slonopotam
Aug13-09, 03:06 PM
calculate
\iint_{M}^{}rot(\vec{F})\vec{dS}
where
\vec{F}=(y^2z,zx,x^2z^2)
M is a part of z=x^2+y^2 which is located in 1=x^2+y^2
and its normal vector points outside
i am used to solve it like this
\iint_{M}^{}rot\vec{F}\vec{dS}=\iint_{D}\frac{rot\ vec{F}\cdot \vec{N}}{|\vec{N}\cdot\vec{K}|}dxdy
\vec{N}=(2x,2y,-1)
\iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{(-x,2xz^2-y^2,z-yz) \cdot (2x,2y,-1)}{1}dxdy
now i convert into polar coordinates
x^2+y^2=r
is this method ok?
\iint_{M}^{}rot(\vec{F})\vec{dS}
where
\vec{F}=(y^2z,zx,x^2z^2)
M is a part of z=x^2+y^2 which is located in 1=x^2+y^2
and its normal vector points outside
i am used to solve it like this
\iint_{M}^{}rot\vec{F}\vec{dS}=\iint_{D}\frac{rot\ vec{F}\cdot \vec{N}}{|\vec{N}\cdot\vec{K}|}dxdy
\vec{N}=(2x,2y,-1)
\iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{(-x,2xz^2-y^2,z-yz) \cdot (2x,2y,-1)}{1}dxdy
now i convert into polar coordinates
x^2+y^2=r
is this method ok?