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.

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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.

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.

Compute the derivative \( \frac{df}{dx} \) for \( f(x) = 12x^2 \): \[ \frac{df}{dx} = 24x \]. Substitute this into the surface area integral.

Set up the two surface area integrals: - For \([-4,4]\): \[ S_1 = \int_{-4}^4 2\pi x \sqrt{1 + (24x)^2} \, dx \] - For \([0,4]\): \[ S_2 = \int_0^4 2\pi x \sqrt{1 + (24x)^2} \, dx \].

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.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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].