Grade 12 · Washer volumes
Washer volumes: A medium hollow cone
Washer volumes: investigate a medium hollow cone with outer radius factor a = 2; inner radius factor b = 0.5.
Predict what will happen, press Play, then pause and explain what changed. Every control also works without playback.
Watch the relationship
PausedHD animation studio
From experiment to screen.
Present a crisp Canvas scene, save a full-HD image, or capture your model as a silent video.
Understand what you are seeing
The idea behind the motion.
Rotating a region between two radius curves around an axis produces annular cross-sections. Each washer's area is the outer disk area minus the inner disk area. Squaring the difference of radii would incorrectly remove the cross term and describe a different area. This investigation starts with Outer radius factor a = 2; Inner radius factor b = 0.5. Predict the result before playing, then change one parameter while keeping the other fixed to test the reason for the change.
A relationship to keep
R=ax, r=bx; V(T)=π(a²−b²)T³/3
Read the symbols alongside the explanation. A diagram shows the relationship; the assumptions tell you when it applies.
- STEP 1
Set up the mathematical model
Read the outer radius ax and inner radius bx at the moving x position. The starting case is “A medium hollow cone.”
- STEP 2
Follow the changing quantity
Use π(R²−r²) for each cross-section while the accumulated interval grows from zero to two.
- STEP 3
Explain and test the result
Integrate the squared radii to obtain the cubic volume formula. Compare a solid cone with a hollow one using the same outer radius.
Your turn to explain
Make a prediction. Test your reasoning.
Keep Outer radius factor a = 2; Inner radius factor b = 0.5. Pause the timeline at 100%. Given upper x bound = 2, calculate outer radius, cross-section area, accumulated volume. Show the substitution into the displayed formula.
Use the values specified in the question. Reset restores the initial values for this investigation.
Compare your explanation
At x=2, π(R²−r²)=π[(2)²−(0.5)²](2)²=47.12389. Integrating from 0 to 2 gives π[2²−0.5²](2)³/3=31.415927. Results: Outer radius: 4; Cross-section area: 47.124; Accumulated volume: 31.416. Decimal values are rounded; retain the original parameters when checking.
Work through a full lesson
Connect the animation to a worked example and practice questions.