Curve of intersection between a sphere and a paraboloid

Question

I am working on a problem about linear differential forms and I have to calculate the following integral: $$\int_\gamma{x\space dx+y\space dy+z\space dz}$$ where $\gamma$ is the curve of intersection between the paraboloid $z=x^2+y^2$ and the sphere $x^2+(y-1)^2+z^2=1\space$. I only have to consider the part of $\gamma\space$ contained in the semi-space $\{x\ge0\}\space$. The form $\omega=x\space dx+y\space dy+z\space dz\space$ is such that the rotor of the associated force field is the vector $(0,0,0)$. This can help me solve the integral by just finding a primitive $f$ of the form (the existence of $f$ is guaranteed by the fact that the rotor is $(0,0,0)$ and by the domain of the form being $\mathbb{R}^2$), and by calculating the difference between the value of $f$ at the "arrival" point of the curve and the value at the "start". The only thing is that I can't find $\gamma$. How can I get the 1-variable vector function that describes $\gamma$ ? Thank you

Answer 1

Please try this: the equation defining the sphere can be written as $$ x^2+y^2-2y+1+z^2=1\,. $$ Putting in $z=x^2+y^2$ gives $$ z(1+z)-2y=0\,. $$ Therefore, $y$ is a function of $z:$ $y=\frac{z(1+z)}{2}$ and $x$ also: $$ x=\pm\sqrt{-z^2-y^2+2y}=\pm\sqrt{-z^2-\frac{z^2(1+z)^2}{4}+z(1+z)}=\pm\frac{1}{2}\sqrt{-z^2-2z^3-z^4 +4z}\,. $$