Understanding Stoke's Theorem: Does Surface Integral Depend on Shape?

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Stokes' Theorem states that the surface integral of the curl of a vector field over a surface is equal to the line integral of the vector field along the boundary curve of that surface. The integral's value is determined solely by the boundary curve, meaning it remains constant regardless of the surface shape, as long as the boundary remains unchanged. Examples provided include a paraboloid, a hemisphere, and a half cone, all sharing the same boundary curve. Consequently, the surface integral does not depend on the specific shape of the surface but rather on the boundary it encloses. This understanding clarifies the relationship between surface integrals and their boundaries in the context of Stokes' Theorem.
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i'm trying to understand stoke's theorem and am having trouble seeing whether the surface integral for a given surface changes with any change in its shape, or if it only changes depending on the cross sectional area perpendicular to the direction of the vector field. can anybody help me out?
 
Mathematics news on Phys.org
Stokes' Theorem equates the flux integral of the curl of a vector field \vec{F} over an orientated piecewise-smooth surface S to the line integral of \vec{F} along the simple, closed, piecewise-smooth curve C which is the boundary of the surface S, symbolically

\iint_S (\vec{\nabla}\times\vec{F})\cdot d\vec{S} = \oint_C \vec{F}\cdot d\vec{r}​

The value of the integral is fixed by the value of the line integral on the righthand side which could only change if the boundary curve C changes, hence the integral is unchanged whether you integrate over the paraboliod S_1: z=4-x^2-y^2,z\geq 0 or the hemisphere S_2: x^2+y^2+z^2=4,z\geq 0 or the half cone S_3: (z-4)^2=4(x^2+y^2),0\leq z\leq 4 since these all have as their boundary curve C the circle C:x^2+y^2=4.
 
perfect just what i wanted to hear! thanks
 
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