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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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 #52 : Stokes' Theorem

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

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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Example Question #61 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

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 #61 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

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 #62 : 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:

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

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

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

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

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 #63 : Surface Integrals

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

Possible Answers:

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

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

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

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

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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Calculus 3 : Calculus 3

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Example Question #55 : Stokes' Theorem

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