Basic Annulus Calculation

Find the area of an annulus with outer radius $R = 5$ cm and inner radius $r = 3$ cm.

Identify the radii: Outer radius $R = 5$ cm, Inner radius $r = 3$ cm = $R = 5$, $r = 3$

Apply the annulus formula: $A = \pi(R^2 - r^2)$ = $A = \pi(5^2 - 3^2)$

Calculate the squares: $5^2 = 25$ and $3^2 = 9$ = $A = \pi(25 - 9)$

Subtract and multiply by pi: $A = \pi \times 16 = 16\pi$ = $A = 16\pi \approx 50.27$ cm$^2$

Answer: The area of the annulus is $16\pi \approx 50.27$ cm$^2$

Real-World Application: Pipe Cross-Section

A water pipe has an outer diameter of 10 cm and the pipe wall is 1 cm thick. Find the cross-sectional area of the pipe material.

Find the outer radius: Outer diameter = 10 cm, so $R = 10 \div 2 = 5$ cm = $R = 5$ cm

Find the inner radius: Wall thickness = 1 cm, so $r = R - 1 = 5 - 1 = 4$ cm = $r = 4$ cm

Apply the annulus formula: $A = \pi(R^2 - r^2) = \pi(5^2 - 4^2)$ = $A = \pi(25 - 16)$

Calculate the final area: $A = \pi \times 9 = 9\pi$ = $A = 9\pi \approx 28.27$ cm$^2$

Answer: The cross-sectional area of the pipe material is $9\pi \approx 28.27$ cm$^2$

Working with Diameters

A metal washer has outer diameter 24 mm and inner diameter 12 mm. What is the area of the washer?

Convert to radii: Outer: $R = 24 \div 2 = 12$ mm, Inner: $r = 12 \div 2 = 6$ mm = $R = 12$, $r = 6$

Apply the annulus formula: $A = \pi(R^2 - r^2) = \pi(12^2 - 6^2)$ = $A = \pi(144 - 36)$

Calculate the final area: $A = \pi \times 108 = 108\pi$ = $A = 108\pi \approx 339.29$ mm$^2$

Answer: The area of the washer is $108\pi \approx 339.29$ mm$^2$

Mistake: Subtracting radii instead of areas: $A = \pi(R - r)^2$

Why: This calculates the area of a circle with radius $(R - r)$, not the ring between two circles.

Correct: Always subtract the areas: $A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$

Mistake: Confusing diameter with radius

Why: Problems often give diameters. Using diameter instead of radius gives an area 4 times too large.

Correct: Always convert diameter to radius first: $r = d \div 2$

Mistake: Forgetting that $R$ must be larger than $r$

Why: If you accidentally swap the radii, you get a negative area, which is impossible.

Correct: Always identify which circle is outer (larger $R$) and which is inner (smaller $r$)

Designing a Circular Walkway

Landscape architects use annulus calculations when designing circular paths around fountains or gardens.

A circular fountain has radius 4 meters. A walkway 2 meters wide surrounds it. The walkway area is $\pi(6^2 - 4^2) = 20\pi \approx 62.83$ m$^2$.

CD and DVD Storage Capacity

The recordable area of optical discs is an annulus because the center has a hole and non-recordable zone.

A standard DVD has inner recording radius 24 mm and outer recording radius 58 mm. The data area is $\pi(58^2 - 24^2) = 2788\pi \approx 8,759$ mm$^2$.

Q: What if I only know the diameters?

A: Divide each diameter by 2 to get the radii. If outer diameter is $D$ and inner diameter is $d$, then $A = \pi\left(\left(\frac{D}{2}\right)^2 - \left(\frac{d}{2}\right)^2\right) = \frac{\pi}{4}(D^2 - d^2)$.

Q: Can the inner circle be empty (a hole)?

A: Yes! The formula works whether there's an actual inner circle or just a hole. The annulus is the ring-shaped region regardless of what's inside.

Q: What's the difference between an annulus and a washer?

A: They're the same shape! 'Annulus' is the mathematical term, while 'washer' is the common name for the physical object (like the metal rings used with bolts).