= Solution
On the circular arc, $z=0$ and $dz=0$, so the <line integral> reduces to
$$
\int(x^2\,dx+y^2\,dy)
=\left[\frac{x^3+y^3}{3}\right]_{(0,0)}^{(1,1)}=\frac23.
$$
This <exact differential> evaluation applies to the specified arc; for example its <parametric equation> is $(x,y,z)=(1-\cos\theta,\sin\theta,0)$, $0\leq\theta\leq\pi/2$. On the vertical segment $x=y=1$, only $dz$ contributes, giving $\int_0^1(\lambda+1)\,dz=\lambda+1$. The whole path therefore gives
$$
\boxed{\lambda+\frac53}.
$$
To answer the final conservativeness requirement, compute the <curl>:
$$
\nabla\times\mathbf M=((2-\mu)y,\ (1-2\lambda)x,\ 0)^T.
$$
It vanishes identically exactly when
$$
\boxed{\mu=2,\qquad\lambda=\frac12}.
$$
On the <simply connected> whole space this condition is sufficient for a <conservative vector field>. Explicitly a <vector-field potential> is
$$
\phi=\frac12x^2z+y^2z+\frac{x^3+y^3}{3}+C.
$$
Both path calculations then give $13/6$, equal to $\phi(1,1,1)-\phi(0,0,0)$ by the <fundamental theorem for line integrals>. Equality of just these two path values would only force $\mu=2\lambda+1$; it is the <curl> calculation that establishes full <path independence>.
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