Difference of Squares, Not Square of a Difference

You'll learnto build a washer as two disc integrals subtracted, to see why the width of a ring is not the radius of a disc, to assign the outer and inner radii by testing a point, and to split where a region stops touching the axis.

When the region does not touch the axis, revolving it leaves a hole and every slice is a washer. Its area is the big circle minus the small one. That is not π times the square of the difference of the radii: for an outer radius of 5 and an inner radius of 3 the first gives 16π and the second gives 4π. The two agree only when the hole has no size at all — which is exactly the case where the washer was a disc.

8 STEPS · 6 QUICK CHECKS · π(R² − r²), AND NEVER π(R − r)² · v1

A = π(R² − r²) ⇒ two disc integrals subtracted ⇒ R − r is a width, not a radius

Before you start

What you're looking at

A region that never reaches the line it is being spun about, and the ring one slice of the resulting solid makes. The ring's outer edge comes from the far boundary and its inner edge from the near one.

The question

How far is each of the two boundaries from the axis, and what is the area of the ring between the circles they sweep?

Watch for

The hole is removed by subtracting its area, never by shortening the outer radius. Those are different operations and they give different answers.

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