Presentation is loading. Please wait.

Presentation is loading. Please wait.

Shell Method for Volumes

Similar presentations


Presentation on theme: "Shell Method for Volumes"— Presentation transcript:

1 Shell Method for Volumes
Lesson 6-3a Shell Method for Volumes

2 Quiz AP Problem: The area of the region enclosed by the graphs of y = x and y = x² - 3x + 3 is Reading questions: a) What is the area of a “shell”? b) If a disc or washer is perpendicular to the axis of rotation, then the shell is …. ? 4/3 sq units A = 2πrh parallel to the axis of rotation

3 Objectives Find volume of revolved area solids using cylindrical shells

4 Vocabulary None new

5 Volume = ∑ Area • thickness (∆variable)
Shell Method An alternative to discs and washers is the shell method. The cylinder is rolled out and represents the area. Again the variable of integration can be either x or y. Volume = ∑ Area • thickness (∆variable) Sometimes it gives us an easy approach to solving the problem. dx 2πr h h circumference dx Volume = ∑ Cylinder Area • thickness = 2π ∫ r(x) h(x) dx x = a x = b

6 Comparison of Disk/Washers and Shell Methods
For the disc method: Representative rectangle is always perpendicular to the axis of revolution Rectangle represents the radius of the disc and its thickness Area is circle (or washer) = πr² (or π[R² - r²]) For the shell method: Representative rectangle is always parallel to the axis of revolution Rectangle represents the height and thickness of the cylinder Area is side of the cylinder rolled out = 2πrh Usually r is either x or y and h is a function of the variable of integration

7

8 6- 3 Example 1 ∆Volume = Area • Thickness Area = cylinders! = 2πrh
r = x ∆Volume = Area • Thickness Area = cylinders! = 2πrh = 2π (x) (x-½) = 2π x1/2 Thickness = ∆x X ranges from 1 out to 4 Area = L∙W Volume = ∫ (2π x1/2) dx x = 1 x = 4 = 2π ∫ (x1/2) dx = 2π ((⅔)x3/2) | = 2π (((⅔)(8)) – (2/3)) = (4π/3) (8 – 1) = 28π/3 = x = 1 x = 4 8

9 6- 3 Example 2 ∆Volume = Area • Thickness Area = cylinders! = 2πrh
= 2π (y) (e-y²) = 2π y e-y² Thickness = ∆y X ranges from 0 out to 1 Volume = ∫ (2π y e-y²) dy y = 0 y = 1 let u = -y², so du = -2ydy form: eu du = 2π ∫ (y e-y²) dy = -2π (e-y²) | = -2π ((e-1) – (1)) = -2π ( = π = 3.972 y = 0 y = 1 9

10 Summary & Homework Summary: Homework:
Shell method is sometimes easier to use than either disc or washers Discs or washers are always perpendicular to the axis of rotation Shells are always parallel to the axis of rotation Always use the method that makes the problem simpler to do Homework: pg 458 – 459 Day 1: 3, 6, 9, Day 2: 4, 7, 11, 38


Download ppt "Shell Method for Volumes"

Similar presentations


Ads by Google