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Calculus 3 : Calculus 3

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Example Questions

Example Question #25 : Surface Integrals

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x - y + 2z)}dx+{(y^2 - x^4)}dz$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

$\int \int_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to utilize Stokes' theorem, note its form

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

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

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x - y + 2z)}dx+{(y^2 - x^4)}dz\\&\text{Meaning that}\\&F_x=x - y + 2z;F_y=0;F_z=y^2 - x^4\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=2y-[0]=2y \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=2-[-4x^3]=4x^3 + 2 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[-1]=1\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(2y)\widehat{i}+(4x^3 + 2)\widehat{j}+(1)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #26 : Surface Integrals

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(y^{(x + z)})}dx$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

$\int \int_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to utilize Stokes' theorem, note its form

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

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

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

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

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Example Question #3861 : Calculus 3

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3)}dx+{(y)}dy+{(y + x^3)}dz$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

$\int \int_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to utilize Stokes' theorem, note its form

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

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

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

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=1-[0]=1 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[3x^2]=-3x^2 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(-3x^2)\widehat{j}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #3862 : Calculus 3

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(y^3 - 6z)}dx+{(x)}dz$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

$\int \int_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-7)\widehat{j}+(-3y^2)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(7)\widehat{i}-(3y^2)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(7)\widehat{j}+(3y^2)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(7)\widehat{i}+(3y^2)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(7)\widehat{i}+(3y^2)\widehat{j}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-7)\widehat{j}+(-3y^2)\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

In order to utilize Stokes' theorem, note its form

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

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

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(y^3 - 6z)}dx+{(x)}dz\\&\text{Meaning that}\\&F_x=y^3 - 6z;F_y=0;F_z=x\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=-6-[1]=-7 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[3y^2]=-3y^2\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-7)\widehat{j}+(-3y^2)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #29 : Surface Integrals

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(9 - z^3)}dx+{(cos{(3z)})}dy$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

$\int \int_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-9zsin{(3z)})\widehat{i}+(6z^2)\widehat{j}]\cdot\overrightarrow{A}}$

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

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-9zsin{(3z)})\widehat{i}+(6z^2)\widehat{k}]\cdot\overrightarrow{A}}$

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

Correct answer:

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

Explanation:

In order to utilize Stokes' theorem, note its form

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

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

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(9 - z^3)}dx+{(cos{(3z)})}dy\\&\text{Meaning that}\\&F_x=9 - z^3;F_y=cos{(3z)};F_z=0\\&\text{From this we can derive our curl vectors:}\end{align*}$

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

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Example Question #30 : Surface Integrals

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(5y + xz)})}dx$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

$\int \int_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5xcos{(5y + xz)})\widehat{j}+(-5cos{(5y + xz)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5xzcos{(5y + xz)})\widehat{j}+(-cos{(5y + xz)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5xcos{(5y + xz)})\widehat{j}+(-cos{(5y + xz)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xcos{(5y + xz)})\widehat{j}+(-5cos{(5y + xz)})\widehat{k}]\cdot\overrightarrow{A}}$

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

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xcos{(5y + xz)})\widehat{j}+(-5cos{(5y + xz)})\widehat{k}]\cdot\overrightarrow{A}}$

Explanation:

In order to utilize Stokes' theorem, note its form

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

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

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(5y + xz)})}dx\\&\text{Meaning that}\\&F_x=sin{(5y + xz)};F_y=0;F_z=0\\&\text{From this we can derive our curl vectors:}\end{align*}$

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

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Example Question #31 : Surface Integrals

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(ln{(yz^5 + 5)})}dx$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

$\int \int_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}$

Possible Answers:

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[({(1yz)}/{(yz+ 1)})\widehat{j}+(z^5/{(yz^5)})\widehat{k}]\cdot\overrightarrow{A}}$

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

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

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

Correct answer:

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

Explanation:

In order to utilize Stokes' theorem, note its form

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

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

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(ln{(yz^5 + 5)})}dx\\&\text{Meaning that}\\&F_x=ln{(yz^5 + 5)};F_y=0;F_z=0\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}={(5yz^4)}/{(yz^5 + 5)}-[0]={(5yz^4)}/{(yz^5 + 5)} \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[z^5/{(yz^5 + 5)}]=-z^5/{(yz^5 + 5)}\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[({(5yz^4)}/{(yz^5 + 5)})\widehat{j}+(-z^5/{(yz^5 + 5)})\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #32 : Surface Integrals

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x^z)}dy$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

$\int \int_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to utilize Stokes' theorem, note its form

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

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

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

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

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Example Question #31 : Stokes' Theorem

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(3^z)}dx+{(3^{(x + y)})}dz$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

$\int \int_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to utilize Stokes' theorem, note its form

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

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

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

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

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Example Question #34 : Surface Integrals

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(y^z)}dx+{(z^x)}dy+{(x^z)}dz$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

$\int \int_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to utilize Stokes' theorem, note its form

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

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

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

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

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