# Thread: Line Integral

1. ## Line Integral

Hi, could you help me out with this line integral?

Calculate the line integral of the field $f=(y,z,x)$ around the surfaces $x+y=2$ and $x^2+y^2+z^2=2(x+y)$

Thanks!!!

2. Sphere may rewriten
$(x-1)^2+(y-1)^2+z^2=2$
with centre (1,1,0).
Intersection with plane y=2-x gives circle
centre (1,1,0) radius $\sqrt{2}$.
In parametic form it is
$x=1+cost$
$y=1-cost$
$z=\sqrt{2}sint$
t[0,2pi].
$\int{f(x,y,z)ds}=\int{f(x(t),y(t),z(t))\sqrt{(\fra c{dx}{dt})^2+({\frac{dy}{dt}})^2+({\frac{dz}{dt}}) ^2}dt}$
This $\sqrt{}$= $\sqrt{2}$ so we get
$\sqrt{2}\int{f(1+cost,1-cost,\sqrt{2}sint)dt}$ from 0 to $2\pi$.
Please check.

3. I think f is a vector field (y,z,x) and the line integral should be
$\int f.dl$

Originally Posted by zzzoak
Sphere may rewriten
$(x-1)^2+(y-1)^2+z^2=2$
with centre (1,1,0).
Intersection with plane y=2-x gives circle
centre (1,1,0) radius $\sqrt{2}$.
In parametic form it is
$x=1+cost$
$y=1-cost$
$z=\sqrt{2}sint$
t[0,2pi].
$\int{f(x,y,z)ds}=\int{f(x(t),y(t),z(t))\sqrt{(\fra c{dx}{dt})^2+({\frac{dy}{dt}})^2+({\frac{dz}{dt}}) ^2}dt}$
This $\sqrt{}$= $\sqrt{2}$ so we get
$\sqrt{2}\int{f(1+cost,1-cost,\sqrt{2}sint)dt}$ from 0 to $2\pi$.
Please check.