I need some help with the following questions, I have made an attempt at both which is below but don't think its right. Any help in where I'm going wrong would be appreciated.
1)
$$\int_{0}^{2} (3ui+u^2j+(u+2)k * (2ui-3uj+(u-2)k) dx$$
$ $ \begin{array}{ccc} i & j & k \\ u & u^2 & (u+2) \\ 2u & -3u & (u-2) \end{array}
$i(u^2(u-2)-(-3u(u+2)))-j((u^2-2u)-(2u^2+4u))+k((-3u^2)-(2u^3))$ $= -3u^3-u^2+2u$
$$\int_{0}^{2} (3u^3-u^2+2u) dx$$
$$ \frac{-3u^4}{4}-\frac{u^3}{3}+\frac{2u^2}{2}$$
$$ \frac{-3*2^4}{4}-\frac{2^3}{3}+{2^2}=-12-\frac{8}{3}+4=\frac{-32}{3}$$
2)
fo this question i can do a similar question where $s=(t,t^2,t^3)$ but am having trouble with this one. First i tried to just get the values of $x,y,z$ and input them into A. and then continue on as normal. This gave me an answer but it doesn't seem right at all.
The result of the cross product between $\bf{F}$ and $\bf{V}$ is a vector and not a scalar. You have to integrate the components in $i$, $j$, $k$, separately. In addition please replace $K$ by $k$. As the surface integral you can have a look here http://mathinsight.org/surface_integral_vector_field_introduction .