Find the line integrel of vector field \( F=x \vec{i}+y \vec{j}+z \vec{k} \) on the interface of sphere \( x^{2}+y^{2}+z=2 x \) and plane \( z=x \) ?

To evaluate the line integral of the vector field \( \mathbf{F} = x\, \mathbf{i} + y\, \mathbf{j} + z\, \mathbf{k} \) along the curve \( C \), which is the intersection of the sphere and the plane given by: \[ \begin{cases} x^2 + y^2 + z^2 = 2x \\ z = x \end{cases} \] we can proceed as follows: ### Step 1: Simplify the Sphere Equation The equation of the sphere can be rewritten by completing the square for \( x \): \[ x^2 - 2x + y^2 + z^2 = 0 \\ (x - 1)^2 + y^2 + z^2 = 1 \] This represents a sphere centered at \( (1, 0, 0) \) with radius \( 1 \). ### Step 2: Find the Intersection Curve \( C \) Substituting \( z = x \) into the sphere's equation: \[ (x - 1)^2 + y^2 + x^2 = 1 \\ 2x^2 - 2x + y^2 = 0 \\ 2(x^2 - x) + y^2 = 0 \\ 2\left( (x - \frac{1}{2})^2 - \frac{1}{4} \right) + y^2 = 0 \\ 2(x - \frac{1}{2})^2 + y^2 = \frac{1}{2} \] This is the equation of a circle in the plane \( z = x \) with center \( \left(\frac{1}{2}, 0, \frac{1}{2}\right) \) and radius \( \frac{1}{2} \). ### Step 3: Evaluate the Line Integral The key observation is that the vector field \( \mathbf{F} = x\, \mathbf{i} + y\, \mathbf{j} + z\, \mathbf{k} \) is **conservative**. A conservative vector field has the property that its line integral around any closed loop is zero. This is because there exists a scalar potential function \( \phi \) such that \( \mathbf{F} = \nabla \phi \), and for closed curves: \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = 0 \] Since the curve \( C \) is a closed loop (the intersection of the sphere and the plane), the line integral of \( \mathbf{F} \) around \( C \) is: \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = 0 \] ### Final Answer The line integral is zero. \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = 0 \]