Computing a line integral along a circle in $3$-D

Question

Let $C$ be the curve of intersection of the two surfaces $x+y=2 , x^2+y^2+z^2=2(x+y)$ . The curve is to be traversed in clockwise direction as viewed from the origin . The what is the value of $\int_Cydx+zdy+xdz$ ? I am not even able to parametrize the curve of intersection . Please help . Thanks in advance

Answer 1

I think a curve along the intersection of the two surfaces can be thought of as consisting of two curves $C_1$ and $C_2$ with parameterizations $r_1: \left [ \ 0, 2 \right ] \to \mathbb{R^3}$ where $$r_1(x) := \left ( x, 2 - x, \sqrt{4x - 2x^2} \right)$$ and $r_2: \left [ \ 0, 2 \right ] \to \mathbb{R^3}$ where $r_2(x) := \left ( 2-x, x, -\sqrt{4x-2x^2} \right )$ for $x \in \left [ \ 0, 2 \ \right ]$. You can obtain this relation by simply writing $y$ in terms of $x$ to find a relation between $x$ and $z$ for the points on the curve of intersection. This parameterization can then possibly help you solve the integral.

See that both $C_1$ and $C_2$ in $\mathbb{R^3}$ lie on the plane $x+y=2$ and are traversed in clockwise direction when seen from the origin.

Alternatively, see that the surface $S$ with parameterization $r: \left [0, 2 \right ] × \left [ -1, 1 \right ] \to \mathbb{R^3}$ where $$r\left ( x, t \right ) := \left ( x, 2 - x, t\sqrt{4x - 2x^2} \right )$$ lies on the plane $x+y = 2$ and the curve of intersection is the boundary of this surface. Then you can easily compute the line integral using stokes thoerem.

Answer 2

Note that your integral is equivalent to:

$$ \int_C \vec{F} \cdot d\vec{r}, $$

with $\vec{F}=(y,z,x)$.

Now, lets have a look at $C$: it is the intersection between the plane $x+y=2$, and the sphere $(x-1)^2+(y-1)^2+z^2=2$. In other words $C$ is an ellipse. Most importantly, it is a closed curve that belongs to the plane $x+y=2$.

It follows by the Stokes theorem that $$ \oint_C \vec{F} \cdot d\vec{r}= \iint_S \nabla \times \vec{F} \; d\vec{S} =\iint_S \nabla \times \vec{F} \cdot \vec{n} \;dS, $$

with $\nabla \times \vec{F}=(-1,-1,-1)$ and $\vec{n}=\frac{1}{\sqrt{2}}(1,1,0)$, which yields

$$ \oint_C \vec{F} \cdot d\vec{r}= -\frac{2}{\sqrt{2}} \iint_S dS = -\frac{2}{\sqrt{2}} \;Area(S). $$

Since the border of $S$ is the ellipse $(x-1)^2+\frac{z^2}{2}=1$, we can conclude that $Area(S)=\pi\sqrt{2}$, and therefore

$$ \oint_C \vec{F} \cdot d\vec{r} = -2\pi. $$

Answer 3

Hint:

In spherical coordinates,

$$r\cos(\phi)\sin(\theta)+r\cos(\phi)\cos(\theta)=2,\\ r^2=2(r\cos(\phi)\sin(\theta)+r\cos(\phi)\cos(\theta))=4.$$

Then from the first equation,

$$\cos(\phi)=\frac1{\sin(\theta)+\cos(\theta)}$$and$$\sin(\phi)=\sqrt{1-\cos^2(\phi)}.$$

This gives a representation of $(x,y,z)$ as a function of $\theta$.