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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #75 : Stokes' Theorem

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^zdy

2xe^zdy

2xe^zdz

x^2e^zdz

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

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

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

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 #72 : Stokes' Theorem

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:

ysin(z)dx

-ycos(z)dx

sin(z)dx

cos(z)dx

ycos(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

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

Report an Error

Example Question #77 : Stokes' Theorem

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:

sin(8xz)dy

cos(8xz)dy

sin(8xz)dz

cos(8xz)dx

cos(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 : Stokes' Theorem

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:

2ln(y^2)dy

ln(y^2)dz

ln(y^2)dy

2ln(y^2)dz

ln(y^2)dx

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 #71 : 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

y^2z^3dx

6y^2z^3dy

6y^2z^3dz

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

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

Report an Error

Example Question #80 : Stokes' Theorem

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:

sin(ze^x)dy

ze^xsin(ze^x)dx

ze^xsin(ze^x)dy

e^xsin(ze^x)dz

zsin(ze^x)dy

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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Example Question #81 : Stokes' Theorem

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:

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

Possible Answers:

$6xe^{y^3}dy$

$-e^{y^3}dz$

$3y^2e^{y^3}dy$

$e^{y^3}dx$

$3y^2e^{y^3}dx$

Correct answer:

$e^{y^3}dx$

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

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

And it can be inferred from this that

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

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

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

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

and

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

Report an Error

Example Question #81 : Stokes' Theorem

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:

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

Possible Answers:

z^4dy

z^4dx

z^4dz

$\frac{1}{4}z^4dy$

$-\frac{1}{4}z^4dx$

Correct answer:

$\frac{1}{4}z^4dy$

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

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

And it can be inferred from this that

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

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

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

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

and

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

Report an Error

Example Question #83 : Stokes' Theorem

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:

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

Possible Answers:

cos(4x)dy

sin(4x)dy

-cos(4x)dz

cos(4x)dz

-cos(4x)dy

Correct answer:

cos(4x)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

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

And it can be inferred from this that

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

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

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

F_x=0;F_y=0;F_z=cos(4x)

and

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

Report an Error

Example Question #84 : Stokes' Theorem

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:

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

Possible Answers:

$\frac{z^5}{10}dy$

$-\frac{z^5}{10}dy$

$\frac{z^5}{2}dx$

$\frac{z^5}{10}dx$

$-\frac{z^5}{2}dy$

Correct answer:

$\frac{z^5}{10}dx$

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

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

And it can be inferred from this that

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

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

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

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

and

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

Report an Error

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #75 : Stokes' Theorem

Example Question #72 : Stokes' Theorem

Example Question #77 : Stokes' Theorem

Example Question #71 : Stokes' Theorem

Example Question #71 : Surface Integrals

Example Question #80 : Stokes' Theorem

Example Question #81 : Stokes' Theorem

Example Question #81 : Stokes' Theorem

Example Question #83 : Stokes' Theorem

Example Question #84 : Stokes' Theorem

All Calculus 3 Resources

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