Stokes' Theorem - Calculus 3

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Question

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

Considering the integral $\int \int_{S} (4x\widehat{i}-4y\widehat{j})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Answer

In order to utilize Stokes' theorem, note its form

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

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

$curl(\overrightarrow{F})=4x\widehat{i}-4y\widehat{j}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=4x$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=-4y$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=4xydz$

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Question

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

Considering the integral $\int \int_{S} (3y^2\widehat{k})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Answer

In order to utilize Stokes' theorem, note its form

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

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

$curl(\overrightarrow{F})=3y^2\widehat{k}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=3y^2$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=-y^3dx$

Compare your answer with the correct one above

Question

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

Considering the integral $\int \int_{S} (-x^2e^z\widehat{i}+2xe^z\widehat{k})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Answer

In order to utilize Stokes' theorem, note its form

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

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

$curl(\overrightarrow{F})=-x^2e^z\widehat{i}+2xe^z\widehat{k}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=-x^2e^z$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=2xe^z$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=x^2e^zdy$

Compare your answer with the correct one above

Question

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

Considering the integral $\int \int_{S} (ycos(z)\widehat{j}-sin(z)\widehat{k})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Answer

In order to utilize Stokes' theorem, note its form

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

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

$curl(\overrightarrow{F})=ycos(z)\widehat{j}-sin(z)\widehat{k}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=ycos(z)$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=-sin(z)$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=ysin(z)dx$

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Question

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

Considering the integral $\int \int_{S} (-8xcos(8xz)\widehat{i}+8zcos(8xz)\widehat{k})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Answer

In order to utilize Stokes' theorem, note its form

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

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

$curl(\overrightarrow{F})=-8xcos(8xz)\widehat{i}+8zcos(8xz)\widehat{k}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=-8xcos(8xz)$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=8zcos(8xz)$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=sin(8xz)dy$

Compare your answer with the correct one above

Question

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

Considering the integral $\int \int_{S} (\frac{2}{y}\widehat{i})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Answer

In order to utilize Stokes' theorem, note its form

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

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

$curl(\overrightarrow{F})=\frac{2}{y}\widehat{i}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=\frac{2}{y}$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

(Note that ; both results are valid)

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=ln(y^2)dz$

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Question

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

Considering the integral $\int \int_{S} (3y^2z^2\widehat{j}-2yz^3\widehat{k})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Answer

In order to utilize Stokes' theorem, note its form

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

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

$curl(\overrightarrow{F})=3y^2z^2\widehat{j}-2yz^3\widehat{k}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=3y^2z^2$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=-2yz^3$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=y^2z^3dx$

Compare your answer with the correct one above

Question

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

Considering the integral $\int \int_{S} (-e^xcos(ze^x)\widehat{i}+ze^xcos(ze^x)\widehat{k})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Answer

In order to utilize Stokes' theorem, note its form

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

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

$curl(\overrightarrow{F})=-e^xcos(ze^x)\widehat{i}+ze^xcos(ze^x)\widehat{k}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=-e^xcos(ze^x)$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=ze^xcos(ze^x)$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=sin(ze^x)dy$

Compare your answer with the correct one above

Question

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

Considering the integral $\int \int_{S} (-3y^2e^{y^3}\widehat{k})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Answer

In order to utilize Stokes' theorem, note its form

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

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

$curl(\overrightarrow{F})=-3y^2e^{y^3}\widehat{k}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=-3y^2e^{y^3}$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

$F_x=e^{y^3};F_y=0;F_z=0$

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=e^{y^3}dx$

Compare your answer with the correct one above

Question

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

Considering the integral $\int \int_{S} (-z^3\widehat{i})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Answer

In order to utilize Stokes' theorem, note its form

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

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

$curl(\overrightarrow{F})=-z^3\widehat{i}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=-z^3$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

$F_x=0;F_y=\frac{1}{4}z^4;F_z=0$

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=\frac{1}{4}z^4dy$

Compare your answer with the correct one above

Question

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

Considering the integral $\int \int_{S} (4sin(4x)\widehat{j})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Answer

In order to utilize Stokes' theorem, note its form

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

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

$curl(\overrightarrow{F})=4sin(4x)\widehat{j}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=4sin(4x)$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=cos(4x)dz$

Compare your answer with the correct one above

Question

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

Considering the integral $\int \int_{S} (\frac{z^4}{2}\widehat{j})\cdot d\overrightarrow{A}$ , utilize Stokes' Theorem to determine $\overrightarrow{F}\cdot d\overrightarrow{s}$ for an equivalent integral of the form:

$\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}$

Answer

In order to utilize Stokes' theorem, note its form

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

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

$curl(\overrightarrow{F})=\frac{z^4}{2}\widehat{j}$

And it can be inferred from this that

$\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0$

$\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=\frac{z^4}{2}$

$\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0$

A helpful approach can be to look at the right sides of the equations and see what variables are represented compared to what variables a vector component of F is being derived for. Doing this and integrating, we can infer that

$F_x=\frac{z^5}{10};F_y=0;F_z=0$

and

$\overrightarrow{F}\cdot d\overrightarrow{s}=\frac{z^5}{10}dx$

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Question

Let S 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:

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

Answer

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

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

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

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

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Question

Let S 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:

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

Answer

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

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

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

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

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Question

Let S 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:

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

Answer

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

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

Meaning that

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

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

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Question

Let S 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:

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

Answer

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

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

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

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

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Question

Let S 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:

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

Answer

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

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

Meaning that

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

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

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Question

Let S 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:

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

Answer

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

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

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

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

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Question

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

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

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

Answer

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

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

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

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

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Question

Let S 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:

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

Answer

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

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

Meaning that

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

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

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