Suggested languages for you:

Americas

Europe

Q. 13

Expert-verifiedFound in: Page 1140

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Why is $dA$ in Green’s Theorem replaced by $dS$ in Stokes’ Theorem?

$dA$refers to ordinary area while $dS$refers to the surface area. So,$dA$ in Green's Theorem is replaced by $dS$ in Stokes' Theorem because .

$dA$ in Green’s Theorem replaced by $dS$ in Stokes’ Theorem.

Green's Theorem states that;"Let $R$ be a region in the plane with smooth boundary curve $C$ oriented counterclockwise by $\mathbf{r}\left(t\right)=\u27e8\left(x\right(t),y(t\left)\right)\u27e9$ for $a\le t\le b$.If a vector field $\mathbf{F}(x,y)=>">{F}_{1}(x,y),{F}_{2}(x,y)$ is defined by $R$, then, ${\int}_{C}\mathbf{F}\xb7d\mathbf{r}={\iint}_{R}\left(\frac{\partial {F}_{2}}{\partial x}-\frac{\partial {F}_{1}}{\partial y}\right)dA."$Stokes' Theorem states that;"Let be $S$ an oriented, smooth or piecewise-smooth surface bounded by a curve $C$. Suppose that $\mathit{n}$ is an oriented unit normal vector of $S$ and $C$ has a parametrization that traverses $C$ in the counterclockwise direction with respect to $\mathit{n}$.If a vector field $\mathbf{F}(x,y,z)={F}_{1}(x,y,z)\mathbf{i}+{F}_{2}(x,y,z)\mathbf{j}+{F}_{3}(x,y,z)\mathbf{k}$is defined on $S$, then, ${\int}_{C}\mathbf{F}(x,y,z)\xb7d\mathbf{r}={\iint}_{S}curl\mathbf{F}(x,y,z)\xb7\mathbf{n}dS"$.

Both the notations $dA$ and $dS$ refer to area.In the case of Green's Theorem, $dA$ refers to the ordinary area.In the case of Stokes' Theorem, $dS$ refers to the surface area. In this case, $dS$ keeps track of any distortion to the area of the surface $S$ caused by transforming a patch of the $uv-$plane to parametrize $S$. Hence, $dA$ in Green's Theorem can be replaced by $dS$ in Stokes' Theorem.

94% of StudySmarter users get better grades.

Sign up for free