How Do Gauss' and Stokes' Theorems Apply to These Integral Problems?

  • Thread starter Thread starter superpig10000
  • Start date Start date
  • Tags Tags
    Gauss Theorem
superpig10000
Messages
7
Reaction score
0
Hi guys,

I am having trouble with this "simple" problem involving these two theorems:

Find the value of the integral (A dot da) over the surface s, where A = xi - yj + zk and S is the closed surface defined by the cylinder c^2 = x^2 + y^2. The top and bottom of the cylinder are z= 0 and z=d.

From common sense, integrating circular layers from z=0 to z=d should give the volume of a cylinder. The book doesn't have any sample problem so I don't know which theorem to apply, and how.

Here's a more complicated question:

Find the value of the integral (curl A da) over the surface s, where A = yi + zj + xk and S is the closed surface defined by the paraboloid z=1-x^2-y^2 where z >=0

I appreciate any help.
 
Physics news on Phys.org
"Find the value of the integral (A dot da) over the surface s, where A = xi - yj + zk and S is the closed surface defined by the cylinder c^2 = x^2 + y^2. The top and bottom of the cylinder are z= 0 and z=d."
Find div A. (It is a constant.) Then just multiply by the volume of a cylinder.
 
"Find the value of the integral (curl A da) over the surface s, where A = yi + zj + xk and S is the closed surface defined by the paraboloid z=1-x^2-y^2 where z >=0"
By either the div theorem or Stokes' theorem, the integral of curl over a closed surface=0. Prove it.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Back
Top