riaology
Unit 5 · Lesson 2 of 4

Stack disks and washers to build volume.

A hollow pipe demonstrates why an annular cross-section uses squared radii. The same slicing idea works when radius changes along the axis.

Change the input. Follow the mathematics.

What you'll be able to do
  • Set up the model and calculate its solid volume.
  • Connect the equation to the changing visual.
  • Explain the assumptions and identify a misleading interpretation.
Predict before exploring

What will change if you increase outer radius r? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Solid of revolution illustrated with a cross-section and a thin ringSlice area × axial thickness
Axial position (m) versus Cross-section area (m²). Washer area. Selected solid volume: 28.27433 m³. Subtract the inner disk area, not the radii. A variable-radius solid uses the same integral with R(x) and r(x).004.7121.59.4253Cross-section area (m²)Axial position (m)
Washer area

solid volumer=1 m,V=3π(22−12)=28.2743 m3r=1\,\mathrm m,\quad\mathcal V=3\pi(2^2-1^2)=28.2743\,\mathrm{m^3}

Subtract the inner disk area, not the radii. A variable-radius solid uses the same integral with R(x) and r(x).

Assumptions, units, and limits

A length-3 m hollow cylinder is the simplest washer solid. r is a fixed fraction of R; r=0r=0 recovers disks.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

01

From the situation to an equation

A thin slice contributes cross-section area times axial thickness. For rotation, subtract inner disk area from outer disk area. Express the radii and bounds in the same slice variable before integrating.

V=∫03π(R2−r2) dx=3π(R2−r2)\mathcal V=\int_0^3\pi(R^2-r^2)\,dx=3\pi(R^2-r^2)
02

Keep the reasoning visible

dV=π[R(x)2−r(x)2] dxd\mathcal V=\pi[R(x)^2-r(x)^2]\,dx

Subtract the inner disk area, not the radii. A variable-radius solid uses the same integral with R(x) and r(x).

03

What makes this model useful?

A length-3 m hollow cylinder is the simplest washer solid. r is a fixed fraction of R; r=0r=0 recovers disks.

Subtracting radii and squaring the difference does not give an annular area. Orientation determines whether slices use x or y.

A worked example

Calculate the starting scenario

  1. Use outer radius r = 2 m; inner radius fraction = 0.5 .
  2. r=1 m,V=3π(22−12)=28.2743 m3r=1\,\mathrm m,\quad\mathcal V=3\pi(2^2-1^2)=28.2743\,\mathrm{m^3}
  3. Subtract the inner disk area, not the radii. A variable-radius solid uses the same integral with R(x) and r(x). The checked quantity is solid volume.
Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the solid volume using outer radius r 2 m, inner radius fraction 0.5 . State the meaning of its sign and units.

Checked quantity: solid volume · m³. Follow the definition for finite-cutoff and numerical-model questions.

Hint

A thin slice contributes cross-section area times axial thickness. For rotation, subtract inner disk area from outer disk area. Express the radii and bounds in the same slice variable before integrating.

Explore the visual solution
Solid of revolution illustrated with a cross-section and a thin ringSlice area × axial thickness
Axial position (m) versus Cross-section area (m²). Washer area. Selected solid volume: 28.27433 m³. Subtract the inner disk area, not the radii. A variable-radius solid uses the same integral with R(x) and r(x).004.7121.59.4253Cross-section area (m²)Axial position (m)
Washer area

solid volumer=1 m,V=3π(22−12)=28.2743 m3r=1\,\mathrm m,\quad\mathcal V=3\pi(2^2-1^2)=28.2743\,\mathrm{m^3}

Subtract the inner disk area, not the radii. A variable-radius solid uses the same integral with R(x) and r(x).

Assumptions, units, and limits

A length-3 m hollow cylinder is the simplest washer solid. r is a fixed fraction of R; r=0r=0 recovers disks.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

A thin slice contributes cross-section area times axial thickness. For rotation, subtract inner disk area from outer disk area. Express the radii and bounds in the same slice variable before integrating.

V=∫03π(R2−r2) dx=3π(R2−r2)\mathcal V=\int_0^3\pi(R^2-r^2)\,dx=3\pi(R^2-r^2)
Calculate and explain

Apply the idea to a new scenario

For outer radius 3 m and inner-radius fraction 1/3, find the length-3 m cylinder volume.

Checked quantity: solid volume · m³. Follow the definition for finite-cutoff and numerical-model questions.

Hint

A thin slice contributes cross-section area times axial thickness. For rotation, subtract inner disk area from outer disk area. Express the radii and bounds in the same slice variable before integrating.

Explore the visual solution
Solid of revolution illustrated with a cross-section and a thin ringSlice area × axial thickness
Axial position (m) versus Cross-section area (m²). Washer area. Selected solid volume: 75.39822 m³. Subtract the inner disk area, not the radii. A variable-radius solid uses the same integral with R(x) and r(x).0012.5661.525.1333Cross-section area (m²)Axial position (m)
Washer area

solid volumer=1 m,V=3π(32−12)=75.3982 m3r=1\,\mathrm m,\quad\mathcal V=3\pi(3^2-1^2)=75.3982\,\mathrm{m^3}

Subtract the inner disk area, not the radii. A variable-radius solid uses the same integral with R(x) and r(x).

Assumptions, units, and limits

A length-3 m hollow cylinder is the simplest washer solid. r is a fixed fraction of R; r=0r=0 recovers disks.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

A thin slice contributes cross-section area times axial thickness. For rotation, subtract inner disk area from outer disk area. Express the radii and bounds in the same slice variable before integrating.

V=∫03π(R2−r2) dx=3π(R2−r2)\mathcal V=\int_0^3\pi(R^2-r^2)\,dx=3\pi(R^2-r^2)
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Solid of revolution illustrated with a cross-section and a thin ringSlice area × axial thickness
Axial position (m) versus Cross-section area (m²). Washer area. Selected solid volume: 28.27433 m³. Subtract the inner disk area, not the radii. A variable-radius solid uses the same integral with R(x) and r(x).004.7121.59.4253Cross-section area (m²)Axial position (m)
Washer area

solid volumer=1 m,V=3π(22−12)=28.2743 m3r=1\,\mathrm m,\quad\mathcal V=3\pi(2^2-1^2)=28.2743\,\mathrm{m^3}

Subtract the inner disk area, not the radii. A variable-radius solid uses the same integral with R(x) and r(x).

Assumptions, units, and limits

A length-3 m hollow cylinder is the simplest washer solid. r is a fixed fraction of R; r=0r=0 recovers disks.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. A thin slice contributes cross-section area times axial thickness. For rotation, subtract inner disk area from outer disk area. Express the radii and bounds in the same slice variable before integrating.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

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