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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

## Discussion

## Study Guides

## Practice Questions

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## Table of Contents

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### Revolving Solids Washer Disk Methods

- Disk method: , derived from
- Use when shape is simple with no holes
- Washer method:
- Use when shape has a hole
- Find outer radius
*f(x)*and inner radius*g(x)* - Apply formula directly

### Revolving Solids Washer Disk Methods

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

### Revolving Solids Washer Disk Methods

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 Washer Disk Methods 0:11
- Explanation
- Formula
- Example 1 3:42
- Example 2 6:54
- Example 3 9:29
- Example 4 12:16
- Example 5 15:35

0 answers

Post by Hana Woldselasie on December 17, 2014

yes but the negative part is below the x-axis so you dont need it for that problem since one of the bounds was the x-axis itself.

1 answer

Last reply by: Hana Woldselasie

Wed Dec 17, 2014 10:24 PM

Post by Jingwei Xie on April 21, 2014

in example 5, why is x=y^2+1 the same thing with y=(x-1)^(1/2)? is it supposed to have a plus and minus sign?