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Q. 13

Expert-verified
Found in: Page 1140

### Calculus

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 .

See the step by step solution

## Step 1. Given Information

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

## Step 2. Understanding both the Theorems

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

## Step 3. Understanding the notations

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.