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Calculus 3 : Stokes' Theorem

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Example Question #51 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #52 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #53 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #54 : Stokes' Theorem

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

Possible Answers:

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(x^yln{(x)})\widehat{i}+(10x - 4)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4 - 10x)\widehat{j}+(-x^yln{(x)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(10x - 4)\widehat{i}+(x^ylog{(x)})\widehat{j}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4 - 10x)\widehat{j}+(-x^yln{(x)})\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

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

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[10x - 4]=4 - 10x \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[x^yln{(x)}]=-x^yln{(x)}\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4 - 10x)\widehat{j}+(-x^yln{(x)})\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #55 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #56 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #57 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #58 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #59 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #60 : Stokes' Theorem

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x^{(sin{(y)} + 1)})}dx+{(yz^3)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x^{(sin{(y)} + 1)})}dx+{(yz^3)}dz\\&\text{Meaning that}\\&F_x=x^{(sin{(y)} + 1)};F_y=0;F_z=yz^3\\&\text{From this we can derive our curl vectors:}\end{align*}$

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

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Calculus 3 : Stokes' Theorem

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Example Question #51 : Stokes' Theorem

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