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.