How Do You Calculate a Line Integral in Physics for a Given Parametric Curve?

[tronter]

Homework Statement

Let [tex] x = \cos^{3} t [/tex] and [tex] y = \sin^{3}t [/tex] ([tex] 0 \leq t \leq 2 \pi [/tex]). Also [tex] \rho(x,y) = k [/tex].

Find [tex] I_0 = \int_{C} (x^{2} + y^{2}) \ dm [/tex]

Homework Equations

The Attempt at a Solution

So [tex] m = \int_{C} k \ ds = 3k \int_{0}^{2 \pi} \cos t \sin t \ dt [/tex].

Then [tex] dm = 3k \cos t \sin t \ dt [/tex]

So does [tex] I_0 = 3k\int_{0}^{2 \pi} \left( \cos^{6} t + \sin^{6} t \right)(\cos t \sin t) \ dt = 0 [/tex]?

Is this correct?

 

[anathema]

Thank you for your post. Your approach is correct, but there is a small error in your final integral. The correct integral should be:

I_0 = 3k \int_{0}^{2 \pi} \left( \cos^{6} t + \sin^{6} t \right) \cos t \sin t \ dt

= 3k \int_{0}^{2 \pi} \left( \cos^{7} t \sin t + \sin^{7} t \cos t \right) \ dt

= 3k \int_{0}^{2 \pi} \left( \frac{1}{8} \cos^{8} t + \frac{1}{8} \sin^{8} t \right) \ dt

= \frac{3 \pi k}{4}

Therefore, the value of I_0 is not equal to 0, as you had assumed. I hope this helps. Let me know if you have any further questions.

 

[ogb p]

No, your solution is not correct. The line integral in this case is a path integral, which means it is an integral over a specific path or curve. In this case, the path is defined by the parametric equations x = cos^3(t) and y = sin^3(t) where t ranges from 0 to 2pi.

To find the line integral, we first need to find the differential element ds, which represents a small segment along the curve. This can be found using the Pythagorean theorem as:

ds = sqrt(dx^2 + dy^2) = sqrt((-3sin^2(t)cos(t))^2 + (3cos^2(t)sin(t))^2) = 3sqrt(sin^4(t) + cos^4(t))dt

Next, we can use this ds to rewrite the line integral as:

I_0 = k * int_C (x^2 + y^2)ds = 3k * int_0^2pi (cos^6(t) + sin^6(t))sqrt(sin^4(t) + cos^4(t))dt

This integral is not zero and would need to be solved using appropriate integration techniques. It is important to note that the value of I_0 will depend on the value of k.