So I have to calculate line integral of a vector field: $\overrightarrow{F}(x,y,z)=(x+y,y+z,x+y)$
And I want to calculate the Line integral:
$$\int_K \overrightarrow{F}(x,y,z)d\overrightarrow{r}$$
And K is defined as intercutting of these two surfaces: $$x^2+y^2=1, z=x^2+y^2-2y$$
I did do a sketch, and is seems like it some sort of elipse or circle.
But how can I get its equation?
I thougth maybe something like this this:
$x^2=1-y^2 \rightarrow z=1- y^2+y^2-2y=-2y $
But it just gave me normal line, or is this all there is to it?
Are there any tricks to it. Any help with the solution would be appreciated.
You wish to integrate over the set \begin{align} K &= \{(x,y,z) \in \Bbb{R}^3: x^2 + y^2 =1 \quad \text{and} \quad z = x^2 + y^2 - 2y \} \\ &= \{(x,y,z) \in \Bbb{R}^3: x^2 + y^2 =1 \quad \text{and} \quad z = 1 - 2y \} \tag{$*$} \end{align} Notice that if you now define the function $\alpha: [0,2 \pi] \to \Bbb{R}^3$ by the rule \begin{equation} \alpha(t) = \begin{pmatrix} \cos(t) \\ \sin(t) \\ 1 - 2 \sin(t) \end{pmatrix}, \end{equation} then $K = \text{image}(\alpha)$. In words, $\alpha$ is a parametrization for $K$. I hope you see how I obtained such a parametrization from ($*$). Hence, \begin{align} \int_K \vec{F}(x,y,z) \cdot d \vec{r} &= \int_0^{2\pi} \left(\vec{F}(\alpha(t)) \cdot \dfrac{d \alpha}{dt} \right) \, dt \end{align}
(the $\cdot$ being a dot product of vectors in $\Bbb{R}^3$) I'll let you substitute the formulas and perform the computation. I got $-3 \pi$ as the answer... there may/may not have been a computation error, but this is the general process of how to approach these questions.