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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3:11 minutes

Problem 6.6.25b

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

Verified step by step guidance

Identify the formula for the surface area of a solid of revolution about the x-axis: $$ S = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$

Given the curve is $y = \cos x$ on the interval $0 \leq x \leq \frac{\pi}{2}$, substitute $y$ into the formula: $$ S = \int_0^{\frac{\pi}{2}} 2\pi \cos x \sqrt{1 + \left(\frac{d}{dx}(\cos x)\right)^2} \, dx $$

Compute the derivative $\frac{dy}{dx}$: $$ \frac{d}{dx}(\cos x) = -\sin x $$

Substitute the derivative back into the integral: $$ S = \int_0^{\frac{\pi}{2}} 2\pi \cos x \sqrt{1 + (-\sin x)^2} \, dx = \int_0^{\frac{\pi}{2}} 2\pi \cos x \sqrt{1 + \sin^2 x} \, dx $$

Use a calculator or software to approximate the value of the integral, which will give the surface area of the solid formed by revolving the curve about the x-axis.

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Key Concepts

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

Surface Area of a Solid of Revolution

The surface area generated by revolving a curve around an axis is found using an integral formula. For a curve y = f(x) revolved about the x-axis, the surface area is given by S = ∫ 2πy √(1 + (dy/dx)²) dx over the interval. This formula accounts for the circumference of circular slices and the curve's slope.

Derivative of the Function y = cos x

To apply the surface area formula, the derivative dy/dx is needed. For y = cos x, the derivative is dy/dx = -sin x. This derivative measures the slope of the curve at each point, which affects the length of the surface element in the integral.

Numerical Approximation Using Calculators or Software

When the integral for surface area cannot be solved analytically or is complex, numerical methods like Simpson's rule or trapezoidal rule are used. Calculators or software approximate the integral value by evaluating the function at discrete points, providing an accurate estimate of the surface area.

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Related Practice

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

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.

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

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Textbook Question

9–20. Arc length calculations Find the arc length of the following curves on the given interval.

x = 2y−4, for −3≤y≤4 (Use calculus, but check your work using geometry.)

Textbook Question

9–20. Arc length calculations Find the arc length of the following curves on the given interval.

x = 2e^√2y + 1/16e^−√2y, for 0 ≤ y ≤ ln²/√2

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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

Key Concepts

Surface Area of a Solid of Revolution

Derivative of the Function y = cos x

Numerical Approximation Using Calculators or Software

Watch next

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