# What is the area of the figure?

## given by $x = {\cos}^{3} \left(t\right) , y = {\sin}^{3} \left(t\right) , 0 \le t \le 2 \pi$

Jun 14, 2018

#### Explanation:

I don't know if it's correct. Being a student this is all what I can do(atleast for now)
$x = {\cos}^{3} t , y = {\sin}^{3} t$
$\implies \frac{y}{x} = {\sin}^{3} \frac{t}{\cos} ^ 3 t = {\tan}^{3} t \implies \tan t = {\left(\frac{y}{x}\right)}^{\frac{1}{3}}$

also,
$y = {\sin}^{3} t \implies \mathrm{dy} = 3 {\sin}^{2} t \cos t \mathrm{dt}$
similarly, $\mathrm{dx} = - 3 {\cos}^{2} t \sin t \mathrm{dt}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \tan t = - {\left(\frac{y}{x}\right)}^{\frac{1}{3}}$ (on solving)

$\implies \frac{\mathrm{dy}}{y} ^ \left(\frac{1}{3}\right) = \frac{\mathrm{dx}}{x} ^ \left(\frac{1}{3}\right)$

now integrate to find the equation and thus the graph

$- \int \frac{\mathrm{dy}}{y} ^ \left(\frac{1}{3}\right) = \int \frac{\mathrm{dx}}{x} ^ \left(\frac{1}{3}\right)$
$\implies y = \frac{x}{k {x}^{\frac{2}{3}} - 1} ^ \left(\frac{3}{2}\right)$ ($k$ is constant)

putting different values of K will give the graph
graph{x/(x^(2/3)-1)^(3/2) [-10, 10, -5, 5]}

Jun 14, 2018

We need to integrate as $4 {\int}_{\frac{\pi}{2}}^{0} y \left(t\right) \cdot x ' \left(t\right) \mathrm{dt}$ to get the area. Taking the absolute value of the equation, the integral should be $\frac{24}{35}$

#### Explanation:

When you have a parametric set of equations and want to find the area under the curve, we need to define a function relative to $t$ to integrate. $x = {\cos}^{3} \left(t\right) , y = {\sin}^{3} \left(t\right)$

What we want to do in the cartesian system is if $y = F \left(x\right)$:

$\text{Area} = {\int}_{a}^{b} F \left(x\right) \mathrm{dx}$

$\text{Area} = {\int}_{a}^{b} y \mathrm{dx}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = - 3 {\cos}^{2} \left(t\right) \sin \left(t\right)$

color(blue)(dx=-3cos^2(t)sin(t)dt

"Area"=int_a^bsin^3(t)color(blue)((-3cos^2(t)sin(t)dt)

$\text{Area} = - 3 {\int}_{a}^{b} {\sin}^{4} \left(t\right) {\cos}^{2} \left(t\right) \mathrm{dt}$

NOTE that this is only applicable when $x$ is non-negative between bounds $a$ and $b$. Because there are no biases or gains to the sine and cosine terms, we can assume symmetry about the $x$ and $y$ axes, meaning we only need to integrate from $\frac{\pi}{2}$ to $0$, since that is the region where $x$ is positive, and then multiply that area by 4.

$\text{Area} = 4 \cdot \left(- 3 {\int}_{\frac{\pi}{2}}^{0} {\sin}^{4} \left(t\right) {\cos}^{2} \left(t\right) \mathrm{dt}\right)$

Now, we integrate:

$\text{Area} = - 12 {\int}_{\frac{\pi}{2}}^{0} {\sin}^{4} \left(t\right) {\cos}^{2} \left(t\right) \mathrm{dt}$

skipping integration steps here for time's sake

$\text{Area} = - \frac{6}{35} {\sin}^{5} \left(t\right) \left(5 \cos \left(2 t\right) + 9\right)$ from $\frac{\pi}{2}$ to $0$

$\textcolor{g r e e n}{\text{Area} = \frac{24}{35}}$