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Calculus 3 : Stokes' Theorem

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Example Questions

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

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

From what we're told

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

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

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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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Example Question #31 : Surface Integrals

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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Example Question #36 : Surface Integrals

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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Example Question #32 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #33 : Surface Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(yz)})}dx+{(x^3)}dy+{(z + 5)}dz\\&\text{Meaning that}\\&F_x=sin{(yz)};F_y=x^3;F_z=z + 5\\&\text{From this we can derive our curl vectors:}\end{align*}$

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

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Example Question #34 : Surface Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #35 : Surface Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(cos{(xy^3z^2)})}dx+{(5y)}dy\\&\text{Meaning that}\\&F_x=cos{(xy^3z^2)};F_y=5y;F_z=0\\&\text{From this we can derive our curl vectors:}\end{align*}$

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

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