$\oint_{c} xy ds$ where $C$ is the intersection of the surfaces $x^2+y^2=4$ and $ y+z=8 $
Hi. I tried to resolve the issue below using the definition of line integral, but I couldn’t solve it. Could someone appreciate my resolution and help me finish it? Below follows what I did:
$$\oint_{c} xy ds = {\int }_{C}f\left(x,y,z\right)ds= \int_{a}^{b}f\left(\text{r}\left(t\right)\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}+{\left({z}^{\prime }\left(t\right)\right)}^{2}}dt.$$
However, I could not continue.
Thanks!
Hint:
You will need to find the parametrization $\textbf{r}(t)$ for the intersection.
First, in order to be on the surface $x^2+y^2=4$, it must look like $x=...$ and $y=...$ (review parametrization of circles).
Then, in order to be on the surface $y+z=8$, i.e. $z=8-y$, we get $z=...$ (simply substitute what you got for y=).
So then $\textbf{r}(t)=...$