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

(i)·1

Setting up an area

RR is bounded by the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 and the line yb−xa=1\dfrac yb-\dfrac xa=1. Show that its area is −12ab+∫−a0baa2−x2 dx-\dfrac12ab+\displaystyle\int_{-a}^0\dfrac ba\sqrt{a^2-x^2}\,dx.

Hint

  1. Where does the line cut the axes? Find the two points where it meets the ellipse.

  2. RR is a quadrant of the ellipse with a triangle removed. Which two shapes?

  3. Rewrite the ellipse as y=…y=\dots and take out the constant ba\frac ba.

Answer · (i)·1

A=−ab/2+∫−a0(b/a)a2−x2 dxA=-ab/2+\int_{-a}^0(b/a)\sqrt{a^2-x^2}\,dx.

(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

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?
(ii)

Volume of revolution

Find the volume generated when RR is rotated completely about the yy-axis, giving your answer in the form ka2bka^2b where kk is a constant.

Explore this idea

Hint

  1. Spinning the quadrant about the yy-axis gives half an ellipsoid. Spinning the triangle gives a cone.

  2. Volume of RR = (half-ellipsoid) − (cone).

  3. For the half-ellipsoid use π∫0bx2 dy\pi\int_0^bx^2\,dy with x2=a2(1−y2b2)x^2=a^2\left(1-\frac{y^2}{b^2}\right).

Answer · (ii)

V=πa2b/3V=\pi a^2b/3; k=π/3k=\pi/3.

a,b>0a,b>0.

Explore the idea

Half an ellipsoid with a cone hollowed out

With a = 3 and b = 2. The outer dome is half an ellipsoid and the cone sits inside it. The region R is what lies between.

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

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.