Question

What is the area of the figure?

given by `x=cos^3(t), y=sin^3(t), 0<=t<=2pi`

Answer 1

Please see below

Explanation:

I don't know if it's correct. Being a student this is all what I can do(atleast for now)
`x=cos^3t,y=sin^3t`
`=>y/x=sin^3t/cos^3t=tan^3t=>tant=(y/x)^(1/3)`

also,
`y=sin^3t=>dy=3sin^2tcostdt`
similarly, `dx=-3cos^2tsintdt`
`=>dy/dx=-tant=-(y/x)^(1/3)` (on solving)

`=>dy/y^(1/3)=dx/x^(1/3)`

now integrate to find the equation and thus the graph

`-intdy/y^(1/3)=intdx/x^(1/3)`
`=>y=x/(kx^(2/3)-1)^(3/2)` (`k` is constant)

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

Answer 2

We need to integrate as `4int_(pi/2)^(0)y(t)*x'(t)dt` to get the area. Taking the absolute value of the equation, the integral should be `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(t),y=sin^3(t)`

What we want to do in the cartesian system is if `y=F(x)`:

`"Area"=int_a^bF(x)dx`

`"Area"=int_a^bydx`

`(dx)/(dt)=-3cos^2(t)sin(t)`

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

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

`"Area"=-3int_a^bsin^4(t)cos^2(t)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 `pi/2` to `0`, since that is the region where `x` is positive, and then multiply that area by 4.

`"Area"=4*(-3int_(pi/2)^0sin^4(t)cos^2(t)dt)`

Now, we integrate:

`"Area"=-12int_(pi/2)^0sin^4(t)cos^2(t)dt`

skipping integration steps here for time's sake

`"Area"=-6/35sin^5(t)(5cos(2t)+9)` from `pi/2` to `0`

`color(green)("Area"=24/35)`