For more information, please see full course syllabus of Calculus AB

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For more information, please see full course syllabus of Calculus AB

For more information, please see full course syllabus of Calculus AB

### Revolving Solids Cylindrical Shells Method

- : derived from surface area of cylinder =
*r*corresponds to*x**h*corresponds to*y*- Use when shape has a hole
- More difficult to setup, but most effective
- Very similar to washer method

### Revolving Solids Cylindrical Shells Method

Find the volume of the solid that results from revolving the area bounded by y = x

^{2}, y = 0, and x = 4 around the y-axis.- Find the lower bound of the integral (we're already handed the upper bound for x)
- x
^{2}= 0 - x = 0
- V = 2 π∫
_{0}^{4}x (x^{2}) dx - V = 2π[(x
^{4})/4] |_{0}^{4} - V = 2π[(4
^{4})/4]

V = 128π

Find the volume of the solid that results from revolving the area bound by y = [1/x], x = 1, and x = 2 around the y-axis.

- V = 2π∫
_{1}^{2}x[[1/x] − 0] dx - V = 2π∫
_{1}^{2}1 dx - V = 2πx |
_{1}^{2} - V = 2π(2 − 1)

V = 2π

Find the volume of the solid that results from revolving the area bound by y = sinx

^{2}, y = 0, and x = 1 around the y-axis.- x
^{2}= 0 - x = 0
- V = 2 π∫
_{0}^{1}x[sinx^{2}] dx - V = 2 π∫
_{0}^{1}x sinx^{2}dx - u = x
^{2} - du = 2x dx
- V = 2 π[1/2] ∫
_{0}^{1}sinu du - V = π(−cosx
^{2}) |_{0}^{1} - V = π(−cos1 + cos0)
- V = π(1 − cos1)

V = π(1 − cos1)

Find the volume of the solid that results from revolving the area bound by y = [1/2] e

^{x2}, y = 0, x = 0, and x = 1 around the y-axis.- V = 2 π∫
_{0}^{1}x [1/2] e^{x2}dx - u = x
^{2} - du = 2x dx
- V = 2π[1/2] [1/2] ∫
_{0}^{1}e^{u}du - V = [(π)/2] e
^{u}|_{0}^{1} - V = [(π)/2] e
^{x2}|_{0}^{1} - V = [(π)/2] (e
^{1}− e^{0})

V = [(π)/2] (e

^{1}− 1)Find the volume of the solid that results from revolving the area bound by y = √{1 − x

^{2}} and y = 0 from x = 0 to x = 1 around the y-axis.- The function is undefined for x < −1 and x > 1
- Let's find the points where the two curves intersect
- x-axis is y = 0
- √{1 − x
^{2}} = 0 - 1 − x
^{2}= 0 - x
^{2}= 1 - x = −1, 1
- The intersections and all values of x in between are defined
- V = (2π) ∫
_{0}^{1}x √{1 − x^{2}} dx - u = 1 − x
^{2} - du = −2x dx
- V = 2π[(−1)/2] ∫
_{0}^{1}u^{[1/2]}du - V = −π[(u
^{[3/2]})/([3/2])] |_{0}^{1} - V = [(−2π)/3] (1 − x
^{2})^{[3/2]}|_{0}^{1} - V = [(−2π)/3] ( (1 − 1)
^{[3/2]}− 1^{[3/2]}) - We've found the volume of a half-sphere of radius 1

V = [(2 π)/3]

Find the volume of the solid that results from revolving the area bound by y = x

^{3}, y = 1, and x = 2 around the y-axis.- Find the x value where the two curves intersect for the other bound of the integral
- x
^{3}= 1 - x = 1
- V = 2π∫
_{1}^{2}x[x^{3}− 1] dx - V = 2π∫
_{1}^{2}x^{4}− x dx - V = 2π([(x
^{5})/5] − [(x^{2})/2]) |_{1}^{2} - V = 2π([32/5] − 2 − ([1/5] − [1/2]))

V = [(47π)/5]

Find the volume of the solid that results from revolving the area in between by y = [10 lnx/(x

^{2})] and y = 0 from x = 1 to x = 6 around the y-axis.- V = 2 π∫
_{1}^{6}x [[10 lnx/(x^{2})]] dx - V = 20 π∫
_{1}^{6}[lnx/x] dx - u = lnx
- du = [1/x] dx
- V = 20 π∫
_{1}^{6}u du - V = 20π[(u
^{2})/2] |_{1}^{6} - V = 20π[((lnx)
^{2})/2] |_{1}^{6} - V = 10 π((ln6)
^{2}− (ln1)^{2})

V = 10 π(ln6)

^{2}Find the volume of the solid that results from revolving the area bound by y = [1/(√{4 − x

^{2}})], x = 0, x = 1 and y = 0 around the y-axis.- V = 2 π∫
_{0}^{1}x[[1/(√{4 − x^{2}})]] dx - u = 4 − x
^{2} - du = −2x dx
- V = 2 π[(−1)/2] ∫
_{0}^{1}[1/(√u)] du - V = −π∫
_{0}^{1}u^{−[1/2]}du - V = −π[1/([1/2])] u
^{[1/2]}|_{0}^{1} - V = −2 π(4 − x
^{2})^{[1/2]}|_{0}^{1} - V = −2π(√3 − √4)

V = 2π(2 − √3)

Find the volume of the solid that results from revolving the area bound by y = 2x, y = 0, and x = 4 around the x-axis.

- Let's rewrite this in terms of x, since we're rotating about the x-axis
- y = 2x
- x = [1/2] y
- Find the bounds of the integral
- [1/2] y = 4
- y = 8
- V = 2 π∫
_{0}^{8}y[4 − [1/2] y] dy - V = 2 π∫
_{0}^{8}4y − [1/2] y^{2}dy - V = 2 π(2y
^{2}− [(y^{3})/6]) |_{0}^{8} - V = 2 π(2(64) − [(8
^{3})/6] − 0) - V = 2 π(128 − [512/6])

V = [(256π)/3]

Find the volume of the solid that results from revolving the area bound by x = y

^{3}and x = y^{2}around the x-axis.- y
^{3}= y^{2} - y = 0, 1
- V = 2 π∫
_{0}^{1}y[y^{2}− y^{3}] dy - V = 2 π∫
_{0}^{1}y^{3}− y^{4}dy - V = 2 π([(y
^{4})/4] − [(y^{5})/5]) |_{0}^{1} - V = 2 π([1/4] − [1/5] − 0)

V = [(π)/10]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Revolving Solids Cylindrical Shells Method

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Revolving Solids: Cylindrical Shells Method 0:09
- Volume Of A Solid
- Formula
- Example 1 2:56
- Example 2 7:28
- Example 3 11:39
- Example 4 17:36
- Example 5 21:45

0 answers

Post by Pedro Valdericeda on July 15, 2014

In the last examaple it is not 2x-x the height?

0 answers

Post by Arshin Jain on March 15, 2014

Correction: Example 1 - It is x^2 instead of (2/3)x^3. [7:05 min].

0 answers

Post by William Dawson on March 24, 2013

the explanation at 16:00 of why (3-x) works for the radius when x=6, which would make that radius -3, was an inadequate explanation.