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Calculus 3 : Surface Integrals

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

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

Possible Answers:

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

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

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

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

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

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

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

From what we're told

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

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

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Example Question #21 : Stokes' Theorem

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

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

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

Possible Answers:

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

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

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

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

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

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 #22 : Stokes' Theorem

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

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

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

Possible Answers:

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

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

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

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

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

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 #23 : Stokes' Theorem

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

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

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

Possible Answers:

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

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

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

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

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

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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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

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 #21 : 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:

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

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

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

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

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

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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Calculus 3 : Surface Integrals

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