Determine whether the following statements are true and give an explanation or counterexample. 


a. If the curve y=f(x) on the interval [a, b] is revolved about the y-axis, the area of the surface generated is ∫f(b)f(a) 2πf(y)√1+f′(y)^2 dy.

Verified step by step guidance

1. Step 1: Understand the problem context. The problem involves finding the surface area generated by revolving a curve about the y-axis. The curve is given by y = f(x) on the interval [a, b]. We need to verify if the given integral expression for the surface area is correct.
2. Step 2: Recall the formula for the surface area when revolving a curve about the y-axis. If the curve is expressed as x = g(y) (i.e., x as a function of y), then the surface area S is given by: \(S = \int_{c}^{d} 2\pi x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy\) where [c, d] is the interval for y.
3. Step 3: Note that the problem gives the curve as y = f(x), so to use the formula revolving around the y-axis in terms of y, we need to express x as a function of y, i.e., find the inverse function x = f^{-1}(y). Then, the derivative \(\frac{dx}{dy}\) is the derivative of the inverse function.
4. Step 4: Compare the given integral \(\int_{f(b)}^{f(a)} 2\pi f(y) \sqrt{1 + (f'(y))^2} \, dy\) with the correct formula. Notice that the integrand uses \(f(y)\) and \(f'(y)\), but since f is originally a function of x, \(f(y)\) and \(f'(y)\) do not make sense unless f is inverted. This suggests the given integral is not correctly formulated.
5. Step 5: Conclusion: The given integral is not the correct formula for the surface area generated by revolving y = f(x) about the y-axis. The correct approach requires expressing x as a function of y and using the formula involving \(x\) and \(\frac{dx}{dy}\). Therefore, the statement is false.

Key Concepts

Surface Area of Revolution

The surface area generated by revolving a curve around an axis is found by integrating the circumference of circular slices times the arc length element. For revolution about the y-axis, the formula involves integrating 2π times the radius (x or function of y) multiplied by the differential arc length along y.

Parametrization and Variable of Integration

When revolving a curve about the y-axis, the integral is typically expressed in terms of y, requiring the function to be rewritten as x = g(y). The limits of integration correspond to y-values, and the derivative inside the integral must be with respect to y, not x.

Arc Length Element in Terms of y

The differential arc length element ds is given by √(1 + (dx/dy)^2) dy when integrating with respect to y. Using f'(y) instead of dx/dy or mixing variables leads to incorrect formulas, so careful differentiation and substitution are essential.