Rotate the region between a bell curve and a line

(i) Sketch y=2/[1+(x−2)2]y=2/[1+(x-2)^2] and y=1y=1 on the same diagram, showing clearly the coordinates of the stationary point and intersection points. marks

(ii) Find the exact area of region R enclosed by the curves and x=1,x=3x=1,x=3. marks

(iii) Find the volume generated when R is rotated 360∘360^\circ about the yy-axis. marks

[2011/MI/Prelim/II/2]

Total marks: 9

Solution

If this part uses an earlier result, revisit (ii) Area between curves or open all working together.

(iii)

Volume of revolution

Find the volume generated when RR is rotated through 360∘360^\circ about the yy-axis.

Explore this idea

Hint

  1. Picture the solid. RR sits between x=1x=1 and x=3x=3, so spinning it about the yy-axis leaves a hole.

  2. For washers, slice horizontally. Each radius must be written as xx in terms of yy.

  3. Outer radius from the right branch, inner radius from the left. Then integrate π(R2−r2)\pi(R^2-r^2) over yy.

Answer · (iii)

Volume =4π(π−2)≈14.3=4\pi(\pi-2)\approx14.3 cubic units (3 s.f.).

Explore the idea

Open interactive exploration

Spin R about the y-axis, then slice the solid

Sweep R around the y-axis to build the solid. Then cut it into washers (horizontal slices) or shells (vertical tubes) and add up their volumes.

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

Takeaway

Connect the parts: area × distance travelled

By the symmetry from (i), the centre of RR is at x=2x=2. Each turn carries that centre around a circle of length 2π×22\pi\times2. Multiply by the area from (ii) and you get the volume from (iii):

V=2π⋅2⋅(π−2)=4π(π−2)V=2\pi\cdot2\cdot(\pi-2)=4\pi(\pi-2)

This is Pappus’s centroid theorem. It is not on the syllabus, but it is a quick independent check of your answer.