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


d. Let f(x)=12x^2.. The area of the surface generated when the graph of f on [−4, 4] is revolved about the y-axis is twice the area of the surface generated when the graph of f on [0, 4] is revolved about the y-axis.

Verified step by step guidance

1. Identify the function and the interval: We have \( f(x) = 12x^2 \) and two intervals: \([-4,4]\) and \([0,4]\). We are revolving the graph about the y-axis to find the surface area generated.
2. Recall the formula for the surface area of a curve revolved about the y-axis: \[ S = \int_a^b 2\pi x \sqrt{1 + \left(\frac{df}{dx}\right)^2} \, dx \]. Here, \(x\) is the radius from the y-axis, and \(f(x)\) is the height.
3. Compute the derivative \( \frac{df}{dx} \) for \( f(x) = 12x^2 \): \[ \frac{df}{dx} = 24x \]. Substitute this into the surface area integral.
4. Set up the two surface area integrals: - For \([-4,4]\): \[ S_1 = \int_{-4}^4 2\pi x \sqrt{1 + (24x)^2} \, dx \] - For \([0,4]\): \[ S_2 = \int_0^4 2\pi x \sqrt{1 + (24x)^2} \, dx \].
5. Analyze the integrand's behavior over \([-4,4]\): Since \(x\) appears as a factor and \(x\) is negative on \([-4,0)\), the integrand is negative there, so the integral over \([-4,4]\) is not simply twice the integral over \([0,4]\). This suggests the statement is false because the surface area integral is not symmetric about zero due to the \(x\) term.

Key Concepts

Surface Area of Revolution

The surface area generated by revolving a curve around an axis is found using an integral formula involving the function, its derivative, and the radius from the axis of revolution. For revolution about the y-axis, the radius is the x-value, and the formula accounts for the curve's length and distance from the axis.

Symmetry of Functions and Intervals

When a function is even (symmetric about the y-axis), the graph on [−a, a] is symmetric. This symmetry affects integrals over symmetric intervals, often allowing simplification by doubling the integral from [0, a]. However, for surface areas revolving around an axis, the radius term may break this symmetry.

Effect of Radius in Surface Area Integrals

In surface area calculations revolving around the y-axis, the radius is the x-coordinate, which is negative on [−a, 0] and positive on [0, a]. Since radius appears as an absolute value or squared term, the contribution from negative x-values may differ, impacting whether the surface area over [−a, a] is exactly twice that over [0, a].