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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #3851 : Calculus 3

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x + y + z)}dx+{(x + y + z)}dz\\&\text{Meaning that}\\&F_x=x + y + z;F_y=0;F_z=x + y + z\\&\text{From this we can derive our curl vectors:}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=1-0=1 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=1-1=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-1=-1\\&\text{This in turn allows us to set up our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(-1)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #3852 : Calculus 3

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

From what we're told

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

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

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Example Question #3853 : Calculus 3

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=tan{(x + y)}^2 + 1-[0]=tan{(x + y)}^2 + 1 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=cos{(z + 5)}-[tan{(x + y)}^2 + 1] \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(tan{(x + y)}^2 + 1)\widehat{i}+(cos{(z + 5)} - tan{(x + y)}^2 - 1)\widehat{j}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #18 : Stokes' Theorem

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

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

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

Possible Answers:

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

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

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

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

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

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

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

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

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

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Example Question #3854 : Calculus 3

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x^2 - 3y)}dx+{(2z)}dy$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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Example Question #20 : Stokes' Theorem

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(y - y^3 + 3)}dx+{(z + z^2)}dy$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

From what we're told

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

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

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Example Question #21 : Stokes' Theorem

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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Example Question #22 : Stokes' Theorem

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

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

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

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(9)\widehat{i}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(9)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(27)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(243)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(27)\widehat{i}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(9)\widehat{i}]\cdot\overrightarrow{A}}$

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

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

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=9-[0]=9 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[0]=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(9)\widehat{i}]\cdot\overrightarrow{A}}\end{align*}$

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Example Question #23 : Stokes' Theorem

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(log{(cos{(e^{(x^3)})})})}dx+{(sin{(tan{(cos{(y^3 - y)})})})}dy+{(3x - 5y)}dz$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

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

Possible Answers:

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5)\widehat{i}+(3)\widehat{j}]\cdot\overrightarrow{A}}$

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-5)\widehat{i}+(-3)\widehat{j}]\cdot\overrightarrow{A}}$

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

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-5)\widehat{i}+(-3)\widehat{j}]\cdot\overrightarrow{A}}$

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

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

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=-5-[0]=-5 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[3]=-3 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-5)\widehat{i}+(-3)\widehat{j}]\cdot\overrightarrow{A}}\end{align*}$

Not as bad as it looked, actually.

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Example Question #24 : Stokes' Theorem

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

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

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

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5cos{(x + 5)}e^{(5y)})\widehat{i}+(sin{(x + 5)}e^{(5y)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5cos{(x + 5)}e^{(5y)})\widehat{i}+(sin{(x + 5)}e^{(5y)})\widehat{j}]\cdot\overrightarrow{A}}$

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

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

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

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5cos{(x + 5)}e^{(5y)})\widehat{i}+(sin{(x + 5)}e^{(5y)})\widehat{j}]\cdot\overrightarrow{A}}$

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

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

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

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