Line integral of a vector field

Question

$$\int_{\gamma}ydx+zdy+xdz$$ given that $\gamma$ is the intersection of $x+y=2$ and $x^2+y^2+z^2=2(x+y)$ and its projection in the $xz$ plane is taken clockwise.

In my solution, I solved the non-linear system of equations and I found that $x^2+y^2+z^2 = 4$. Given the projection in the $xz$ plane is taken clockwise, the parametrization is: $\gamma(t) = (2\sin t, 2 - 2\sin t,2\cos t)$, $0<t<2\pi$. But when I evaluate this integral, I keep getting wrong results:

$\int_{0}^{2\pi}[(2-2\cos t)2\cos t+2\cos t(-2\cos t)+2\sin t(-2\sin t)]dt$, I've only susbstituted $x,y,z$ and $dx,dy,dz$; which leads me to $-8\pi$, but the answer on my textbook is $-2\pi \sqrt{2}$.

Is there any conceptual mistake in my solution?

Answer 1

Let us just rewrite some equations first

$$\begin{align} x^2+y^2+z^2&=2(x+y)\\ (x^2-2x+1)+(y^2-2y+1)+z^2-2&=0 \\ (x-1)^2+(y-1)^2+z^2&=2 \\ \end{align}$$

So this is a sphere of radius $R=\sqrt{2}$ centered at $(1,1,0)$. It's parametric equations are

$$\begin{align} x&=1+\sqrt{2}\sin u \cos v \\ y&=1+\sqrt{2}\sin u \sin v \\ z&=\sqrt{2} \cos u \end{align}, \qquad 0 \le u \le \pi, \quad 0 \le v \lt 2\pi$$

Now, put these parametric equations into $x+y=2$ to get

$$\begin{align} x+y=2+\sqrt{2} &\sin u (\cos v + \sin v)=2 \\ &\sin u (\cos v + \sin v) =0 \\ \end{align}$$

and hence according to the range of $u$ and $v$ we have

$$\begin{array}{} u=0 & \text{Not acceptable as it just corresponds to a special point $(1,1,\sqrt{2})$} \\ v=\frac{3\pi}{4}, \frac{7\pi}{4} & \text{You can choose any of them as they just effect the orientation of the intersection curve} \end{array}$$

and finally by choosing $v=\frac{7\pi}{4}$ which gives a clock orientation we get the parametric equations as

$$\boxed{ \begin{align} x&=1+\sin u \\ y&=1-\sin u \\ z&=\sqrt{2} \cos u \end{align}, \qquad 0 \le u \lt 2\pi }$$

so your parametric equations were wrong. Here is a picture which helps you to visualize better

Answer 2

Alternatively, you can use the Stokes theorem here, it simplifies calculations. Let $\vec{F}=(y,z,x)$. Since $\gamma$ is a closed curve, the following equality holds: $$ \oint_{\gamma}\vec{F}\cdot d\vec{r} = \iint_{S}\nabla\times\vec{F}\cdot d\vec{S} = \iint_{S}\nabla\times\vec{F}\cdot\vec{n}\;dS, $$ and this is not a bad idea because $\nabla\times\vec{F}$ simplifies to $-(1,1,1)$ and $\vec{n}$ is easy to compute as the surface $S$ is part of the plane $x+y=2$, in other words $\vec{n}=\frac{1}{\sqrt{2}}(1,1,0)$ (note that it is correctly oriented). Therefore: $$ \iint_{S}\nabla\times\vec{F}\cdot\vec{n}\;dS = -\sqrt{2}\;Area(S)=-\sqrt{2}\pi2, $$ since $S$ is a circle with radius $\sqrt{2}$ (indeed, the center of the sphere ,$(1,1,0)$, lies in the plane $x+y=2$, so the intersection of the plane and the sphere is a circle).