Question

Find the centroid of the plane region R enclosed by the loop curve \[y^2=x^3-x^4\].

Answer 1

the y value will be \(y_c=0\) and the x-coord is
\[ x_c= \frac{\int \int x \ dA}{\int \int dA} \]

The denominator is the area of the enclosed region

Answer 2

How would we go about approaching the problem? Would we take an integral of our equation to determine the area first? And then substitute it in the equation you gave above ot solve for the x coordinate?

Answer 3

you can start by working out the area.

but before that, you really need to know what it looks like, in terms of a plot.

Answer 4

The integral of the upper half, doubled, gives the total area:
\[A= 2 \int_0^1 \sqrt{x^3 - x^4} \ dx \\ = \int_0^1 x \sqrt{x(1-x)} \ dx \]
let \( x= \sin^2 t \)
\[ dx = 2 \sin t \cos t \ dt\]
lower bound: t= arcsin(0) = 0
upper bound: t= arcsin(1) = pi/2

\[ A= 2 \int_0^{\frac{\pi}{2}} \sin^2 t \sqrt{\sin^2 t (1- \sin^2 t)} 2 \sin t \cos t \ dt \\

= 4 \int_0^{\frac{\pi}{2}} \sin^4 t - \sin^6 t \ dt \]

to integrate, see
https://en.wikipedia.org/wiki/List_of_definite_integrals#Definite_integrals_involving_trigonometric_functions

the top integral is similar

Answer 5

Ok, thank you!