Solution
= Solution
On the specified <curve> $\mathbf r(t)=(t,t,t^2)$, the <tangent vector> is $(1,1,2t)$ and
$$
\mathbf F(\mathbf r(t))=(\mu t^2,\ t-\lambda t^2,\ t^2+(\lambda-2-\mu)t).
$$
Their <dot product> is $t+(\lambda-\mu-4)t^2+2t^3$. Hence
$$
\boxed{\int\mathbf F\cdot d\mathbf r=\int_0^1[t+(\lambda-\mu-4)t^2+2t^3]dt=\frac{\lambda-\mu-1}{3}.}
$$