Divergence Theorem: Volume and Surface Integral Solutions

  • Thread starter Thread starter Phymath
  • Start date Start date
AI Thread Summary
The discussion focuses on applying the divergence theorem to a vector field in spherical coordinates, specifically analyzing the volume and surface integrals of a hemisphere. The user initially calculates the volume integral and finds a result of 10Rπ. However, they encounter a discrepancy when performing the surface integral, leading to confusion over the correct differential area element. Another participant points out that the user incorrectly used R^3 instead of the correct R^2 for the surface integral. This correction highlights the importance of accurately applying the divergence theorem in spherical coordinates.
Phymath
Messages
183
Reaction score
0
ok well basicly use the divergence thrm on this..
<br /> \vec{v} = rcos\theta \hat{r} + rsin\theta\hat{\theta} + rsin\theta cos \phi \hat{\phi}

so i did (remember spherical coords) i get..
5cos \theta - sin\phi
taking that over the volume of a hemisphere resting on the xy-plane i get 10R \pi

however with the surface intergral i use d\vec{a} = R^3 sin\theta d\theta d\phi \hat{r} where R is the radius of the sphere doing that intergral gives me something obvioiusly R^3 which is not what I'm getting in the volume intergral any help or explanations (if its i need to take the surface of the bottom of the hemi sphere aka the xy-plane in a circle that will suck but let me know!
 
Physics news on Phys.org
however with the surface intergral i use d\vec{a} = R^3 sin\theta d\theta d\phi \hat{r} where R is the radius of the sphere doing that intergral
Your surface integral is wrong.
d\vec{a} = R^2 sin\theta d\theta d\phi \hat{r} not R3.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top