# Can someone just check if I did this line integral correctly?

1. Nov 7, 2013

### ainster31

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

$$\int _{ 0 }^{ 2\pi }{ xdx } \\ =-\int _{ 0 }^{ 2\pi }{ sintcostdt } \\ =0$$

It feels wrong.

2. Nov 7, 2013

### brmath

I got zero also using a different method. Let's review the steps. You said
x = cos t, dx = -sint dt and t now runs from 0 to 2$\pi$.

Since cost and sint are both periodic in 2$\pi$ you got zero.

What may feel wrong is that $\int_0^{2\pi}x dx$ is not zero. However, this is not the same thing as integrating x around a circle. Basically you are running cos(t) around that circle, and it's going to start and end with the same value.

3. Nov 8, 2013

### HallsofIvy

Staff Emeritus
Notice that x is positive on the right side of the circle, negative on the left side of the circle. The two halves cancel, leaving 0.