Question

Find the area enclosed by the curve

$x=t^2-2t$, $y=\sqrt{t}$ and the $y-$axis.

Answer 1

Step 1. The area is enclosed by a parametric curve.

The area enclosed by a parametric curve $A={\int }_a^{b}g\left(t\right)f\text{'}\left(t\right)dt$

Take $\mathrm{x}=\mathrm{f}\left(\mathrm{t}\right)\mathrm{and}\mathrm{y}=\mathrm{g}\left(\mathrm{t}\right)$

Now to find the limit $\mathrm{a}$ and $\mathrm{b}$ for which the curve intersects the $y-$axis, i.e. $x=0$

$\begin{array}{rcl}\mathrm{x}& =& {\mathrm{t}}^2-2\mathrm{t}\\ & \Rightarrow & {\mathrm{t}}^2-2\mathrm{t}=0\\ & \Rightarrow & \mathrm{t}(\mathrm{t}-2)=0\\ & \Rightarrow & \mathrm{t}=0,2\end{array}$

Step 2. Integrate the equation to obtain the area $A$.

Then area

$$\begin{gathered} \mathrm{A}={\int }_0^2\left(\sqrt{\mathrm{t}}\right)(2\mathrm{t}-2)\mathrm{dt}\mathrm{where}\mathrm{f}\text{'}\left(\mathrm{t}\right)=2\mathrm{t}-2 \\ ={\int }_0^2(2t^{\frac{3}{2}}-2t^{\frac{1}{2}})dt \end{gathered}$$:

$\begin{array}{rcl}& =& {{\left[\frac{4}{5}{\mathrm{t}}^{\frac{5}{2}}-\frac{4}{3}{\mathrm{t}}^{\frac{3}{2}}\right]}^2}_0\\ & =& \left[\frac{4\times 4\sqrt{2}}{5}-\frac{4\times 2\sqrt{2}}{3}\right]-0\\ & =& \frac{8\sqrt{2}}{15}\\ & \approx & 0.7542\mathrm{square}\mathrm{units}\end{array}$

Hence, the area enclosed by the curve is approximately $0.7542$ square units.