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)·2 Trigonometric substitution or open all working together.

(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.

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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.

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Open interactive exploration

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.