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Calculus 3 : Stokes' Theorem

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Example Question #3871 : Calculus 3

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(x^3)})}dx+{(6z + 3x^2)}dy+{(y^4)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(6 - 6x)\widehat{j}+(-4y^3)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(6)\widehat{j}+(-6)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-6x)\widehat{i}+(-4y^3)\widehat{j}+(6)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4y^3 - 6)\widehat{i}+(6x)\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4y^3 - 6)\widehat{i}+(6x)\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(x^3)})}dx+{(6z + 3x^2)}dy+{(y^4)}dz\\&\text{Meaning that}\\&F_x=sin{(x^3)};F_y=6z + 3x^2;F_z=y^4\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=4y^3-[6] \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[0]=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=6x-[0]=6x\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4y^3 - 6)\widehat{i}+(6x)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #3872 : Calculus 3

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(y + z)})}dx+{(xy^2)}dy+{(x^z)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(x^{(z - 1)}z - y^2)\widehat{i}+(cos{(y + z)})\widehat{j}+(-cos{(y + z)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(cos{(y + z)} - x^{(z - 1)}z)\widehat{j}+(y^2 - cos{(y + z)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-y^2)\widehat{j}+(x^{(z - 1)}z)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2xy - x^zlog{(x)})\widehat{i}+(x^zlog{(x)})\widehat{j}+(-2xy)\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(cos{(y + z)} - x^{(z - 1)}z)\widehat{j}+(y^2 - cos{(y + z)})\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(y + z)})}dx+{(xy^2)}dy+{(x^z)}dz\\&\text{Meaning that}\\&F_x=sin{(y + z)};F_y=xy^2;F_z=x^z\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=cos{(y + z)}-[x^{(z - 1)}z] \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=y^2-[cos{(y + z)}]\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(cos{(y + z)} - x^{(z - 1)}z)\widehat{j}+(y^2 - cos{(y + z)})\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #3873 : Calculus 3

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3)}dx+{(y^4)}dy+{(x^2 + y^3)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4y^3)\widehat{i}+(-3x^2)\widehat{j}+(3x^2 - 4y^3)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(3y^2)\widehat{i}+(-2x)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2x - 3y^2)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2x)\widehat{i}+(-3y^2)\widehat{j}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(3y^2)\widehat{i}+(-2x)\widehat{j}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3)}dx+{(y^4)}dy+{(x^2 + y^3)}dz\\&\text{Meaning that}\\&F_x=x^3;F_y=y^4;F_z=x^2 + y^3\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=3y^2-[0]=3y^2 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[2x]=-2x \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(3y^2)\widehat{i}+(-2x)\widehat{j}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #3874 : Calculus 3

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(log{(2x + x^5)})}dx+{(cos{(y^4)})}dy+{(x^2y - 2x)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2xy - x^2 - 2)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(x^2)\widehat{i}+(2 - 2xy)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2xy - 2)\widehat{i}+(-x^2)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-4y^3sin{(y^4)})\widehat{i}+(-{(5x^4 + 2)}/{(2x + x^5)})\widehat{j}+({(5x^4 + 2)}/{(2x + x^5)} + 4y^3sin{(y^4)})\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(x^2)\widehat{i}+(2 - 2xy)\widehat{j}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(log{(2x + x^5)})}dx+{(cos{(y^4)})}dy+{(x^2y - 2x)}dz\\&\text{Meaning that}\\&F_x=log{(2x + x^5)};F_y=cos{(y^4)};F_z=x^2y - 2x\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=x^2-[0]=x^2 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[2xy - 2]=2 - 2xy \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(x^2)\widehat{i}+(2 - 2xy)\widehat{j}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #41 : Stokes' Theorem

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(log{(yz)})}dx+{(e^{(xy)})}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1/y - 1/z)\widehat{i}+(ye^{(xy)} - xe^{(xy)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =0}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(ye^{(xy)})\widehat{i}+(1/y - xe^{(xy)})\widehat{j}+(-1/z)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xe^{(xy)})\widehat{i}+(1/z - ye^{(xy)})\widehat{j}+(-1/y)\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xe^{(xy)})\widehat{i}+(1/z - ye^{(xy)})\widehat{j}+(-1/y)\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(log{(yz)})}dx+{(e^{(xy)})}dz\\&\text{Meaning that}\\&F_x=log{(yz)};F_y=0;F_z=e^{(xy)}\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=xe^{(xy)}-[0]=xe^{(xy)} \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=1/z-[ye^{(xy)}] \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[1/y]=-1/y\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xe^{(xy)})\widehat{i}+(1/z - ye^{(xy)})\widehat{j}+(-1/y)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #42 : Surface Integrals

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(xyz)}dx+{(3y^3)}dy+{(x + y + z)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(xy - 1)\widehat{j}+(-xz)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(9y^2 - 1)\widehat{i}+(1 - yz)\widehat{j}+(yz - 9y^2)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xz - xy)\widehat{i}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(xz - 1)\widehat{j}+(-xy)\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(xy - 1)\widehat{j}+(-xz)\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(xyz)}dx+{(3y^3)}dy+{(x + y + z)}dz\\&\text{Meaning that}\\&F_x=xyz;F_y=3y^3;F_z=x + y + z\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=1-[0]=1 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=xy-[1] \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[xz]=-xz\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(xy - 1)\widehat{j}+(-xz)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #43 : Surface Integrals

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x + y^3z)}dx+{(2y - x^2)}dy\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2)\widehat{i}+(-1)\widehat{j}+(-1)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2x)\widehat{i}+(3y^2z)\widehat{j}+(-y^3)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(3y^2z - y^3)\widehat{i}+(2x)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(y^3)\widehat{j}+(- 2x - 3y^2z)\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(y^3)\widehat{j}+(- 2x - 3y^2z)\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x + y^3z)}dx+{(2y - x^2)}dy\\&\text{Meaning that}\\&F_x=x + y^3z;F_y=2y - x^2;F_z=0\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=y^3-[0]=y^3 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=-2x-[3y^2z]=- 2x - 3y^2z\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(y^3)\widehat{j}+(- 2x - 3y^2z)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #44 : Surface Integrals

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x - y + z)}dx+{(x + y + z)}dy+{(x - y - z)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-2)\widehat{i}+(2)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2)\widehat{i}+(-2)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-2)\widehat{i}+(-2)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =0}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-2)\widehat{i}+(2)\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x - y + z)}dx+{(x + y + z)}dy+{(x - y - z)}dz\\&\text{Meaning that}\\&F_x=x - y + z;F_y=x + y + z;F_z=x - y - z\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=-1-[1]=-2 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=1-[1]=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=1-[-1]=2\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-2)\widehat{i}+(2)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #45 : Surface Integrals

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(y^x)}dx+{(z^y)}dy+{(x^z)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-yz^{(y - 1)})\widehat{i}+(-x^{(z - 1)}z)\widehat{j}+(-xy^{(x - 1)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(z^ylog{(z)} - x^zlog{(x)})\widehat{i}+(x^zlog{(x)} - y^xlog{(y)})\widehat{j}+(y^xlog{(y)} - z^ylog{(z)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(x^{(z - 1)}z)\widehat{i}+(xy^{(x - 1)})\widehat{j}+(yz^{(y - 1)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xy^{(x - 1)})\widehat{i}+(yz^{(y - 1)})\widehat{j}+(x^{(z - 1)}z)\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-yz^{(y - 1)})\widehat{i}+(-x^{(z - 1)}z)\widehat{j}+(-xy^{(x - 1)})\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(y^x)}dx+{(z^y)}dy+{(x^z)}dz\\&\text{Meaning that}\\&F_x=y^x;F_y=z^y;F_z=x^z\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[yz^{(y - 1)}]=-yz^{(y - 1)} \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[x^{(z - 1)}z]=-x^{(z - 1)}z \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[xy^{(x - 1)}]=-xy^{(x - 1)}\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-yz^{(y - 1)})\widehat{i}+(-x^{(z - 1)}z)\widehat{j}+(-xy^{(x - 1)})\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #46 : Surface Integrals

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3sin{(x)})}dx+{(y^2 - cos{(y)}sin{(y)})}dy+{(x + y + z)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2y - cos{(y)}^2 + sin{(y)}^2 - 1)\widehat{i}+(1 - 3x^2sin{(x)} - x^3cos{(x)})\widehat{j}+(x^3cos{(x)} - 2y + 3x^2sin{(x)} + cos{(y)}^2 - sin{(y)}^2)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(1)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =0}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(-1)\widehat{j}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(-1)\widehat{j}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3sin{(x)})}dx+{(y^2 - cos{(y)}sin{(y)})}dy+{(x + y + z)}dz\\&\text{Meaning that}\\&F_x=x^3sin{(x)};F_y=y^2 - cos{(y)}sin{(y)};F_z=x + y + z\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=1-[0]=1 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[1]=-1 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(-1)\widehat{j}]\cdot\overrightarrow{A}}\end{align*}$

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Calculus 3 : Stokes' Theorem

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