Two reciprocal curves, and a solid with a changing floor

C_1

Total marks: 6

Solution

(a)

Area between curves

Curve $C_1$ is $y=\dfrac1x$ and curve $C_2$ is $y=\dfrac1{\sqrt{2-x^2}}$ . Region $Q$ is bounded by $x=\dfrac1{\sqrt2}$ and the two curves. Find the exact area of $Q$ .

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Hint

Where do the curves meet? That gives the right-hand edge of $Q$ .
On the interval between $x=\frac1{\sqrt2}$ and that point, which curve is on top?
Integrate (top − bottom). One of the two integrals is an arcsine.

Answer · (a)

=\frac12\ln2-\pi/12

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A thin sliver between two curves

Q is only about 0.08 square units, so small errors matter. Add strips and watch the sum home in on the exact area.

Number of stripsn = 4

Try an example

Where does one strip’s area come from?

1 of 4

Click a strip in the graph to inspect it. The darker rectangle is this strip. Its red marks use the left edge at x* = 0.70711.

Zoomed-in rectangle: width and height are scaled independently. The sample sets its upper and lower values across the whole strip.

Sample x*: 0.70711
Width Δx: 0.073223
Upper value: 1.4142
Lower value: 0.8165
Height: upper − lower: 0.59772
Strip area: 0.043767

0.073223 ×(1.4142 − 0.8165)≈ 0.043767

Add this contribution to those of the other 3 strips to get the sum below. Displayed values are rounded.

Strip width

0.073

0.293 ÷ 4

Sum of strips

0.10683

Exact, ½ ln 2 − π/12

0.08477

Error

+0.02206

26.02% too big

How the sum improves as n grows

The vertical scale fits n ≥ 3; earlier values may lie outside the view.

C₁, y = 1/x. It is the upper boundary on this interval
C₂, y = 1/√(2−x²). It is the lower boundary
Rectangles whose height is C₁ − C₂ at the chosen sample point
Exact area ½ ln 2 − π/12 ≈ 0.08477

Key idea

The integrand $\frac1x-\frac1{\sqrt{2-x^2}}$ shrinks to 0 at $x=1$ , so $Q$ tapers to a point. The sum of strips settles on $\frac12\ln2-\frac\pi{12}\approx0.0848$ .

Ideas to try

Look at the right-hand end. The curves meet at (1, 1), so the strips shrink to nothing there.
Compare left-edge and right-edge sums at n = 5. Which one overshoots?
Drag n to 40 and read the error.

(b)

Volume of revolution

The region in the first quadrant bounded by $y=5$ and the curves $C_1$ and $C_2$ is rotated about the $x$ -axis through $2\pi$ radians. Find the volume of the solid generated.

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Hint

Where does $y=5$ meet each curve? Those are the ends of the region.
Spun about the $x$ -axis, each slice is a washer: outer radius 5, inner radius equal to the lower curve.
The lower curve changes at $x=1$ , so split the integral there, or subtract from a cylinder.

Answer · (b)

=77.8

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Spin the region about the x-axis

The flat top at y = 5 becomes the outside of a cylinder. The two curves carve out the hollow underneath. Sweep it round, then cut the solid into washers.

Open the interactive view to explore this geometry in 3D.

Takeaway

Two routes to the same volume

You can integrate $\pi(5^2-y^2)$ directly, or take the whole cylinder and subtract what lies under the curves. Both methods add the same washers; split the inner boundary at $x=1$ in either calculation.

Subtracting from a cylinder is often neater when the outer boundary is a constant.

Two reciprocal curves, and a solid with a changing floor

Solution

Area between curves

Find where the curves meet

Decide which curve is on top

Set up the integral

Integrate each term

Substitute the limits

A thin sliver between two curves

Where does one strip’s area come from?

Volume of revolution

Find the ends of the region

Describe the floor

Subtract from a cylinder

Evaluate the two pieces

Exact value (optional)

Combine

Spin the region about the x-axis

Two routes to the same volume