Question

Show that the line integral of x^2*y^3 + x^3*y^2, over the curve C denoted by x = cos^3 (t), and y = sin^3 (t) is zero.

Answer 1

This is the integral \[\int\limits_{0}^{2\pi} x^2y^3 + x^3y^2 ds\]

Answer 2

The parameters
x = \[\cos^3 (\theta)\]
y = \[\sin^3 (\theta)\]

Answer 3

I need to show that the integral is zero. I am supposed to use a symmetry agrument. Using odd functions to show that the symmetry cancels out the integral.

Answer 4

Let t=theta
Start by differentiating your parameterizations.
dx/dt=3cos^2(t)(-sin(t))
dy/dt=3sin^2(t)(cos(t))
Once you have this. Find ds. \[\sqrt{(dx/dt)^2+(dy/dt)^2}\]
Then take that, multiply it by the integrand (after replacing x and y) then integrate as normal. However, I do not know how to do it using a symmetry argument other than cos(t) is even and sin(t) is odd :P Sorry I can't do more than that^^

Answer 5

Thanks for ur response but
if I could show that the integrand is an odd function, then i could show that the integral evaluates to zero.