Finding the mass of an wire using a line integral I am having a hard time evaluating this line integral. I am tempted to change my $x$ value to something like $\cos(u)$ and parameterize. Heres the question. 
Find the mass of a wire lying along the first octant of the curve $C$ the intersection
of the elliptic paraboloid $z = 2 − x^2 − y^2$
and parabolic cylinder $z = x^2$
between $(0,
1, 0)$ and $(1, 0, 1)$ if the density of the wire at position $(x, y, z)$ is $\rho(x, y, z) = xy$.
Choose
$x = t$, $y = \sqrt{1-t^2}$ and $z = t^2$
where $0 ≤ t ≤ 1$ as a parametric equation of $C$.
 A: Solving the system $
\begin{cases}
z=2-x^2-y^2\\
z=x^2
\end{cases}$
yields $\;x^2=2-x^2-y^2\; \Rightarrow\; x^2+\frac{y^2}{2}=1.$
In other words the projection of the curve in the $xy$ plane is the ellipse parametrized by
$$
\begin{cases}
x(t)=\cos t\\
y(t)=\sqrt{2}\sin t
\end{cases}
$$
with $t\in [0,2\pi]$. So $C$ can be parametrized by
$$
\begin{cases}
x(t)=\cos t\\
y(t)=\sqrt{2}\sin t\\
z(t)=x^2(t)=\cos^2t
\end{cases}
$$
It follows that the mass equals
$$
m=\int_{(0,1,0)}^{(1,0,1)} xy\; dr = \int_{t_1}^{2\pi} \sqrt{2}\sin t \cos t \sqrt{\sin^2t+2\cos^2t+4\sin^2t\cos^2t}\; dt
$$
where $(x(t_1),y(t_1),z(t_1))=(0,1,0)$. Such a $t_1$ does not exist. Also, the parametrization you proposed does not satisfy the equations. Are you sure about the equations in the question?
A: If you want to use $x = t$ as the starting point of your parameterization, then $z = t^2$ and $t^2 = 2 - t^2 - y^2 \implies y = \sqrt {2-2t^2}$
I would also note that $(0,1,0)$ does not lie on the intersection of these two surfaces.
