Alright, I'm new here, so hopefully I can get the problem and my work across well.

 

Textbook Problem: A vector field D = a_R(cos²(Phi))/R³ exists in the region between two spherical shells defined by R=1 and R=2. Find Integral D^.ds and Intregral Gradient^.Ddv. Note the vector (a_R = R).

 

** Integral D^.ds **

ds_R = RR²sin(Theta)d(Theta)d(Phi) **From textbook

D^.ds_R = (cos²(Phi)/R³) R² sin(Theta) d(Theta) d(Phi)

Intregral (from 0 to 2pi) Integral (from 0 to pi) (1/R) cos²(Phi) sin(Theta) d(Theta) d(Phi) **Double integral for surface area

Intregral (from 0 to 2pi) (1/R) cos²(Phi) (-cos(Theta) | (from 0 to pi)) d(Phi)

** -cos(pi) = 1; -cos(0) = -1; 1 - (-1) = 2

Intregral (from 0 to 2pi) (2/R)cos²(Phi) d(Phi)

(2/R) ((sin(Phi)cos(Phi)/2 + Phi/2) | (from 0 to 2pi))

** sin(2pi) = 0; sin(0) = 0; 2pi/2 = pi; 0/2 = 0; pi - 0 = pi

2pi/R

** Evaluate for R = 1 to R = 2

pi - 2pi = -pi

 

** Intregral Gradient^.Ddv **

Gradient = R partial/partial® + Theta (1/R) partial/partial(Theta) + Phi (1/(Rsin(Theta)) partial/partial(Theta)

Gradient^.D = partial/partial® (cos²(Phi))/R³

Gradient^.D = -3cos²(Phi)/R⁴

Integral (from 1 to 2) Integral (from 0 to 2pi) Integral (from 0 to pi) [-3cos²(Phi)/R⁴]R²sin(Theta)d(Theta)d(Phi)dR

Integral (from 1 to 2) Integral (from 0 to 2pi) Integral (from 0 to pi) -3/R² cos²(Phi) sin(Theta) d(Theta) d(Phi) dR ** Triple integral for volume

Integral (from 1 to 2) Integral (from 0 to 2pi) -3/R² cos²(Phi) (-cos(Theta) | (from 0 to pi)) d(Phi) dR

Integral (from 1 to 2) Integral (from 0 to 2pi) -6/R² cos²(Phi) d(Phi) dR **Used evaluation of (-cos(Theta)) = 2 from above

Integral (from 1 to 2) -6/R² ((sin(Phi)cos(Phi)/2 + Phi/2) | (from 0 to 2pi)) dR

Integral (from 1 to 2) -6pi/R² dR **Used evaluation of ((sin(Phi)cos(Phi)/2 + Phi/2) | (from 0 to 2pi)) = pi from above

6pi/R | (from 1 to 2)

6pi/2 = 3pi; 6pi/1 = 6pi; 3pi - 6pi = -3pi

 

I can't find any error I have made in the above problem, but by divergence theorem, these two should be equal. Sorry for the lack of symbols. I couldn't figure out how to add them. Thanks in advance for any help.

 

argh, this is extremely hard to read.

use

[math]\phi  \Phi \int\limits_a^b \Delta \frac{numerator}{denominator}\sin{(x^2)}\nabla \partial \mathbf{D}\mathcal{R}\hat{a}[/math]

And so on to do symbols.

Producing this type of thing:

[math]\phi \Phi \int\limits_a^b \Delta \frac{numerator}{denominator}\sin{(x^2)}\nabla \partial \mathbf{D} \mathcal{R}\hat{a}[/math]

 

There's a full tutorial for using latex in this forum stickied in the maths forum in case you haven't used latex before.

Edited October 11, 201114 yr by Schrödinger's hat

Textbook Problem: A vector field [math]D = \hat{R}(cos^2{(\phi)})/R^3[/math] exists in the region between two spherical shells defined by R=1 and R=2. Find [math]\int D.ds[/math] and [math]\int \nabla.Ddv[/math]. Note the vector (a_R = R).

 

** [math] \int D.ds [/math]

ds_R = [math] \hat {R} R^2 sin( \theta ) d \theta d \phi [/math]

D^.ds_R = [math] (cos^2( \phi )/R^3) R^2 sin( \theta ) d \theta d \phi [/math]

[math] \int \limits_0^{2\pi} \int \limits_0^\pi \frac {1} {R} cos^2( \phi ) sin( \theta ) d \theta d \phi [/math]

[math] \int \limits_0^{2\pi} \frac {1} {R} cos^2 ( \phi ) (-cos ( \theta ) | from 0 to \pi ) d \phi [/math]

** [math] -cos ( \pi ) = 1; -cos (0) = -1; 1 - (-1) = 2 [/math]

[math] \int \limits_0^{2\pi} \frac {2} {R} cos^2 ( \phi ) d \phi [/math]

[math] \frac {2} {R} ( \frac {sin ( \phi ) cos( \phi )} {2} + \frac { \phi } {2} | from 0 to 2\pi) [/math]

** [math] sin(2\pi) = 0; sin(0) = 0; \frac {2\pi} {2} = \pi; \frac {0} {2} = 0; \pi - 0 = \pi [/math]

[math] \frac {2\pi} {R} [/math]

** Evaluate for R = 1 to R = 2

[math] \pi - 2\pi = -\pi [/math]

 

** [math] \int \nabla D dv [/math]

[math] \nabla = \hat {D} \frac {\partial} {\partial R} + \hat {\theta} \frac {1} {R} \frac {\partial} {\partial \theta} + \hat {\phi} \frac {1} {R sin \theta} \frac {\partial} {\partial \theta} [/math]

[math] \nabla \hat {D} = \frac {\partial} {\partial R} \frac {cos^2( \phi )} {R^3} [/math]

[math] \nabla \hat {D} = \frac {-3cos^2( \phi )} {R^4} [/math]

[math] \int \limits_1^2 \int \limits_0^{2\pi} \int \limits_0^\pi \frac {-3cos^2( \phi )} {R^4}R^2 sin( \theta ) d \theta d \phi dR [/math]

[math] \int \limits_1^2 \int \limits_0^{2\pi} \int \limits_0^\pi \frac {-3} {R^2} cos^2 ( \phi ) sin( \theta ) d \theta d \phi dR [/math]

[math] \int \limits_1^2 \int \limits_0^{2\pi} \frac {-3} {R^2cos^2} ( \phi ) (-cos ( \theta) | (from 0 to \pi)) d \phi dR [/math]

[math] \int \limits_1^2 \int \limits_0^{2\pi} \frac {-6} {R^2}cos^2( \phi ) d \phi dR [/math] **Used evaluation of [math] (-cos( \theta )) = 2 [/math] from above

[math] \int \limits_1^2 \frac {-6} {R^2} ( \frac {sin ( \phi ) cos( \phi )} {2} + \frac {\phi} {2}) | (from 0 to 2\pi)) dR [/math]

[math] \int \limits_1^2 \frac {-6\pi} {R^2}dR [/math] **Used evaluation of [math] (\frac {sin ( \phi ) cos ( \phi )} {2} + \frac {\phi} {2} | (from 0 to 2\pi)) = \pi [/math] from above

[math] \frac {6\pi} {R} | (from 1 to 2) [/math]

[math] \frac {6\pi} {2} = 3\pi; \frac {6\pi} {1} = 6\pi; 3\pi - 6\pi = -3\pi [/math]

 

Gotta run to class now, hopefully this is better! Thanks Schrodinger's hat for the quick tutorial.

Textbook Problem: A vector field [math]D = \hat{R}(cos^2{(\phi)})/R^3[/math] exists in the region between two spherical shells defined by R=1 and R=2. Find [math]\int D.ds[/math] and [math]\int \nabla.Ddv[/math]. Note the vector (a_R = R).

 

 

I lack the patience to go through your calculation in detail, but you should realize that the generalized Stokes theorem (in the form that elementary texts call the divergence theorem) states that

 

[math]\int D \cdot ds[/math] = [math]\int \nabla \cdot Ddv[/math]