frac{dx}{dy}\) of the function \(x = \sqrt{4y}\).$$ \frac{dx}{dy} = \frac{d}{dy}\sqrt{4y} = \frac{1}{2}(4y)^{-\frac{1}{2}}(4) = \frac{2}{\sqrt{4y}} $$

Step 3: Plug the function and its derivative into the surface area formula

Now we plug the rewritten function \(x = f(y) = \sqrt{4y}\) and its derivative \(\frac{dx}{dy} = \frac{2}{\sqrt{4y}}\) into the surface area formula.$$S = 2\pi \int_a^b x\sqrt{1+\left(\frac{dx}{dy}\right)^2} dy = 2\pi\int_a^b \sqrt{4y}\sqrt{1+\left(\frac{2}{\sqrt{4y}}\right)^2} dy $$

Step 4: Evaluate the definite integral

We need to evaluate the integral on the interval given in the initial equation \(y=\frac{x^2}{4}\) and corresponding to \(2\le x\le 4\). We will substitute limits of \(y\) as the lower limit and upper limit of the integral according to the given interval. First, let's find the lower and upper limits of \(y\), For \(x=2\), $$y = \frac{2^2}{4} = 1$$ For \(x=4\), $$y = \frac{4^2}{4} = 4$$ So, our integral is: $$S = 2\pi\int_1^4 \sqrt{4y}\sqrt{1+\left(\frac{2}{\sqrt{4y}}\right)^2} dy $$ Now, we can simplify the expression inside the integral: $$S = 2\pi\int_1^4 \sqrt{4y}\sqrt{1+\frac{4}{4y}} dy = 2\pi\int_1^4 \sqrt{4y}(1+\frac{1}{y}) dy $$ Next, we evaluate the integral: $$S = 2\pi\left[\int_1^4 2y dy + \int_1^4 2 dy\right]$$ Evaluate the two separate integrals: $$S = 2\pi\left[\left[y^2\right]_1^4 + 2\left[y\right]_1^4\right]$$ Substitute the limits and compute the final result: $$S = 2\pi\left[(4^2 - 1^2) + 2(4 - 1)\right] = 2\pi(15+6) = 42\pi$$ So the area of the surface generated when the curve \(y=\frac{x^2}{4}\) is revolved about the y-axis for \(2\le x\le 4\) is \(42\pi\) square units.