# Work done?

• June 17th 2007, 04:43 AM
m777
Work done?
Hello,
please try to solve this question.
Thanks
• June 17th 2007, 06:02 AM
CaptainBlack
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