What is the area of the figure?

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

2 Answers
Jun 14, 2018

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]}

Jun 14, 2018

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.

DesmosDesmos

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)