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For more information, please see full course syllabus of College Calculus: Level I

For more information, please see full course syllabus of College Calculus: Level I

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### Volume by Method of Cylindrical Shells

- I highly recommend starting with a sketch of the functions involved and drawing in a sample radius!
- I recommend memorizing the formulas but also understanding the geometry that leads to the formulas.
- If an interval is not given, you may need to set the two functions equal in order to determine the interval involved.
- Sometimes, geometrical considerations can help you to check your results or even provide a non-calculus way to finish a problem.

### Volume by Method of Cylindrical Shells

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

### Volume by Method of Cylindrical Shells

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
- Important Equations 0:50
- Equation 1: Rotation about y-axis
- Equation 2: Rotation about y-axis (2 curves)
- Equation 3: Rotation about x-axis
- Lecture Example 1 8:57
- Lecture Example 2 14:26
- Lecture Example 3 18:15
- Additional Example 4
- Additional Example 5

0 answers

Post by Meghan Brunet on April 25, 2013

"=...=" is mind blowing and not very helpful.

0 answers

Post by Joanna Nicholson on February 16, 2013

This is a very good question, I asked it myself. I wish someone would answer the questions....

0 answers

Post by Jacqueline Measer on July 11, 2012

Why do you draw two different cylindrical shells? In class only one was drawn.