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.

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.