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.

Desmos

#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)#