# Line integrals for a circle

Say i want to find

$$\int_CF\cdot d\mathbf{r}$$

for $\mathbf{F}=x\,\mathbf{i} + zy\,\mathbf{j} + zx^2\,\mathbf{k}$

Where C is the circle $x^2+y^2=16$ lying in the place $z=3$ with clockwise orientation.

So i have parametrised getting $x(t)=rCost, y(t)=-rsint, z(t)=3$

getting $\mathbf{r(t)}=4Cos(t)\,\mathbf{i}-12Sin(t)\mathbf{j}+(4cos(t))^2(3)\mathbf{k}$

so we have $\mathbf{r'(t)}=-4Sin(t)\,\mathbf{i}-12Cos(t)\mathbf{j}+(-96SintCost)\mathbf{k}$

which could also be written as $\mathbf{r'(t)}=-4Sin(t)\,\mathbf{i}-12Cos(t)\mathbf{j}+(-48Sin(2t))\mathbf{k}$

I need to find $|\mathbf{r'(t)}|$ but I am completely confused is i have done everything right so far as squaring -48 or 96 is going to give me a very messy answer surely??

I know that $cos^2(x)+Sin^2(x)=1$ so i can use this for finding my absolute value, but what about the $-96^2$?

Could someone please tell me if this is correct so far?? Any help appreciated,

Many thanks

Let's start simplifying as much as possible, using the plane $z=3$ in the first place: $$\mathbf{F}=x\,\mathbf{i} + 3 y\,\mathbf{j} + 3 x^2\,\mathbf{k}$$ The integral is over an inner product $F\cdot d\mathbf{r}$ and at a circle with radius $4$ lying in the plane $z=3$, so there is no $\mathbf{k}$ in the expression for $d\mathbf{r}$: $$d\mathbf{r} = 4 \left[\,-\sin(t) \,\mathbf{i} + \cos(t) \,\mathbf{j}\,\right]\,dt$$ Hence the integral becomes (guess why the term with $\mathbf{k}$ in the force $F$ is cancelled) : $$\int_C F\cdot d\mathbf{r} = \int_0^{2\pi} \left[4\cos(t) \,\mathbf{i} + 3\cdot4 \sin(t)\,\mathbf{j}\right] \cdot \left[-4 \sin(t) \,\mathbf{i} + 4 \cos(t) \,\mathbf{j}\right]\,dt$$ Can you proceed from here?