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I have a problem in which I'm given a force field $\vec{F}(x,y,z)=x\hat{i}+y\hat{j}+ 3\hat{k}$ and a path $\vec{r}(t)=4cos(t)\hat{i}+4sin(t)\hat{j}+3t\hat{k}$ over the interval $0\le t\le 2\pi$. I need to find the work done by the force over the path. My first thought is to take the line integral of the vector field over the path:

$$W=\int_C\vec{F} \cdot d\vec{r} $$

Where C is the path $\vec{r}(t)$ over the interval $0 \le t \le 2\pi$. So I plug the path into $\vec{F}(x,y,z)$ and the integral becomes $$W=\int_0^{2\pi}\vec{F}(\vec{r}(t))\cdot \vec{r}(t)dt$$ $$W=\int_0^{2\pi}(4cos(t)\hat{i}+4sin(t)\hat{j}+3\hat{k})\cdot(4cos(t)\hat{i}+4sin(t)\hat{j}+3t\hat{k})dt$$ $$W=\int_0^{2\pi}(16cos^2(t)+16sin^2(t)+9t)dt$$ $$W=\int_0^{2\pi}(16+9t)dt$$

From this I get $W = 32\pi + 18\pi^2$, but apparently that's the wrong answer. Can anyone tell me what I'm doing wrong?

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You must have the inner product by the derivative of $\,r\,$ !:$$\int F(r(t))\cdot r'(t)dt$$ and erase that $\,t\,$ from $\,3k\,$ , too. – DonAntonio Mar 28 '13 at 4:06
agreed. Also, you're working too hard. This force field is conservative, you could just find a potential function and use the fundamental theorem of calculus for line integrals! – James S. Cook Mar 28 '13 at 4:07
@JamesS.Cook Ah, that is true, I see that because the curl is 0. Unfortunately though, I don't know how to get the potential field from a force field. – Ataraxia Mar 28 '13 at 4:12
@ZettaSuro just guess for something as simple as the $\vec{F}$ you face. You want $U$ such that $\partial_x U = x$, $\partial_y U = y$ and $\partial_z U = 3$. I bet you can guess $U$ without some formal method. (of course, learn the formal method in due time) – James S. Cook Mar 28 '13 at 4:24
up vote 1 down vote accepted

The integral is over $\vec{F}\cdot d\vec{r}$; you did not compute $d\vec{r}$, which is done as follows:

$$d\vec{r} = \frac{d}{dt}\vec{r}(t) \,dt = -4 \sin{t} \,\vec{i} + 4 \cos{t} \,\vec{j} + 3 \vec{k}$$

$$\int_C \vec{F}\cdot d\vec{r} = \int_C \vec{F}\cdot \frac{d}{dt}\vec{r}(t) \,dt $$

which you should now be able to do.

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