Evaluate the line integral $$\int_c F\bullet dr$$
where C is given by the vector function r(t).
$$F(x,y,z)=(x+y)\hat{i}+(y-z)\hat{j}+z^2\hat{k}$$
$$r(t)=t^2\hat{i}+t^3\hat{j}+t^2\hat{k}$$
$$0 \le t \le 1$$
$$r'(t)=2t\hat{i}+3t^2\hat{j}+2t\hat{k}$$
$$\int_c F \bullet dr=\int_a^b F(r(t)) \bullet r'(t)dt$$
$$\int_0^1 (t^2\hat{i}+(t^3-t^2)\hat{j} + t^4\hat{k})\bullet (2t\hat{i} +3t^2\hat{j}+2t\hat{k})dt$$
how am I doing?

Question

Evaluate the line integral $$\int_c F\bullet dr$$
where C is given by the vector function r(t).
$$F(x,y,z)=(x+y)\hat{i}+(y-z)\hat{j}+z^2\hat{k}$$
$$r(t)=t^2\hat{i}+t^3\hat{j}+t^2\hat{k}$$
$$0 \le t \le 1$$ 
$$r'(t)=2t\hat{i}+3t^2\hat{j}+2t\hat{k}$$
$$\int_c F \bullet dr=\int_a^b F(r(t)) \bullet r'(t)dt$$
$$\int_0^1 (t^2\hat{i}+(t^3-t^2)\hat{j} + t^4\hat{k})\bullet (2t\hat{i} +3t^2\hat{j}+2t\hat{k})dt$$
how am I doing?

anonymous · Answer

Oops wrong question, it's $$F(x,y,z)=z\hat{i}+y\hat{j}-z\hat{k}$$
$$r(t)=t\hat{i}+sint\hat{j}+cost\hat{k}$$
$$0 \le t \le pi$$

anonymous · Answer

for your i, x+y

anonymous · Answer

alright guess we'll start over hahha continues

anonymous · Answer

would that make 
$$r'(t)=\hat{i}+cost\hat{j}-sint\hat{k}$$
and

anonymous · Answer

$$\int_0^{\pi} (t\hat{i}+sint\hat{j}+cost\hat{k})\bullet (\hat{i}+cost\hat{j}-sint\hat{k})dt$$

kinggeorge · Answer

So far that looks right to me.

anonymous · Answer

thanks

kinggeorge · Answer

You're welcome. 

Once you do the dot product, it looks like it will be very easy to evaluate.

kinggeorge · Answer

So can you show me what you think the dot product will be?

anonymous · Answer

I can't just multiply the i    j and  k   coefficients can I?
$$t+sintcost-sintcost$$

kinggeorge · Answer

That's exactly what it is. You should end up with $$t+\sin(t)\cos(t)-\sin(t)\cos(t)=t$$

anonymous · Answer

Yaaaaayyyyy!!!!! I just need to build confidence i guess.... sigh

kinggeorge · Answer

You'll get used to it.

anonymous · Answer

here is the mistake I see:
x=t   y=sint  z=cost
that would give us 
$$\int_0^{\pi} cost dt+\int_0^{\pi}sintcostdt+\int_0^{\pi}tsint dt$$

turingtest · Answer

$$\vec F(x,y,z)=z\hat{i}+y\hat{j}-z\hat{k}$$$$\vec r(t)=t\hat i+\sin t\hat j+\cos t\hat k$$$$\vec r'(t)=\hat i+\cos t\hat j-\sin t\hat k$$$$\int_0^\pi\vec F(\vec r(t))\cdot\vec r'(t)dt=\int_0^\pi\langle\cos t,\sin t,\cos t\rangle\cdot\langle1,\cos t,-\sin t\rangle dt$$do you see a problem with that?

turingtest · Answer

$$=\int_0^\pi\cos t+\sin t\cos t-\sin t\cos tdt=\int_0^\pi\cos tdt=-2$$

anonymous · Answer

oh I see what I did! It's z y -z and not z y -x

anonymous · Answer

thanks!

turingtest · Answer

welcome!

turingtest · Answer

now I go to bed :)

anonymous · Answer

Have a good night's rest!

turingtest · Answer

likewise!