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Calculus 3 : Calculus 3

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Example Questions

Example Question #3891 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #3892 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #3893 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #3894 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #51 : Surface Integrals

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

Possible Answers:

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

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

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

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

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 #3895 : Calculus 3

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

Possible Answers:

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

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

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

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

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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Example Question #3896 : 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:

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

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

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

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

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 #3897 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #61 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

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 #3899 : Calculus 3

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

Possible Answers:

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

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

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

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

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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Please follow these steps to file a notice:

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A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

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I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

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This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

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Calculus 3 : Calculus 3

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