Work done?

Quote:

Originally Posted by m777

Hello,
please try to solve this question.
Thanks

Why is this a problem?

You have the parametric for fr the curve:

$ \bold{r}(t) = \sin(t)\bold{i} + \cos(t)\bold{j} + \frac{t}{6}\bold{k} $

and for the force:

$ \bold{F}(x\bold{i}+y\bold{j}+x\bold{k}) = 6z \bold{i} + y^2 \bold{j} + 12x \bold{K} $

so:

$ \bold{F}(\bold{r}(t))= t\bold{i} + \cos^2(t)\bold{j}+12\sin(t)\bold{k} $

and:

$ \bold{dr}= \left( \cos(t)\bold{i}-\sin(t)\bold{j}+\frac{1}{6}\bold{k} \right) dt $

Now you can write out the integral:

$ \int_0^{2\pi} \bold{F}(\bold{r}(t)).\bold{dr} $

explicitly, and it is a normal integral.

RonL