Carry negative parameter bounds through the substitution · source reference

Printed question

C: $x=-1/t$, $y=1/\sqrt{4-t^2}$, $-2<t<0$. S is bounded by C, $x=1/\sqrt3,x=1/\sqrt2$ and the $x$-axis.

Show its area is $\int_a^b\frac1{t^2\sqrt{4-t^2}}dt$, determining exact constants $a,b$. Using $t=2\sin\theta$, find the exact area. [6]

[2011/SAJC/Prelim/II/2(b)]

Printed working

When $x=1/\sqrt3,t=-\sqrt3,y=1$; when $x=1/\sqrt2,t=-\sqrt2,y=1/\sqrt2$. $x=-1/t\Rightarrow dx/dt=1/t^2$.

$$\text{Area S}=\int_{1/\sqrt3}^{1/\sqrt2}y dx=\int_{-\sqrt3}^{-\sqrt2}\frac1{\sqrt{4-t^2}}\frac{dx}{dt}dt=\int_{-\sqrt3}^{-\sqrt2}\frac1{\sqrt{4-t^2}}\frac1{t^2}dt.$$

Thus $a=-\sqrt3,b=-\sqrt2$. $t=2\sin\theta$, $dt/d\theta=2\cos\theta$; $t=-\sqrt3\Rightarrow\theta=-\pi/3$; $t=-\sqrt2\Rightarrow\theta=-\pi/4$.

$$\begin{aligned}\text{Area S}&=\int_{-\pi/3}^{-\pi/4}\frac1{\sqrt{4-4\sin^2\theta}}\frac1{2^2\sin^2\theta}2\cos\theta d\theta\\&=\int_{-\pi/3}^{-\pi/4}\frac1{\sqrt{4\cos^2\theta}}\frac1{2^2\sin^2\theta}2\cos\theta d\theta\\&=\frac14\int_{-\pi/3}^{-\pi/4}\csc^2\theta d\theta=-\frac14[\cot\theta]_{-\pi/3}^{-\pi/4}\\&=-\frac14(-1+1/\sqrt3)=\frac14(1-1/\sqrt3)\quad\text{units}^2.\end{aligned}$$

Teacher notes

Express the area of S in Cartesian form first before expressing it in terms of $t$. For a definite integral remember to change limits from $x=\cdots$ to $t=\cdots$.

Printed answer key

$a=-\sqrt3,b=-\sqrt2$; $\frac14(1-1/\sqrt3)$ units$^2$.

Source diagrams in native form