Subtract a triangle from an ellipse quadrant · source reference

Printed question

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

$$-\frac12ab+\int_{-a}^0\frac ba\sqrt{a^2-x^2}dx.$$

Hence, substituting $x=a\sin\theta$, find the exact area in terms of $a,b$. [8]

(ii) Find the volume when R is rotated completely about the $y$-axis, in the form $ka^2b$ with constant $k$. [4]

[2010/MJC/Prelim/II/5]

Printed working

(i)

The ellipse gives $y=\pm b\sqrt{1-x^2/a^2}$.

$$\text{Area R}=\text{area of quadrant}-\text{area of triangle}=\int_{-a}^0b\sqrt{1-x^2/a^2}dx-\frac12ab=\int_{-a}^0b\sqrt{(a^2-x^2)/a^2}dx-\frac12ab=-\frac12ab+\int_{-a}^0\frac ba\sqrt{a^2-x^2}dx\quad\text{(shown)}.$$

$x=a\sin\theta$, $dx/d\theta=a\cos\theta$; $a\sin\theta=0$ gives $\theta=0$; $a\sin\theta=-a$ gives $\theta=-\pi/2$.

$$\begin{aligned}\text{Area R}&=\frac ba\int_{-\pi/2}^0\sqrt{a^2-(a\sin\theta)^2}(a\cos\theta)d\theta-\frac12ab\\&=\frac{a^2b}a\int_{-\pi/2}^0\sqrt{1-\sin^2\theta}\cos\theta d\theta-\frac12ab\\&=ab\int_{-\pi/2}^0\cos^2\theta d\theta-\frac12ab\\&=ab\int_{-\pi/2}^0\frac{1+\cos2\theta}2d\theta-\frac12ab\\&=ab[\theta/2+\sin2\theta/4]_{-\pi/2}^0-\frac12ab\\&=ab[0+0-((1/2)(-\pi/2)+\sin(-\pi)/4)]-\frac12ab\\&=ab\frac\pi4-\frac12ab=\frac{ab}4(\pi-2)\quad\text{units}^2.\end{aligned}$$

(ii)

$x^2=a^2(1-y^2/b^2)$.

$$\begin{aligned}V&=\pi\int_0^bx^2dy-\frac13\pi a^2b\\&=\pi\int_0^b(a^2-a^2y^2/b^2)dy-\frac13\pi a^2b\\&=\pi[a^2y-a^2y^3/(3b^2)]_0^b-\frac13\pi a^2b\\&=\pi[a^2b-a^2b^3/(3b^2)]-\frac13\pi a^2b\\&=\pi[a^2b-a^2b/3]-\frac13\pi a^2b=\pi(2a^2b/3)-\frac13\pi a^2b=\frac13\pi a^2b.\end{aligned}$$

Thus $k=\pi/3$.

Teacher notes

Area of quadrant $=\int_{-a}^0y dx$, with $y=b\sqrt{1-x^2/a^2}\ge0$.

Integration by substitution: $\int f(x)dx=\int g(\theta)(dx/d\theta)d\theta$; change definite limits from $x$ to $\theta$.

Volume of quadrant rotated about $y$-axis is $\pi\int_0^bx^2dy$. Volume of triangle rotated about $y$-axis is a cone, radius $a$ units, height $b$ units.

The solution’s first question line misprints the line as the ellipse equation; the question booklet and diagram correctly state $y/b-x/a=1$. The solution graph also places an ellipse equation by the straight line. The native article retains the correct question and labels the source misprint here.

Printed answer key

(i) $ab(\pi-2)/4$ units$^2$; (ii) $\frac13\pi a^2b$ units$^3$.

Source diagrams in native form

Mathematical review

Assume positive semi-axes a,b as encoded by the labelled ellipse. The source solution repeats the ellipse equation where it calls the boundary a line; use y/b−x/a=1 from the original question/diagram.