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Calculus 3 : Surface Integrals

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Example Questions

Example Question #71 : Stokes' Theorem

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

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =0}$

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #72 : Stokes' Theorem

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(4y + 4xz)}dx+{(xy)}dy+{(2z + 2x^2z)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #71 : Surface Integrals

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:

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

Possible Answers:

4xydy

2x^2y^2dz

2x^2y^2dx

4xydz

4xydx

Correct answer:

4xydz

Explanation:

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

F_x=0;F_y=0;F_z=4xy

and

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

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Example Question #71 : Surface Integrals

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:

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

Possible Answers:

-y^3dy

-y^3dx

-y^3dz

$\frac{1}{3}y^2dy$

$\frac{1}{3}y^2dz$

Correct answer:

-y^3dx

Explanation:

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

From what we're told

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

And it can be inferred from this that

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

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

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

F_x=-y^3;F_y=0;F_z=0

and

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

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Example Question #72 : Surface Integrals

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:

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

Possible Answers:

2xe^zdx

x^2e^zdz

2xe^zdz

x^2e^zdy

2xe^zdy

Correct answer:

x^2e^zdy

Explanation:

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

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

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

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

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

F_x=0;F_y=x^2e^z;F_z=0

and

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

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Example Question #3911 : Calculus 3

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:

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

Possible Answers:

ycos(z)dx

cos(z)dx

ysin(z)dx

-ycos(z)dx

sin(z)dx

Correct answer:

ysin(z)dx

Explanation:

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

From what we're told

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

And it can be inferred from this that

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

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

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

F_x=ysin(z);F_y=0;F_z=0

and

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

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Example Question #3912 : Calculus 3

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:

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

Possible Answers:

cos(8xz)dz

cos(8xz)dy

sin(8xz)dy

cos(8xz)dx

sin(8xz)dz

Correct answer:

sin(8xz)dy

Explanation:

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

From what we're told

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

And it can be inferred from this that

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

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

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

F_x=0;F_y=sin(8xz);F_z=0

and

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

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Example Question #71 : Surface Integrals

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:

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

Possible Answers:

ln(y^2)dz

ln(y^2)dx

ln(y^2)dy

2ln(y^2)dy

2ln(y^2)dz

Correct answer:

ln(y^2)dz

Explanation:

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

From what we're told

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

And it can be inferred from this that

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

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

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

F_x=0;F_y=0;F_z=ln(y^2)

(Note that ln(y^2)=2ln(y) ; both results are valid)

and

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

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Example Question #72 : Surface Integrals

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:

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

Possible Answers:

6xy^2z^3dx

6y^2z^3dz

6y^2z^3dy

y^2z^3dx

6y^2z^3dx

Correct answer:

y^2z^3dx

Explanation:

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

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

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

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

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

F_x=y^2z^3;F_y=0;F_z=0

and

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

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Example Question #73 : Surface Integrals

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:

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

Possible Answers:

e^xsin(ze^x)dz

zsin(ze^x)dy

ze^xsin(ze^x)dy

sin(ze^x)dy

ze^xsin(ze^x)dx

Correct answer:

sin(ze^x)dy

Explanation:

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

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

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

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

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

F_x=0;F_y=sin(ze^x);F_z=0

and

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

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Calculus 3 : Surface Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #71 : Stokes' Theorem

Example Question #72 : Stokes' Theorem

Example Question #71 : Surface Integrals

Example Question #71 : Surface Integrals

Example Question #72 : Surface Integrals

Example Question #3911 : Calculus 3

Example Question #3912 : Calculus 3

Example Question #71 : Surface Integrals

Example Question #72 : Surface Integrals

Example Question #73 : Surface Integrals

All Calculus 3 Resources

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