Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

9. Graphical Applications of Integrals

Introduction to Volume & Disk Method

Calculus

9. Graphical Applications of Integrals

Introduction to Volume & Disk Method

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1:49 minutes

Problem 6.6.23d

Textbook Question

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

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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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].

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Textbook Question

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.

Textbook Question

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

b. If f is not one-to-one on the interval [a, b], then the area of the surface generated when the graph of f on [a, b] is revolved about the x-axis is not defined.

Textbook Question

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

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

Textbook Question

Consider the following curves on the given intervals.

b. Use a calculator or software to approximate the surface area.

y=cos x, for 0≤x≤π/2; about the x-axis

Textbook Question

Consider the following curves on the given intervals.

a. Write the integral that gives the area of the surface generated when the curve is revolved about the given axis.

y=tan x , for 0≤x≤π/4; about the x-axis

Textbook Question

Consider the following curves on the given intervals.

b. Use a calculator or software to approximate the surface area.

y=tan x , for 0≤x≤π/4; about the x-axis