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College Calculus: Level I Volume by Method of Disks and Washers

Section 7: Application of Integrals, part 1: Lecture 2 | 24:22 min

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Lecture Comments (9)

	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.

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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 πr². The volume would be πr² * thickness
So the volume of a tiny slice would be πr² 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 = πy² dx = πx² dx
Total volume from x = 0 to x = 4 is V
V = ∫₀⁴ πx² dx
V = π[(x³)/3] |₀⁴
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∫₀⁶ 4² dy
V = π16x |₀⁶
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] sec² 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∫₀¹ x² dy
y = lnx
e^y = e^lnx
x = e^y
V = π∫₀¹ e^2y dy
u = 2y
du = 2 dy
V = [(π)/2] ∫₀¹ e^u du
V = [(π)/2] e^u |₀¹
V = [(π)/2] e^2y |₀¹
V = [(π)/2] (e² − e⁰)

V = [(π)/2] (e² − 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 u² du
V = π[(u³)/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∫₀² (e^x)² − 1² dx
V = π∫₀² e^2x − 1 dx
V = π([1/2] e^2x − x) |₀²
V = π([1/2] e⁴ − 2 − [1/2] e⁰ + 0)
V = π([1/2] e⁴ − [5/2])

V = [(π)/2] (e⁴ − 5)

Find the volume of the solid that results from revolving y = x² + 1 from x = 0 to x = 1 around the x-axis.

V = π∫₀¹ (x² + 1)² dx
V = π∫₀¹ x⁴ + 2x² + 1 dx
V = π([(x⁵)/5] + [(2x³)/3] + x) |₀¹
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 − x²} from x = −2 to x = 2 around the x-axis.

V = π∫₋₂² (√{4 − x²})² dx
V = π∫₋₂² 4 − x² dx
V = π(4x − [(x³)/3]) |₋₂²
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∫₁³ [1/x] dx
V = πlnx |₁³
V = π(ln3 − ln1)
V = π(ln3 − 0)

V = πln3

The area bounded by y = x³ and y = x is rotated about the x-axis. Find the volume of the resulting solid.

Find their points of intersection
x³ = x
x = 0, 1, −1
V = π(− ∫₋₁⁰ (x)² − (x³)² dx) + π∫₀¹ (x)² − (x³)² dx
We can use symmetry here instead of using two separate integrals
V = 2 π∫₀¹ (x)² − (x³)² dx
V = 2 π([(x³)/3] − [(x⁷)/7]) |₀¹
V = 2π([1/3] − [1/7] − 0)

V = [(8π)/21]

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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

Important Equations

Lecture Example 1

Lecture Example 2

Lecture Example 3

Additional Example 4

Additional Example 5

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

College Calculus 1 Online Course

Section 1: Overview of Functions
	Review of Functions	26:29
	Compositions of Functions	12:29
Section 2: Limits
	Average and Instantaneous Rates of Change	20:59
	Limit Investigations	22:37
	Algebraic Evaluation of Limits	28:19
	Formal Definition of a Limit	23:39
	Continuity and the Intermediate Value Theorem	19:09
Section 3: Derivatives, part 1
	Limit Definition of the Derivative	22:52
	The Power Rule	26:01
	The Product Rule	14:54
	The Quotient Rule	19:17
	Applications of Rates of Change	17:43
	Trigonometric Derivatives	26:58
	The Chain Rule	23:47
	Inverse Trigonometric Functions	27:05
	Equation of a Tangent Line	15:52
Section 4: Derivatives, part 2
	Implicit Differentiation	30:05
	Higher Derivatives	13:16
	Logarithmic and Exponential Function Derivatives	17:42
	Hyperbolic Trigonometric Function Derivatives	14:30
	Related Rates	29:05
	Linear Approximation	23:52
Section 5: Application of Derivatives
	Absolute Minima and Maxima	18:57
	Mean Value Theorem and Rolle's Theorem	20:00
	First Derivative Test, Second Derivative Test	27:11
	L'Hopital's Rule	23:09
	Curve Sketching	40:16
	Applied Optimization	25:37
	Newton's Method	25:13
Section 6: Integrals
	Approximating Areas and Distances	36:50
	Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus	22:02
	Substitution Method for Integration	23:19
Section 7: Application of Integrals, part 1
	Area Between Curves	19:59
	Volume by Method of Disks and Washers	24:22
	Volume by Method of Cylindrical Shells	30:29
	Average Value of a Function	16:31
Section 8: Extra
	Graphs of f, f', f''	23:58
	Slope Fields for Differential Equations	18:32
	Separable Differential Equations	17:04

Related Books

	Calculus by Larson, 9th Edition Authors: Ron Larson, Bruce H. Edwards ISBN: 0547167024 Publisher: Brooks Cole Year: 2009 The book is filled with examples and detailed explanations. Capstone exercises are also included for each section and synthesize the main concepts into a single example.
	Calculus by Larson, 9th Edition, Solutions Manual, Volume 1 Author: Ron Larson ISBN: 547213093 Publisher: Brooks Cole Year: 2008 This solutions manual includes worked out solutions to every odd-numbered exercise in Calculus of a Single Variable, 9th edition.
	Calculus by Larson, 9th Edition, Solutions Manual, Volume 2 Authors: Ron Larson, Bruce H. Edwards ISBN: 547213107 Publisher: Brooks Cole Year: 2009 This solutions manual includes worked out solutions to every odd-numbered exercise in Multivariable Calculus, 9th edition.

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Professor Switkes

Volume by Method of Disks and Washers

Slide Duration:

Table of Contents

Section 1: Overview of Functions

Review of Functions

26m 29s

Intro