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Calculus 3 : Stokes' Theorem

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Example Question #1 :年代urface Integrals

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(y^3z^4)}dx$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

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

Possible Answers:

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

$\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(12y^3z^3)\widehat{j}+(9y^2z^4)\widehat{k}]\cdot\overrightarrow{A}$

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

$\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(12y^3z^3)\widehat{j}+(-9y^2z^4)\widehat{k}]\cdot\overrightarrow{A}$

$\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(12y^3z^3)\widehat{i}+(-9y^2z^4)\widehat{j}]\cdot\overrightarrow{A}$

Correct answer:

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

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 functionFover an oriented surface年代is equivalent to the functionFitself integrated over the boundary curve,C, of年代.

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

$\overrightarrow{F}\cdot d\overrightarrow{s}={(y^3z^4)}dx$

Meaning that

F_x=y^3z^4;F_y=0;F_z=0

From this we can derive our curl vectors

$\begin{align*}&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-0=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=4y^3z^3-0=4y^3z^3 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-3y^2z^4=-3y^2z^4\end{align*}$

This allows us to set up our surface integral

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

Report an Error

Example Question #2 :年代urface Integrals

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 functionFover an oriented surface年代is equivalent to the functionFitself integrated over the boundary curve,C, of年代.

Note that

From what we're told

$\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(xy)})}dx+{(x^2y)}dz$

Meaning that

$F_x=sin{(xy)};F_y=0;F_z=x^2y$

From this we can derive our curl vectors

$\begin{align*}&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=x^2-0=x^2 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-2xy=-2xy \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-xcos{(xy)}=-xcos{(xy)}\end{align*}$

This allows us to set up our surface integral

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

Report an Error

Example Question #3 :年代urface Integrals

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 functionFover an oriented surface年代is equivalent to the functionFitself integrated over the boundary curve,C, of年代.

Note that

From what we're told

$\overrightarrow{F}\cdot d\overrightarrow{s}={(e^{(xy)})}dx+{(sin{(xy)})}dy+{(yz)}dz$

Meaning that

$F_x=e^{(xy)};F_y=sin{(xy)};F_z=yz$

From this we can derive our curl vectors

$\begin{align*}&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=z-0=z \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-0=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=ycos{(xy)}-xe^{(xy)}\end{align*}$

This allows us to set up our surface integral

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

Report an Error

Example Question #4 :年代urface Integrals

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 functionFover an oriented surface年代is equivalent to the functionFitself integrated over the boundary curve,C, of年代.

Note that

From what we're told

$\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(y)})}dx+{(sin{(4x)})}dy$

Meaning that

$F_x=sin{(y)};F_y=sin{(4x)};F_z=0$

From this we can derive our curl vectors

$\begin{align*}&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-0=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-0=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=4cos{(4x)}-cos{(y)}\end{align*}$

This allows us to set up our surface integral

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

Report an Error

Example Question #5 :年代urface Integrals

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(3y)}dx+{(4x + 5y)}dz$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

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

Possible Answers:

$\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5)\widehat{i}+(-4)\widehat{j}+(-3)\widehat{k}]\cdot\overrightarrow{A}$

$\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-5)\widehat{i}+(-4)\widehat{j}+(-3)\widehat{k}]\cdot\overrightarrow{A}$

$\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-5)\widehat{i}+(4)\widehat{j}+(-3)\widehat{k}]\cdot\overrightarrow{A}$

$\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5)\widehat{i}+(4)\widehat{j}+(-3)\widehat{k}]\cdot\overrightarrow{A}$

$\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5)\widehat{i}+(4)\widehat{j}+(3)\widehat{k}]\cdot\overrightarrow{A}$

Correct answer:

$\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5)\widehat{i}+(-4)\widehat{j}+(-3)\widehat{k}]\cdot\overrightarrow{A}$

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 functionFover an oriented surface年代is equivalent to the functionFitself integrated over the boundary curve,C, of年代.

Note that

From what we're told

$\overrightarrow{F}\cdot d\overrightarrow{s}={(3y)}dx+{(4x + 5y)}dz$

Meaning that

F_x=3y;F_y=0;F_z=4x + 5y

From this we can derive our curl vectors

$\begin{align*}&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=5-0=5 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-4=-4 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-3=-3\end{align*}$

This allows us to set up our surface integral

$\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5)\widehat{i}+(-4)\widehat{j}+(-3)\widehat{k}]\cdot\overrightarrow{A}$

Report an Error

Example Question #6 :年代urface Integrals

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 functionFover an oriented surface年代is equivalent to the functionFitself integrated over the boundary curve,C, of年代.

Note that

From what we're told

$\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(e^{(y + x^2)})})}dx+{(x + z^3)}dy+{(log{(xy)})}dz$

Meaning that

$F_x=sin{(e^{(y + x^2)})};F_y=x + z^3;F_z=log{(xy)}$

From this we can derive our curl vectors

$\begin{align*}&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=1/y-3z^2 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-1/x=-1/x \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=1-e^{(y + x^2)}cos{(e^{(y + x^2)})}\end{align*}$

This allows us to set up our surface integral

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

Report an Error

Example Question #7 :年代urface Integrals

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

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

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

Possible Answers:

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

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

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

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

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 functionFover an oriented surface年代is equivalent to the functionFitself integrated over the boundary curve,C, of年代.

Note that

From what we're told

$\overrightarrow{F}\cdot d\overrightarrow{s}={(cos{(e^{(x + 2y)})})}dx+{(sin{(e^{(x^2y)})})}dz$

Meaning that

$F_x=cos{(e^{(x + 2y)})};F_y=0;F_z=sin{(e^{(x^2y)})}$

From this we can derive our curl vectors

$\begin{align*}&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=x^2e^{(x^2y)}cos{(e^{(x^2y)})}-0=x^2e^{(x^2y)}cos{(e^{(x^2y)})} \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-2xye^{(x^2y)}cos{(e^{(x^2y)})}=-2xye^{(x^2y)}cos{(e^{(x^2y)})} \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0--2e^{(x + 2y)}sin{(e^{(x + 2y)})}=2e^{(x + 2y)}sin{(e^{(x + 2y)})}\end{align*}$

This allows us to set up our surface integral

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

Report an Error

Example Question #8 :年代urface Integrals

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 functionFover an oriented surface年代is equivalent to the functionFitself integrated over the boundary curve,C, of年代.

Note that

From what we're told

$\overrightarrow{F}\cdot d\overrightarrow{s}={(x + y)}dx+{(y + x^2)}dy+{(y + x^3)}dz$

Meaning that

F_x=x + y;F_y=y + x^2;F_z=y + x^3

From this we can derive our curl vectors

$\begin{align*}&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=1-0=1 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-3x^2=-3x^2 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=2x-1\end{align*}$

This allows us to set up our surface integral

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

Report an Error

Example Question #9 :年代urface Integrals

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

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

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

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-5e^{(5z + x^3)})\widehat{i}+(3x^2e^{(5z + x^3)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-5e^{(5z + x^3)}+3x^2e^{(5z + x^3)})\widehat{k}]\cdot\overrightarrow{A}}$

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

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-5e^{(5z + x^3)})\widehat{j}+(3x^2e^{(5z + x^3)})\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-5e^{(5z + x^3)})\widehat{i}+(3x^2e^{(5z + x^3)})\widehat{k}]\cdot\overrightarrow{A}}$

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 functionFover an oriented surface年代is equivalent to the functionFitself integrated over the boundary curve,C, of年代.

Note that

From what we're told

$\overrightarrow{F}\cdot d\overrightarrow{s}={(e^{(5z + x^3)})}dy$

Meaning that

$F_x=0;F_y=e^{(5z + x^3)};F_z=0$

From this we can derive our curl vectors

$\begin{align*}&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-5e^{(5z + x^3)}=-5e^{(5z + x^3)} \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-0=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=3x^2e^{(5z + x^3)}-0=3x^2e^{(5z + x^3)}\end{align*}$

This allows us to set up our surface integral

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-5e^{(5z + x^3)})\widehat{i}+(3x^2e^{(5z + x^3)})\widehat{k}]\cdot\overrightarrow{A}}$

Report an Error

Example Question #10 :年代urface Integrals

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

Considering the integral $\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(2y + z)}dx+{(x + 5)}dz$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 functionFover an oriented surface年代is equivalent to the functionFitself integrated over the boundary curve,C, of年代.

Note that

From what we're told

$\overrightarrow{F}\cdot d\overrightarrow{s}={(2y + z)}dx+{(x + 5)}dz$

Meaning that

F_x=2y + z;F_y=0;F_z=x + 5

From this we can derive our curl vectors

$\begin{align*}&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-0=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=1-1=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-2=-2\end{align*}$

This allows us to set up our surface integral

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

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