Volume Calculation through Parallel Cross Section and Shell Method, Exams of Calculus

Examples and formulas for calculating the volume of solid figures using the parallel cross section and shell methods. Topics include disc method, cylinder volume, washer method, and cylindrical shell volume. Examples cover cones, spheres, and various regions of revolution.

Typology: Exams

Pre 2010

Uploaded on 08/19/2009

koofers-user-ji6
koofers-user-ji6 🇺🇸

10 documents

1 / 11

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Lecture 22Section 6.2 Volume by Parallel Cross Section
Section 6.3 Volume by the Shell Method
Jiwen He
Test 3
Test 3: Dec. 4-6 in CASA
Material - Through 6.3.
Final Exam
Final Exam: Dec. 14-17 in CASA
Review for Test 3
Review for Test 3 by the College Success Program.
Friday, November 21 2:30–3:30pm in the basement of the library by the
C-site.
Online Quizzes
Online Quizzes are available on CourseWare.
Quiz 1
What is today?
a. Monday
b. Wednesday
c. Friday
d. None of these
1
pf3
pf4
pf5
pf8
pf9
pfa

Partial preview of the text

Download Volume Calculation through Parallel Cross Section and Shell Method and more Exams Calculus in PDF only on Docsity!

Lecture 22Section 6.2 Volume by Parallel Cross Section

Section 6.3 Volume by the Shell Method

Jiwen He

Test 3

  • Test 3: Dec. 4-6 in CASA
  • Material - Through 6.3.

Final Exam

  • Final Exam: Dec. 14-17 in CASA

Review for Test 3

  • Review for Test 3 by the College Success Program.
  • Friday, November 21 2:30–3:30pm in the basement of the library by the C-site.

Online Quizzes

  • Online Quizzes are available on CourseWare.

Quiz 1 What is today?

a. Monday

b. Wednesday

c. Friday

d. None of these

1 Volume by Parallel Cross Section; Discs and

Washers

1.1 Solid of Revolution: Disk Method

Solid of Revolution About the x-Axis: Disk

Cylinder Volume: π[f (x∗ i )]^2 ∆xi Riemann Sum:

π[f (x∗ i )]^2 ∆xi

V =

∫ (^) b

a

π[f (x)]^2 dx = lim ‖P ‖→ 0

π[f (x∗ i )]^2 ∆xi.

Example Example 1. Find the volume of the cone shown in the figure below.

Example Example 2. Find the volume of a sphere of radius r by revolving about the x-axis the region below the graph of

f (x) =

r^2 − x^2 , −r ≤ x ≤ r.

1.2 Solid of Revolution: Washer Method

Solid of Revolution About the x-Axis: Washer

Cylinder Volume:∑ π([f (x∗ i )]^2 − [g(x∗ i )]^2 )∆xi [1ex] Riemann Sum: π([f (x∗ i )]^2 − [g(x∗ i )]^2 )∆xi [1ex]

V =

∫ (^) b

a

π([f (x)]^2 − [g(x)]^2 ) dx = lim ‖P ‖→ 0

π([f (x∗ i )]^2 − [g(x∗ i )]^2 )∆xi

Solid of Revolution About the y-Axis: Washer Cylinder Volume:∑ π([F (y∗ i )]^2 − [G(y∗ i )]^2 )∆yi [1ex] Riemann Sum: π([F (y∗ i )]^2 − [G(y i∗ )]^2 )∆yi [1ex]

V =

∫ (^) d

c

π([F (y)]^2 − [G(y)]^2 ) dy = lim ‖P ‖→ 0

π([F (y∗ i )]^2 − [G(y i∗ )]^2 )∆yi

Example Example 4. Find the volume of the solid generated by revolving the region between y = x^2 and y = 2x about the x-axis.

Example Example 5. Find the volume of the solid generated by revolving the region between y = x^2 and y = 2x about the y-axis.

2 Volume by the Shell Method

2.1 Solid of Revolution: Shell Method

Volume of a Cylindrical Shell

Volume of a Cylindrical Shell

Solid of Revolution About the y-Axis: Shell

V =

∫ (^) b

a

2 π x [f (x) − g(x)] dx = lim ‖P ‖→ 0

2 π x∗ i [f (x∗ i ) − g(x∗ i )]∆xi.

The integrand 2π x [f (x) − g(x)] is the lateral area of the cylinder.

Solid of Revolution About the x-Axis: Shell

V =

∫ (^) d

c

2 π y [F (y) − G(y)] dy = lim ‖P ‖→ 0

2 π y∗ i [F (y∗ i ) − G(y∗ i )]∆yi.

The integrand 2π y [F (y) − G(y)] is the lateral area of the cylinder.

Example

Example 8. Find the volume of the solid generated by revolving about the x-axis the region between y = x^2 and y = 2x.

Example Example 9. A round hole of radius r is drilled through the center of a hemisphere of radius a. Find the volume of the potion of the hemisphere that remains.

Example Example 10. The region Ω between y =

x and y = x^2 , 0 ≤ x ≤ 1, is revolved about the line x = −2. Find the volume of the solid that is generated.