Blizzard_1708

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 (aR = R). ** [math] \int D.ds [/math] dsR = [math] \hat {R} R^2 sin( \theta ) d \theta d \phi [/math] D.dsR = [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.

Alright, I'm new here, so hopefully I can get the problem and my work across well. Textbook Problem: A vector field D = aR(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 (aR = R). ** Integral D.ds ** dsR = RR2sin(Theta)d(Theta)d(Phi) **From textbook D.dsR = (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.