Unlock hundreds more features
Save your Quiz to the Dashboard
View and Export Results
Use AI to Create Quizzes and Analyse Results

OR
Sign inSign in with Facebook
Sign inSign in with Google
Quizzes > Mathematics

Shell Method vs Washer Method Quiz: Volume Practice

Quick quiz to test washer vs shell method choices - plus disk cases. Instant results.

Editorial: Review CompletedCreated By: John ZnamirowskiUpdated Aug 24, 2025
Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art illustration with washer disk shell shapes, quiz on disc washer shell volume methods on coral background

This quiz helps you decide when to use the shell method vs washer method to find volumes of revolution, and when a disk is better. Work through short calculus problems, compare setups, and see where each method fits. For extra practice, try surface area and volume practice in geometry, or refresh basics with mass volume and density quiz.

Which method for finding volumes of solids of revolution uses cylindrical shells?
Washer method
Shell method
Cross-sectional method
Disk method
The shell method involves slicing the solid into cylindrical shells whose volumes are approximated and summed. Each shell has radius equal to the distance from the axis of rotation to the slice, height given by the function, and thickness dx or dy. This approach is especially useful when the axis of rotation is parallel to the slicing direction. .
Which method uses slicing the solid into thin disks perpendicular to the axis of rotation?
Pappus' theorem
Washer method
Disk method
Shell method
The disk method slices a solid perpendicular to the axis into thin disks with radius given by the function value. The volume is found by summing the volumes of these disks, ?[R(x)]^2 dx. This is straightforward when the region has no hole and slices are perpendicular. .
In the disk method, what does R(x) represent?
The outer radius in terms of x
The thickness of the slice
The inner radius in terms of x
The height of the disk
In disk or washer integrals, R(x) denotes the distance from the axis of rotation to the outer curve, forming the radius of each disk. When there is an inner radius, we denote it by r(x). The volume then involves ?[R(x)]^2 for disks or ?[R(x)]^2 - ?[r(x)]^2 for washers. .
What is the typical integrand for the washer method when rotating around the x-axis?
?[f(x)]^2
?R(x)
?[R(x)^2 - r(x)^2]
2? x f(x)
The washer method subtracts the volume of the inner disk from the outer disk, giving ?[R(x)]^2 - ?[r(x)]^2. This accounts for any hole in the solid. Here R(x) is the outer radius and r(x) the inner radius, both measured perpendicular to the axis. .
For the shell method, the volume element includes a factor 2?r. What does r represent?
Volume of each shell
Thickness of the shell
Distance from axis of rotation to the sample rectangle
Height of the shell
In the shell method, r is the radius of each cylindrical shell - i.e., the distance from the axis of rotation to the vertical or horizontal slice. The circumference of the shell is 2?r, and the shell's height is given by the function value. .
What differential slice do you integrate when using cylindrical shells around the y-axis for y = f(x)?
dx
dy
r dr
d?
When rotating the graph of y = f(x) about the y-axis with shells, each vertical slice has thickness dx. The volume element becomes 2?·(radius)·(height)·dx. Using dy would correspond to horizontal slicing instead. .
When the region has a hole upon rotation, which method directly accommodates it?
Shell method
Slicing method
Washer method
Disk method
The washer method subtracts the volume of an inner solid from an outer solid for each slice, naturally handling holes. Disks only work for solids without holes, and shells treat regions differently. Washers are thus ideal for hollow shapes. .
If rotating around the horizontal line y = 1, which adjustment is needed for the radius in disk/washer method?
Integrate with respect to x twice
Use r(x) = f(x) + 1
Use R(x) = |f(x) - 1|
Multiply radius by 2?
When the axis is y = 1 rather than y = 0, the radius becomes the vertical distance |f(x) - 1|. Inner and outer radii both shift by 1 unit. This absolute difference ensures correct geometric distance. .
When might you choose the shell method over the disk method?
When the axis of rotation is horizontal
When only an outer radius is needed
When the solid can be sliced parallel to the axis to avoid inverting functions
When the region has no hole
The shell method is often chosen when slicing parallel to the axis yields simpler expressions than slicing perpendicular. It avoids solving for the inverse function, making integrals easier. Disk method may require inverting a function to describe radius in terms of the slicing variable. .
What is the shell method integral for the region bounded by y = x^2, x = 1, and y = 0, rotated about the y-axis?
2? ??¹ x³ dx
2? ??¹ x² dx
? ??¹ (1 - x²) dx
? ??¹ (1 - x?) dx
Using shells, radius = x, height = f(x) = x², and thickness dx from x=0 to 1. The volume is 2?? x·x² dx = 2?? x³ dx. This accounts for the region's shape and axis location. .
What is the washer method integral for the region between y = x and y = x² rotated about the x-axis from x = 0 to 1?
? ??¹ (x - x²) dx
2? ??¹ x(x - x²) dx
? ??¹ (x? - x²) dx
? ??¹ (x² - x?) dx
Outer radius = x, inner radius = x², so volume = ?? [x² - (x²)²] dx = ?? (x² - x?) dx from 0 to 1. This subtracts the hole's volume from the full disk. .
When rotating around the y-axis using washers with x = f(y), which differential is used?
dx
dr
d?
dy
Using x = f(y) to form washers around the y-axis requires slicing horizontally, so the thickness is dy. The outer and inner radii are given by functions of y. Integrating with respect to x would not match the horizontal slices. .
What is the volume of the solid generated by rotating y = ?x from x = 0 to 4 about the x-axis using the disk method?
8?
4?
2?
16?
Using disks, radius = y = ?x, so area = ?(?x)² = ?x. Integrate from 0 to 4: ??? ?x dx = ?·[x²/2]?? = ?·8 = 8?. .
Which formula represents the volume by the shell method for rotation about the x-axis?
2? ? radius · height dx
? ? (R² - r²) dx
? ? R² dx
? 2? R² dx
The shell method formula is V = 2?? (radius)(height) d(variable). For rotation about the x-axis using horizontal shells, radius is the y-value and height is ?x. This distinguishes it from disk/washer formulas. .
What is the main reason to choose the disk/washer method?
When the function is implicit
When cross-sections perpendicular to the axis are easier to describe
When the region can only be described parametrically
When the axis of rotation is oblique
Disk and washer methods are best when perpendicular slices to the axis yield simple radii. The shapes formed by these slices are disks or washers, making integration straightforward. Implicit or oblique cases may require other techniques. .
What is the volume using the shell method for the region bounded by y = x³ and y = x, from x = 0 to 1, rotated about the y-axis?
2?/15
4?/15
4?/5
8?/15
Radius = x, height = x - x³, so V = 2? ??¹ x(x - x³) dx = 2? ??¹ (x² - x?) dx = 2?(1/3 - 1/5) = 4?/15. This uses standard shell integration. .
The volume of a torus with major radius R and minor radius r is given by which formula?
? R² r²
4/3 ? R³
?² R² r
2?² R r²
By the Pappus centroid theorem or washers, the torus volume is V = (area of circle)·(distance traveled by its centroid) = ?r² · (2?R) = 2?² R r². .
When rotating the region between y = x² and x = y² about the y-axis, which method is preferred?
Shell method
Neither method applies
Washer method
Both are equally simple
The region is described more easily by vertical slices giving shells: radius = x, height = sqrt(x) - x². Washer requires solving for x in terms of y twice. Shell method avoids complicated inverses. .
For the region x = y² from y = 0 to 2 rotated about the x-axis, what is the shell radius in terms of y?
y
2 - y
y²
x
Using horizontal shells for rotation about the x-axis, each shell's radius is the vertical distance from the x-axis to the slice: r = y. The shell height = x? - x?, here simply y² - 0. .
When revolving around the vertical line x = -1, how do you express the shell radius for the function y = sqrt(x)?
x + 1
x - 1
sqrt(x) + 1
|sqrt(x) + 1|
For shells, radius = horizontal distance from the rotation line: r = |x - (-1)| = x + 1. The function value enters the height of each shell. No absolute value needed when x ? 0. .
Which integral gives the volume of the region between y = 1/x and y = 1, from x = 1 to 2, rotated about the x-axis?
? ??² [1/x - 1]² dx
2? ??² x[1 - 1/x] dx
? ??² [1² - (1/x)²] dx
? ??² [1 - 1/x] dx
Using washers: outer radius = 1, inner = 1/x, so V = ??[1² - (1/x)²] dx from 1 to 2. Shell method would be more complex here. .
Which scenario requires splitting into two separate integrals when using the shell method?
Region symmetric about the axis
Region whose boundary changes definition at an interior point
Region with no hole and single axis
Region bounded by a single smooth curve
When the region's boundary or limiting function changes form over the interval, you must split the integral at that point. Each piece has its own height or radius formula. Smooth single-boundary regions need only one integral. .
According to Pappus's centroid theorem, the volume of a solid generated by rotating a plane region about an external axis is equal to what?
The area of the region multiplied by the distance traveled by its centroid
The area times the square of the centroid's distance
The moment of inertia of the region about the axis
The double integral of radius dA
Pappus's theorem states V = A·d, where A is the area of the region and d is the path length of its centroid (usually 2?·distance to axis). It provides a powerful alternative to slicing methods for certain shapes. .
Which integral correctly gives the volume of the solid from rotating the region bounded by y = x³, y = 0, x = 2 about the line y = -1?
2? ??² x(x³ + 1) dx
? ??² [ (x³ + 1)² - 1² ] dx
? ??² [ x? - 1 ] dx
? ??² (x³ - 1)² dx
Using washers: outer radius = distance from y = -1 to y = x³ = x³ + 1, inner radius = distance to y = 0 = 1. The integrand is ?[(x³+1)² - 1²]. .
0
{"name":"Which method for finding volumes of solids of revolution uses cylindrical shells?", "url":"https://www.quiz-maker.com/QPREVIEW","txt":"Which method for finding volumes of solids of revolution uses cylindrical shells?, Which method uses slicing the solid into thin disks perpendicular to the axis of rotation?, In the disk method, what does R(x) represent?","img":"https://www.quiz-maker.com/3012/images/ogquiz.png"}

Study Outcomes

  1. Differentiate Volume Methods -

    Understand the key distinctions between the disc, washer, and shell techniques to choose the appropriate approach for a given solid of revolution.

  2. Set Up Washer Integrals -

    Apply the washer disk shell method by formulating integrals that calculate volume through concentric rings and identifying inner and outer radii.

  3. Construct Shell Integrals -

    Use the shell method explanation to derive lateral cylindrical shells and set up the corresponding volume integrals efficiently.

  4. Analyze Region Boundaries -

    Interpret region boundaries and axis of revolution to decide between disc washer and shell methods for simplified integration.

  5. Compare Method Efficiency -

    Evaluate the computational advantages and drawbacks of disc, washer, and shell approaches to streamline complex volume calculations.

  6. Solve Practice Problems -

    Demonstrate mastery by completing quiz questions that reinforce your ability to apply and transition between disc, washer, and shell techniques.

Cheat Sheet

  1. Disc Method Fundamentals -

    The disc method slices the solid perpendicular to the axis, treating each cross-section as a solid disk with volume V = π ∫[R(x)]² dx (Stewart Calculus, Ch. 6). It shines when there's no hole - just radius R(x) from the curve to the rotation axis. Remember "disk = no hole," so apply when the region touches the axis directly.

  2. Washer Method Expansion -

    The washer disk shell method introduces an inner radius r(x), giving V = π ∫(R(x)² − r(x)²) dx (MIT OpenCourseWare). It handles solids with holes by subtracting the empty core, like calculating a circular ring's volume. Mnemonic: "big circle minus little circle = washer" to recall R² - r².

  3. Shell Method Explanation -

    In the shell method, use cylindrical shells parallel to the axis: V = 2π ∫ x f(x) dx (University of Illinois Calculus Notes). It excels when slicing parallel saves you from complicated inverse functions - just multiply circumference by height times thickness. Think "wrap it up" to recall shells encase volume like layers of an onion.

  4. Choosing the Right Method -

    Comparing disc, washer and shell method lets you pick the simplest integral for a given problem (disc shell and washer method strategy guides from Khan Academy). If the region is bounded away from the axis, washers often beat discs; if solving for x in terms of y is a pain, switch to shells. Always sketch the region and axis to decide quickly.

  5. Mnemonic Tricks & Common Pitfalls -

    A quick memory phrase for the disc washer and shell method: "DWS - Draw, Write, Solve." Draw the region, write the correct radius(s), and solve the integral. Watch for sign errors when revolving around lines y = k or x = h, and always adjust R and r accordingly!

Make a Quiz (FREE) Copy & Edit This Quiz Create a Quiz with AI

Advanced Math Quizzes

Powered by: Quiz Maker