Subtract a triangle from an ellipse quadrant

Ellipse and line bounding R; labels −a, a, b, −b, O, x, ySource-given figure redrawn with D3. Ellipse and line bounding R; labels −a, a, b, −b, O, x, y. Only labels supplied in the original question are included.xyOR−aab−by/b − x/a = 1x²/a² + y²/b² = 1
Question diagram 1
Ellipse and line bounding R; labels −a, a, b, −b, O, x, ySource-given figure redrawn with D3. Ellipse and line bounding R; labels −a, a, b, −b, O, x, y. Only labels supplied in the original question are included.xyOR−aab−by/b − x/a = 1x²/a² + y²/b² = 1

Given ellipse and line enclosing R, labelled semi-axes a and b.

(i) R is bounded by the ellipse x2/a2+y2/b2=1x^2/a^2+y^2/b^2=1 and the line y/b−x/a=1y/b-x/a=1, as shown. Show that its area is

−12ab+∫−a0baa2−x2dx.-\frac12ab+\int_{-a}^0\frac ba\sqrt{a^2-x^2}dx.

Hence, substituting x=asin⁡θx=a\sin\theta, find the exact area in terms of a,ba,b. marks

(ii) Find the volume when R is rotated completely about the yy-axis, in the form ka2bka^2b with constant kk. marks

[2010/MJC/Prelim/II/5]

Total marks: 12

Solution

If this part uses an earlier result, revisit (i)·1 Setting up an area or open all working together.

(i)·2

Trigonometric substitution

Hence, using the substitution x=asin⁡θx=a\sin\theta, find the exact area of RR in terms of aa and bb.

Explore this idea

Hint

  1. Change the variable and the limits: what is θ\theta when x=0x=0 and when x=−ax=-a?

  2. Simplify a2−a2sin⁡2θ\sqrt{a^2-a^2\sin^2\theta} with cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta.

  3. Use cos⁡2θ=12(1+cos⁡2θ)\cos^2\theta=\frac12(1+\cos2\theta) to integrate.

Answer · (i)·2

A=ab(π−2)/4A=ab(\pi-2)/4.

Explore the idea

Open interactive exploration

A quarter-ellipse with a triangle cut away

Drawn with a = 3 and b = 2 (and reflected into x ≥ 0, which keeps the area). The strips fill the gap between the ellipse and the line.

Number of stripsn = 4
Try an example

Where does one strip’s area come from?

1 of 4

The curves meet at this sample, so the strip has zero height and contributes zero. Its red marks use the left edge at x* = 0.

zero heightupper = lowerwidth Δx
Zoomed-in rectangle: width and height are scaled independently. The sample sets its upper and lower values across the whole strip.
Sample x*
0
Width Δx
0.75
Upper value
2
Lower value
2
Height: upper − lower
0
Strip area
0

0.75 ×(2 − 2)≈ 0

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

Strip width
0.750
3.000 ÷ 4
Sum of strips
1.49356
Exact, ab(π − 2)/4
1.71239
Error
−0.21883
12.78% too small

How the sum improves as n grows

  • Quarter-ellipse x²/9 + y²/4 = 1
  • Chord y/2 + x/3 = 1 joining the ends of the arc
  • Rectangles whose height is (ellipse − line)
  • Exact area ab(π − 2)/4 ≈ 1.7124
Key idea

Quarter-ellipse πab4=4.712\frac{\pi ab}{4}=4.712 minus triangle ab2=3\frac{ab}2=3 gives 1.7121.712. The strips settle on exactly that.

Ideas to try
  1. Check the reflection: the region in the question lies in x ≤ 0, and this picture is its mirror image.
  2. Use midpoints and 10 strips. How close is the error?
Takeaway

Check with known solids

Half an ellipsoid has volume 23πa2b\frac23\pi a^2b and a cone of radius aa, height bb has 13πa2b\frac13\pi a^2b. Quoting both from memory gives the answer to (ii) in one line, and it is a good check on the integral.