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

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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

### Volume by Method of Disks and Washers

- 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 Disks and Washers

Find the volume of the solid that results from revolving y = x from x = 0 to x = 4 around the x-axis.

- The area of a slice would be πr
^{2}. The volume would be πr^{2}* thickness - So the volume of a tiny slice would be πr
^{2}dx where dx is the thickness - In this problem, we're rotating around the x-axis, so the radius is just y in this case.
- Volume of slice = πy
^{2}dx = πx^{2}dx - Total volume from x = 0 to x = 4 is V
- V = ∫
_{0}^{4}πx^{2}dx - V = π[(x
^{3})/3] |_{0}^{4} - V = π[64/3] − 0
- We just found the volume of a cone using calculus!

V = [(64 π)/3]

Find the volume of the solid that results from revolving x = 4 from y = 0 to y = 6 around the y-axis.

- V = π∫
_{0}^{6}4^{2}dy - V = π16x |
_{0}^{6} - V = 96π− 0
- Now we found the volume of a cylinder.

V = 96π

Find the volume of the solid that results from revolving y = secx from x = −[(π)/4] to x = [(π)/4] around the x-axis.

- V = π∫
_{−[(π)/4]}^{[(π)/4]}sec^{2}x dx - V = πtanx |
_{−[(π)/4]}^{[(π)/4]} - V = π(tan[(π)/4] − tan[(−π)/4])
- V = π(1 − (−1))

V = 2π

Find the volume of the solid that results from revolving y = lnx from y = 0 to y = 1 about the y-axis.

- V = π∫
_{0}^{1}x^{2}dy - y = lnx
- e
^{y}= e^{lnx} - x = e
^{y} - V = π∫
_{0}^{1}e^{2y}dy - u = 2y
- du = 2 dy
- V = [(π)/2] ∫
_{0}^{1}e^{u}du - V = [(π)/2] e
^{u}|_{0}^{1} - V = [(π)/2] e
^{2y}|_{0}^{1} - V = [(π)/2] (e
^{2}− e^{0})

V = [(π)/2] (e

^{2}− 1)Find the volume of the solid that results from revolving y = [lnx/(√x)] from x = 1 to x = e about the x-axis.

- V = π∫
_{1}^{e}[((lnx)^{2})/x] dx - u = lnx
- du = [1/x] dx
- V = π∫
_{1}^{e}u^{2}du - V = π[(u
^{3})/3] |_{1}^{e} - V = π[((lnx)
^{3})/3] |_{1}^{e} - V = [(π)/3] ( (lne)
^{3}− (ln1)^{3})

V = [(π)/3]

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

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

V = [(π)/2] (e

^{4}− 5)Find the volume of the solid that results from revolving y = x

^{2}+ 1 from x = 0 to x = 1 around the x-axis.- V = π∫
_{0}^{1}(x^{2}+ 1)^{2}dx - V = π∫
_{0}^{1}x^{4}+ 2x^{2}+ 1 dx - V = π([(x
^{5})/5] + [(2x^{3})/3] + x) |_{0}^{1} - V = π([1/5] + [2/3] + 1 − 0 − 0 − 0)

V = [(28π)/15]

Find the volume of the solid that results from revolving y = √{4 − x

^{2}} from x = −2 to x = 2 around the x-axis.- V = π∫
_{−2}^{2}(√{4 − x^{2}})^{2}dx - V = π∫
_{−2}^{2}4 − x^{2}dx - V = π(4x − [(x
^{3})/3]) |_{−2}^{2} - V = π(8 − [8/3] − (−8 + [8/3]))
- V = π(16 − [16/3])
- This is the volume of a sphere of radius 2!

V = [(32π)/3]

Find the volume of the solid that results from revolving y = [1/(√x)] from x = 1 to x = 3 around the x-axis.

- V = π∫
_{1}^{3}[1/x] dx - V = πlnx |
_{1}^{3} - V = π(ln3 − ln1)
- V = π(ln3 − 0)

V = πln3

The area bounded by y = x

^{3}and y = x is rotated about the x-axis. Find the volume of the resulting solid.- Find their points of intersection
- x
^{3}= x - x = 0, 1, −1
- V = π(− ∫
_{−1}^{0}(x)^{2}− (x^{3})^{2}dx) + π∫_{0}^{1}(x)^{2}− (x^{3})^{2}dx - We can use symmetry here instead of using two separate integrals
- V = 2 π∫
_{0}^{1}(x)^{2}− (x^{3})^{2}dx - V = 2 π([(x
^{3})/3] − [(x^{7})/7]) |_{0}^{1} - V = 2π([1/3] − [1/7] − 0)

V = [(8π)/21]

*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 Disks and Washers

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 0:16
- Equation 1: Rotation about x-axis (disks)
- Equation 2: Two curves about x-axis (washers)
- Equation 3: Rotation about y-axis
- Lecture Example 1 6:05
- Lecture Example 2 8:28
- Lecture Example 3 11:55
- Additional Example 4
- Additional Example 5

0 answers

Post by Richard Gregory on October 17, 2013

Hi. I worked example three by converting the graph and co-ordinates into x values and then subtracting the area under the curve from the total disc volume. Is this a valid method? The answer was the same.

0 answers

Post by Kimberly McDevitt on November 19, 2012

There is something wrong with example number 4. The video jams around 1:50 and plays like a broken record. Please fix. Thank you.

0 answers

Post by Daniela Valencia on April 17, 2012

The lecture examples are very simple for a college calculus class. No very helpful.

1 answer

Last reply by: Mohanraj Ponraj

Sat Jan 19, 2013 10:45 AM

Post by Matthew Odunjo on June 21, 2011

please I didnt why the first Big radius is 4 and the small being x

2 answers

Last reply by: Mary Shriver

Sat May 26, 2012 9:54 AM

Post by Ruben Ayllon Ayllon on April 11, 2011

In Lecture Example 1 isn't the answer 128/7 pie. If not could you explain how you got the answer?

0 answers

Post by Fatima Sulaiman on May 4, 2010

how rude. These videos are pretty good except there needs to be 3d computerized examples for the area and volume problems.