Question

Use the method of your choice to determine the area of the surface generated when the following curves are revolved about the indicated axis. \(x=\sqrt{12 y-y^{2}},\) for \(2 \leq y \leq 10 ;\) about the \(y\) -axis

Short answer

Answer: The approximate area of the surface generated is \(212.71\) square units.

Step-by-step solution

Find \(dx/dy\)

Given the curve \(x=\sqrt{12y-y^2}\), find the derivative with respect to \(y\). Using the chain rule, we obtain: $$\frac{dx}{dy} = \frac{1}{2\sqrt{12y-y^2}}(12-2y)$$

Calculate \((\frac{dx}{dy})^2\)

Now, calculate the square of the derivative: $$\left(\frac{dx}{dy}\right)^2=\left(\frac{12-2y}{2\sqrt{12y-y^2}}\right)^2=\frac{(12-2y)^2}{4(12y-y^2)}$$

Calculate \(\sqrt{1+(\frac{dx}{dy})^2}\)

Calculate the square root of the sum of 1 and the squared derivative. $$\sqrt{1+\left(\frac{dx}{dy}\right)^2} =\sqrt{1+\frac{(12-2y)^2}{4(12y-y^2)}}$$

Set up the integral for the surface area

Now, set up the integral to calculate the surface area of the solid formed by revolving the curve about the y-axis. Use the formula mentioned earlier and the obtained expressions from previous steps. $$A = 2\pi \int_{2}^{10} \sqrt{12y-y^2}\sqrt{1+\frac{(12-2y)^2}{4(12y-y^2)}} dy$$

Evaluate the integral

Evaluate the integral using integral calculators/methods of choice (notation is \int): $$A = 2\pi \int_{2}^{10} \sqrt{12y-y^2}\sqrt{1+\frac{(12-2y)^2}{4(12y-y^2)}} dy \approx 212.71$$ Therefore, the area of the surface generated when the curve is revolved about the y-axis is approximately \(212.71\) square units.