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Calculus 3 : Calculus 3

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Example Questions

Example Question #41 : Surface Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #46 : Surface Integrals

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

Possible Answers:

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

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(9y^2 - 1)\widehat{i}+(1 - yz)\widehat{j}+(yz - 9y^2)\widehat{k}]\cdot\overrightarrow{A}}$

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

Correct answer:

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

Explanation:

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

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

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Example Question #41 : Surface Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #48 : Surface Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #41 : Surface Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #42 : Surface Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #3881 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #3882 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #3883 : Calculus 3

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

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 #3884 : Calculus 3

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

Possible Answers:

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

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

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

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

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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Calculus 3 : Calculus 3

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