Let $C$ be part of the intersection curve of the Paraboloid $z=x^2+y^2$ with the plane $2x+2y-z+2=0$ that starts from point $(3,1,10)$ and ends in $(1,3,10)$.
We define $f(x,y,z)=z-y^2-2x-1$.
Calculate $\int_{C}f dl$.
My work:
Finding $C$:
from the plane equation: $z=2x+2y+2$.
Substituting that into the paraboloid equation:
$2x+2y+2=x^2+y^2 \Longrightarrow x^2-2x+y^2-2y=2 \Longrightarrow (x-1)^2+(y-1)^2=4$.
I find this result of getting a circle very weird, because the plane isn't parallel to $z=0$ plane, so I can't see why I received a circle, I expected an ellipse or something. The only thing I can think about is that I received the "Shadow" of the ellipse on the $xy$ plane, but I would appreciate any help understanding what have happened here!
Anyway, I also got stuck here on which parametrization should I choose, if it's $x=r\cos(t), y=r\sin(t),z=r^2$
OR
$x=1+r\cos(t),y=1+r\sin(t),z=?$ Then if I substitute it in the circle's equation I can find $z$. But I'm not sure if I can do that since $C$ isn't all the circle, it's just part of it.
I would appreciate any help, thanks in advance!
Edit After the help from answers:
If I define $\vec r(t)=(1+2cos(t), 1+2sin(t), 4cos(t)+4sin(t)+6)$ to be the vector that draws the circle.
Then $\vec r'(t) = (-2sin(t), -2cos(t), 4cos(t)-4sin(t))$.
Where $\frac{\pi}{2} \ge t \ge 0$.
$f(x,y,z)=z-y^2-2x-1=2x+2y+2-y^2-2x-1=-y^2+2y+1 = 2 - (y-1)^2$
And so my integral:
$$
\begin{split}
\int_C f dl
&= \int_0^{\pi/2}
2\cos(2t)
\sqrt{(2\sin(t))^2 + (2\cos(t))^2 + (4\cos(t)-4\sin(t))^2} dt \\
&= \int_0^{\pi/2}
2\cos(2t)
\sqrt{8 + (16\cos(t)^2 - 16\sin(2t) + 16\sin(t)^2)} dt \\
&= \int_0^{\pi/2} 2\cos(2t)\sqrt{24-16\sin(2t)} dt
\end{split}
$$
I'm having some difficult times deciding how to do this integral
The intersection curve of paraboloid and the slant plane is indeed an ellipse but it is an ellipse in the plane $2x+2y-z+2=0$. When you take the projection of each of the points on the intersection curve in xy-plane, it is a circle. In other words, the $x$ and $y$ coordinates of the points on the intersection curve follows $(x-1)^2 + (y-1)^2 = 4$ and $z$ coordinate can be written from the equation of the plane $z = 2x + 2y + 2$.
For you to visualize this, I will present another way to look at it. Take cylinder $(x-1)^2 + (y-1)^2 = 4$ and paraboloid $z = x^2 + y^2$. Their intersection curve is an ellipse in the plane $z = 2x + 2y + 2$. You can draw this in a 3D tool and see it.
Coming to parametrization, it is easier to parametrize as you mentioned later,
$r(t) = (1 + 2 \cos t, 1 + 2 \sin t, 6 + 4 \cos t + 4 \sin t), a \leq t \leq b$
$f(x,y,z)=z-y^2-2x-1 = 2y + 1 - y^2 = 2 - (y-1)^2$
So $f(r(t)) = 2 - 4 \sin ^2 t = 2 \cos 2 t$
Now find the bounds of $t$ plugging in start and end points which is pretty straightforward in this case. Then evaluate $|r'(t)| \ $ and finally follows the line integral.