The vector field:
$$F = 10xy^2 \hat{i} + (4z-2xy^2) \hat{j} + (5y -2x^2y) \hat{k}$$
The line integral yields $\int_{c} F \cdot dr = \frac{230}{3}$, where $c$ is the curve it goes along. It's just stated that $c$ lies on the plane $x + 5y + z = 4$, which is oriented clockwise as seen from above.
I want to find out the curve, which has to be the projection of the plane $x + 5y + z = 4$ on the $xy$ plane.
The stated solution is $y^2 - x^2 \le \frac{1}{4}$ but I don't understand neither why nor the method used.
Actually, I tried to project the vector perpendicular to the given plane (pointing inwards because of the clockwise direction) onto the xy plane but did not get that.
EDIT
The problem statement
Find the maximum value of $\oint_{c} F \cdot dr$ where $F = 10xy^2 \hat{i} + (4z-2xy^2) \hat{j} + (5y -2x^2y) \hat{k}$ and $c$ is a simple closed curve in the plane $x + 5y + z = 4$ oriented clockwise as seen from high on the z-axis. What curve $c$ gives this maximum?
The maximum value of the closed path integral has been calculated: $\oint_{c} F \cdot dr = \frac{230}{3}$, which occurs when $c$ is the boundary of the plane surface region $S$ (here we recall Stoke's theorem) in $x + 5y + z = 4$ whose projection onto the xy-plane is $y^2 - x^2 \le \frac{1}{4}$
The last bold line is what I don't really see. Maybe the projection is wrong
All stated options for the projections onto the xy-plane are:
a) $x - y \le \frac{1}{4}$
b) $y - x \le \frac{1}{4}$
c) $x^2 + y^2 \le \frac{1}{4}$
d) $y^2 - x^2 \le \frac{1}{4}$
e) $x + y \le \frac{1}{4}$
f) $x^2 - y^2 \le \frac{1}{4}$
This solved problem states that option d) is correct but could be wrong
Thanks