Confused- Integrating a vector field along a curve in 3D.

[joelio36]

Homework Statement

Let f be a vector function, f = (xz, 0, 0), and C a contour formed by the boundary of the surface S

S : x^2 + y^2 + z^2 = R^2 , x ≥ 0, y ≥ 0, z ≥ 0 , and oriented counterclockwise (as seen from the origin).
Evaluate the integral
(Closed integral sign) f · dr , directly as a contour integral. [8 marks]

The Attempt at a Solution

Here's what I tried:

-The shell of the sphere quadrant in spherical co-ords is:
x = R sin(theta)cos(phi)
y = R sin(theta)sin(phi)
z = R cos(theta)

for theta, phi between 0 and pi/2, gives the surface. However, we want to integrate along the boundary, i.e. along the following theta-phi routes:

(0,0)-->(pi/2,0)-->(pi/2,pi/2)-->(0,pi/2).

Now I have the path I want to integrate over.

for dr in the integratal (i.e. dr is the infintiesimal displacement), I used:

(let U be the (x,y,z) function defined above, the position vector for the surface)

dr = dU/d(theta) * d(theta) + dU/d(phi) * d(phi)

I've tried figuring it out from there but it gets very messy. Far too messy for an 8 mark question.

Thanks very much! Joel

Sorry for lack of LATEX use, I'm trying to learn!

 

[LCKurtz]

Homework Statement

Let f be a vector function, f = (xz, 0, 0), and C a contour formed by the boundary of the surface S

S : x^2 + y^2 + z^2 = R^2 , x ≥ 0, y ≥ 0, z ≥ 0 , and oriented counterclockwise (as seen from the origin).
Evaluate the integral
(Closed integral sign) f · dr , directly as a contour integral. [8 marks]

The Attempt at a Solution

Here's what I tried:

-The shell of the sphere quadrant in spherical co-ords is:
x = R sin(theta)cos(phi)
y = R sin(theta)sin(phi)
z = R cos(theta)

for theta, phi between 0 and pi/2, gives the surface. However, we want to integrate along the boundary, i.e. along the following theta-phi routes:

(0,0)-->(pi/2,0)-->(pi/2,pi/2)-->(0,pi/2).

Now I have the path I want to integrate over.

for dr in the integratal (i.e. dr is the infintiesimal displacement), I used:

(let U be the (x,y,z) function defined above, the position vector for the surface)

dr = dU/d(theta) * d(theta) + dU/d(phi) * d(phi)

I've tried figuring it out from there but it gets very messy. Far too messy for an 8 mark question.

Thanks very much! Joel

Sorry for lack of LATEX use, I'm trying to learn!

Remember what you are integrating:

$$\int_C \langle xz, 0, 0\rangle \cdot \langle dx,dy,dz\rangle$$

You only get one term and if you look carefully you will see that only one of the three arcs gives anything nonzero.