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Calculus 3 : Calculus 3

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Example Questions

Example Question #3901 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #3902 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #3903 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #62 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

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 #63 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

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 #3906 : Calculus 3

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

Possible Answers:

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

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

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

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

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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Example Question #71 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #72 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #71 : Stokes' Theorem

Let S be a known surface with a boundary curve, C.

Considering the integral $\int \int_{S} (4x\widehat{i}-4y\widehat{j})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Possible Answers:

2x^2y^2dx

4xydz

4xydx

2x^2y^2dz

4xydy

Correct answer:

4xydz

Explanation:

In order to utilize Stokes' theorem, note its form

$\int \int_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} = \oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}$

The curl of a vector function F over an oriented surface S is equivalent to the function F itself integrated over the boundary curve, C, of S.

Note that

$curl(\overrightarrow{F})=\begin{bmatrix} \widehat{i}& \widehat{j}& \widehat{k}\\ \frac{\delta}{\delta x}&\frac{\delta}{\delta y} &\frac{\delta}{\delta z} \\ F_x&F_y &F_z \end{bmatrix}=(\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z})\widehat{i}+(\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x})\widehat{j}+(\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y})\widehat{k}$

From what we're told

$curl(\overrightarrow{F})=4x\widehat{i}-4y\widehat{j}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=4x$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=-4y$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

F_x=0;F_y=0;F_z=4xy

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=4xydz$

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Example Question #74 : Stokes' Theorem

Let S be a known surface with a boundary curve, C.

Considering the integral $\int \int_{S} (3y^2\widehat{k})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Possible Answers:

-y^3dx

-y^3dy

$\frac{1}{3}y^2dz$

-y^3dz

$\frac{1}{3}y^2dy$

Correct answer:

-y^3dx

Explanation:

In order to utilize Stokes' theorem, note its form

$\int \int_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} = \oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}$

The curl of a vector function F over an oriented surface S is equivalent to the function F itself integrated over the boundary curve, C, of S.

Note that

From what we're told

$curl(\overrightarrow{F})=3y^2\widehat{k}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=3y^2$

F_x=-y^3;F_y=0;F_z=0

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=-y^3dx$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3901 : Calculus 3

Example Question #3902 : Calculus 3

Example Question #3903 : Calculus 3

Example Question #62 : Stokes' Theorem

Example Question #63 : Stokes' Theorem

Example Question #3906 : Calculus 3

Example Question #71 : Stokes' Theorem

Example Question #72 : Stokes' Theorem

Example Question #71 : Stokes' Theorem

Example Question #74 : Stokes' Theorem

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