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Calculus 3 : Stokes' Theorem

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Example Questions

Example Question #3897 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #3898 : Calculus 3

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

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2xe^{(2z)} - 5y)\widehat{i}+(-2xe^{(2z)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5y)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-2xe^{(2z)})\widehat{i}+(-5y)\widehat{j}+(2xe^{(2z)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5x - 2x^2e^{(2z)})\widehat{i}+(2x^2e^{(2z)} - e^{(2z)})\widehat{j}+(e^{(2z)} - 5x)\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5y)\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

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

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=2xe^{(2z)}-[2xe^{(2z)}]=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=5y-[0]=5y\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5y)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #3899 : Calculus 3

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(2z + 3e^{(4y)} + 5x^2)}dx\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-10x)\widehat{j}+(10x)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(12e^{(4y)})\widehat{j}+(-2)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(12e^{(4y)} - 2)\widehat{i}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2)\widehat{j}+(-12e^{(4y)})\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2)\widehat{j}+(-12e^{(4y)})\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(2z + 3e^{(4y)} + 5x^2)}dx\\&\text{Meaning that}\\&F_x=2z + 3e^{(4y)} + 5x^2;F_y=0;F_z=0\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=2-[0]=2 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[12e^{(4y)}]=-12e^{(4y)}\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2)\widehat{j}+(-12e^{(4y)})\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #3900 : Calculus 3

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(8e^{(x)})}dx+{(-9z^2)}dy+{(3y^3)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-18z)\widehat{j}+(-9y^2)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-9y^2)\widehat{j}+(-18z)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-8e^{(x)})\widehat{j}+(8e^{(x)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(18z + 9y^2)\widehat{i}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(18z + 9y^2)\widehat{i}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(8e^{(x)})}dx+{(-9z^2)}dy+{(3y^3)}dz\\&\text{Meaning that}\\&F_x=8e^{(x)};F_y=-9z^2;F_z=3y^3\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=9y^2-[-18z]=18z + 9y^2 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[0]=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(18z + 9y^2)\widehat{i}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #61 : Surface Integrals

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x + y)}dx+{(z^3)}dy+{(y + y^5)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5y^4 - 3z^2 + 1)\widehat{i}+(-1)\widehat{k}]\cdot\overrightarrow{A}}$

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(3z^2)\widehat{j}+(- 5y^4 - 1)\widehat{k}]\cdot\overrightarrow{A}}$

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

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5y^4 - 3z^2 + 1)\widehat{i}+(-1)\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x + y)}dx+{(z^3)}dy+{(y + y^5)}dz\\&\text{Meaning that}\\&F_x=x + y;F_y=z^3;F_z=y + y^5\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=5y^4 + 1-[3z^2] \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[0]=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[1]=-1\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5y^4 - 3z^2 + 1)\widehat{i}+(-1)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #62 : Surface Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=-2x^2y-[0]=-2x^2y \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=-1-[-2xy^2]=2xy^2 - 1 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=3x^2-[2]\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-2x^2y)\widehat{i}+(2xy^2 - 1)\widehat{j}+(3x^2 - 2)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #63 : Surface Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=1-[0]=1 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=-y^3-[0]=-y^3 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[-3y^2z]=3y^2z\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(-y^3)\widehat{j}+(3y^2z)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #64 : Surface Integrals

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(ye^{(5z - z^2)})}dx+{(z + x^3)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(- 3x^2 - ye^{(5z - z^2)}{(2z - 5)})\widehat{j}+(-e^{(5z - z^2)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(e^{(5z - z^2)} + ye^{(5z - z^2)}{(2z - 5)})\widehat{i}+(3x^2)\widehat{k}]\cdot\overrightarrow{A}}$

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

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(- 3x^2 - ye^{(5z - z^2)}{(2z - 5)})\widehat{j}+(-e^{(5z - z^2)})\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(ye^{(5z - z^2)})}dx+{(z + x^3)}dz\\&\text{Meaning that}\\&F_x=ye^{(5z - z^2)};F_y=0;F_z=z + x^3\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=-ye^{(5z - z^2)}{(2z - 5)}-[3x^2]=- 3x^2 - ye^{(5z - z^2)}{(2z - 5)} \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[e^{(5z - z^2)}]=-e^{(5z - z^2)}\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(- 3x^2 - ye^{(5z - z^2)}{(2z - 5)})\widehat{j}+(-e^{(5z - z^2)})\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #65 : Surface Integrals

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x + 3yz^3)}dx+{(-x^2)}dy+{(z + 5)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(3z^3 - 9yz^2)\widehat{i}+(2x)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2x)\widehat{i}+(3z^3)\widehat{j}+(-9yz^2)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(9yz^2)\widehat{j}+(- 2x - 3z^3)\widehat{k}]\cdot\overrightarrow{A}}$

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

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(9yz^2)\widehat{j}+(- 2x - 3z^3)\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x + 3yz^3)}dx+{(-x^2)}dy+{(z + 5)}dz\\&\text{Meaning that}\\&F_x=x + 3yz^3;F_y=-x^2;F_z=z + 5\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=9yz^2-[0]=9yz^2 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=-2x-[3z^3]=- 2x - 3z^3\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(9yz^2)\widehat{j}+(- 2x - 3z^3)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #66 : Surface Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[x^{(e^{(2z)} - 1)}e^{(2z)}]=-x^{(e^{(2z)} - 1)}e^{(2z)} \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[3y^2e^{(x)}]=-3y^2e^{(x)}\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-x^{(e^{(2z)} - 1)}e^{(2z)})\widehat{j}+(-3y^2e^{(x)})\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Calculus 3 : Stokes' Theorem

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