- #1
TheDestroyer
- 402
- 1
Hi guys, I want to solve this integral without using gauss theorem to convert from double to triple integral,
my problem here is to find a general way for setting the differential surface part (ds) from the integral
if [tex]\vec{F}[/tex] was a vector field defined as:
[tex]\vec{F} = x^3\vec{i} + y^3\vec{j} + z^2\vec{k}[/tex]
[tex]\vec{i},\vec{j},\vec{k}[/tex] Unit Vectors for axes [tex]x,y,z[/tex]
Evaluate:
[tex] \oint_{s} \vec{F}\cdot\vec{ds} [/tex]
While s is the area of the volume bounded with equation surfaces:
[tex]z=0, z=1-(x^2+y^2)[/tex]
the answer will be [tex]\frac{5\pi}{6}[/tex]
PLEASE DON'T USE THE THEOREM:
[tex]\int\int\int_{V}\vec{\nabla}\cdot\vec{F}\cdot dV[/tex]
my problem here is to find a general way for setting the differential surface part (ds) from the integral
if [tex]\vec{F}[/tex] was a vector field defined as:
[tex]\vec{F} = x^3\vec{i} + y^3\vec{j} + z^2\vec{k}[/tex]
[tex]\vec{i},\vec{j},\vec{k}[/tex] Unit Vectors for axes [tex]x,y,z[/tex]
Evaluate:
[tex] \oint_{s} \vec{F}\cdot\vec{ds} [/tex]
While s is the area of the volume bounded with equation surfaces:
[tex]z=0, z=1-(x^2+y^2)[/tex]
the answer will be [tex]\frac{5\pi}{6}[/tex]
PLEASE DON'T USE THE THEOREM:
[tex]\int\int\int_{V}\vec{\nabla}\cdot\vec{F}\cdot dV[/tex]