Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Surface Integrals

Study concepts, example questions & explanations for Calculus 3

Create An Account

All Calculus 3 Resources

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

Example Questions

Example Question #41 : Stokes' Theorem

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(x^3)})}dx+{(6z + 3x^2)}dy+{(y^4)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$ $\displaystyle \begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(x^3)})}dx+{(6z + 3x^2)}dy+{(y^4)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4y^3 - 6)\widehat{i}+(6x)\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4y^3 - 6)\widehat{i}+(6x)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(6 - 6x)\widehat{j}+(-4y^3)\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(6 - 6x)\widehat{j}+(-4y^3)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(6)\widehat{j}+(-6)\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(6)\widehat{j}+(-6)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-6x)\widehat{i}+(-4y^3)\widehat{j}+(6)\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-6x)\widehat{i}+(-4y^3)\widehat{j}+(6)\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(x^3)})}dx+{(6z + 3x^2)}dy+{(y^4)}dz\\&\text{Meaning that}\\&F_x=sin{(x^3)};F_y=6z + 3x^2;F_z=y^4\\&\text{From this we can derive our curl vectors:}\end{align*}$ $\displaystyle \begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(x^3)})}dx+{(6z + 3x^2)}dy+{(y^4)}dz\\&\text{Meaning that}\\&F_x=sin{(x^3)};F_y=6z + 3x^2;F_z=y^4\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=4y^3-[6] \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[0]=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=6x-[0]=6x\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4y^3 - 6)\widehat{i}+(6x)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$ $\displaystyle \begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=4y^3-[6] \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[0]=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=6x-[0]=6x\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4y^3 - 6)\widehat{i}+(6x)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

Report an Error

Example Question #41 : Surface Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

Explanation:

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

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=cos{(y + z)}-[x^{(z - 1)}z] \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=y^2-[cos{(y + z)}]\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(cos{(y + z)} - x^{(z - 1)}z)\widehat{j}+(y^2 - cos{(y + z)})\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$ $\displaystyle \begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=cos{(y + z)}-[x^{(z - 1)}z] \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=y^2-[cos{(y + z)}]\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(cos{(y + z)} - x^{(z - 1)}z)\widehat{j}+(y^2 - cos{(y + z)})\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

Report an Error

Example Question #43 : Stokes' Theorem

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3)}dx+{(y^4)}dy+{(x^2 + y^3)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$ $\displaystyle \begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3)}dx+{(y^4)}dy+{(x^2 + y^3)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

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

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

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

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

Correct answer:

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3)}dx+{(y^4)}dy+{(x^2 + y^3)}dz\\&\text{Meaning that}\\&F_x=x^3;F_y=y^4;F_z=x^2 + y^3\\&\text{From this we can derive our curl vectors:}\end{align*}$ $\displaystyle \begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3)}dx+{(y^4)}dy+{(x^2 + y^3)}dz\\&\text{Meaning that}\\&F_x=x^3;F_y=y^4;F_z=x^2 + y^3\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=3y^2-[0]=3y^2 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[2x]=-2x \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(3y^2)\widehat{i}+(-2x)\widehat{j}]\cdot\overrightarrow{A}}\end{align*}$ $\displaystyle \begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=3y^2-[0]=3y^2 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[2x]=-2x \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(3y^2)\widehat{i}+(-2x)\widehat{j}]\cdot\overrightarrow{A}}\end{align*}$

Report an Error

Example Question #3872 : Calculus 3

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(log{(2x + x^5)})}dx+{(cos{(y^4)})}dy+{(x^2y - 2x)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$ $\displaystyle \begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(log{(2x + x^5)})}dx+{(cos{(y^4)})}dy+{(x^2y - 2x)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

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

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-4y^3sin{(y^4)})\widehat{i}+(-{(5x^4 + 2)}/{(2x + x^5)})\widehat{j}+({(5x^4 + 2)}/{(2x + x^5)} + 4y^3sin{(y^4)})\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-4y^3sin{(y^4)})\widehat{i}+(-{(5x^4 + 2)}/{(2x + x^5)})\widehat{j}+({(5x^4 + 2)}/{(2x + x^5)} + 4y^3sin{(y^4)})\widehat{k}]\cdot\overrightarrow{A}}$

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

Correct answer:

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(log{(2x + x^5)})}dx+{(cos{(y^4)})}dy+{(x^2y - 2x)}dz\\&\text{Meaning that}\\&F_x=log{(2x + x^5)};F_y=cos{(y^4)};F_z=x^2y - 2x\\&\text{From this we can derive our curl vectors:}\end{align*}$ $\displaystyle \begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(log{(2x + x^5)})}dx+{(cos{(y^4)})}dy+{(x^2y - 2x)}dz\\&\text{Meaning that}\\&F_x=log{(2x + x^5)};F_y=cos{(y^4)};F_z=x^2y - 2x\\&\text{From this we can derive our curl vectors:}\end{align*}$

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

Report an Error

Example Question #45 : Stokes' Theorem

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

Possible Answers:

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(ye^{(xy)})\widehat{i}+(1/y - xe^{(xy)})\widehat{j}+(-1/z)\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(ye^{(xy)})\widehat{i}+(1/y - xe^{(xy)})\widehat{j}+(-1/z)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1/y - 1/z)\widehat{i}+(ye^{(xy)} - xe^{(xy)})\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1/y - 1/z)\widehat{i}+(ye^{(xy)} - xe^{(xy)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xe^{(xy)})\widehat{i}+(1/z - ye^{(xy)})\widehat{j}+(-1/y)\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xe^{(xy)})\widehat{i}+(1/z - ye^{(xy)})\widehat{j}+(-1/y)\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(log{(yz)})}dx+{(e^{(xy)})}dz\\&\text{Meaning that}\\&F_x=log{(yz)};F_y=0;F_z=e^{(xy)}\\&\text{From this we can derive our curl vectors:}\end{align*}$ $\displaystyle \begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(log{(yz)})}dx+{(e^{(xy)})}dz\\&\text{Meaning that}\\&F_x=log{(yz)};F_y=0;F_z=e^{(xy)}\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=xe^{(xy)}-[0]=xe^{(xy)} \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=1/z-[ye^{(xy)}] \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[1/y]=-1/y\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xe^{(xy)})\widehat{i}+(1/z - ye^{(xy)})\widehat{j}+(-1/y)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$ $\displaystyle \begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=xe^{(xy)}-[0]=xe^{(xy)} \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=1/z-[ye^{(xy)}] \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[1/y]=-1/y\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xe^{(xy)})\widehat{i}+(1/z - ye^{(xy)})\widehat{j}+(-1/y)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

Report an Error

Example Question #46 : Stokes' Theorem

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(xyz)}dx+{(3y^3)}dy+{(x + y + z)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$ $\displaystyle \begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(xyz)}dx+{(3y^3)}dy+{(x + y + z)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(xy - 1)\widehat{j}+(-xz)\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(xy - 1)\widehat{j}+(-xz)\widehat{k}]\cdot\overrightarrow{A}}$

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(xz - 1)\widehat{j}+(-xy)\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(xz - 1)\widehat{j}+(-xy)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(9y^2 - 1)\widehat{i}+(1 - yz)\widehat{j}+(yz - 9y^2)\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(9y^2 - 1)\widehat{i}+(1 - yz)\widehat{j}+(yz - 9y^2)\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(xyz)}dx+{(3y^3)}dy+{(x + y + z)}dz\\&\text{Meaning that}\\&F_x=xyz;F_y=3y^3;F_z=x + y + z\\&\text{From this we can derive our curl vectors:}\end{align*}$ $\displaystyle \begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(xyz)}dx+{(3y^3)}dy+{(x + y + z)}dz\\&\text{Meaning that}\\&F_x=xyz;F_y=3y^3;F_z=x + y + z\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=1-[0]=1 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=xy-[1] \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[xz]=-xz\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(xy - 1)\widehat{j}+(-xz)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$ $\displaystyle \begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=1-[0]=1 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=xy-[1] \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[xz]=-xz\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(xy - 1)\widehat{j}+(-xz)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

Report an Error

Example Question #47 : Stokes' Theorem

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x + y^3z)}dx+{(2y - x^2)}dy\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$ $\displaystyle \begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x + y^3z)}dx+{(2y - x^2)}dy\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

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

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

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

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

Correct answer:

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x + y^3z)}dx+{(2y - x^2)}dy\\&\text{Meaning that}\\&F_x=x + y^3z;F_y=2y - x^2;F_z=0\\&\text{From this we can derive our curl vectors:}\end{align*}$ $\displaystyle \begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x + y^3z)}dx+{(2y - x^2)}dy\\&\text{Meaning that}\\&F_x=x + y^3z;F_y=2y - x^2;F_z=0\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=y^3-[0]=y^3 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=-2x-[3y^2z]=- 2x - 3y^2z\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(y^3)\widehat{j}+(- 2x - 3y^2z)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$ $\displaystyle \begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=y^3-[0]=y^3 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=-2x-[3y^2z]=- 2x - 3y^2z\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(y^3)\widehat{j}+(- 2x - 3y^2z)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

Report an Error

Example Question #41 : Stokes' Theorem

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

Possible Answers:

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

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

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

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

Correct answer:

Explanation:

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

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=-1-[1]=-2 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=1-[1]=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=1-[-1]=2\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-2)\widehat{i}+(2)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$ $\displaystyle \begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=-1-[1]=-2 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=1-[1]=0 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=1-[-1]=2\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-2)\widehat{i}+(2)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

Report an Error

Example Question #49 : Stokes' Theorem

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

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(x^{(z - 1)}z)\widehat{i}+(xy^{(x - 1)})\widehat{j}+(yz^{(y - 1)})\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(x^{(z - 1)}z)\widehat{i}+(xy^{(x - 1)})\widehat{j}+(yz^{(y - 1)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(z^ylog{(z)} - x^zlog{(x)})\widehat{i}+(x^zlog{(x)} - y^xlog{(y)})\widehat{j}+(y^xlog{(y)} - z^ylog{(z)})\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(z^ylog{(z)} - x^zlog{(x)})\widehat{i}+(x^zlog{(x)} - y^xlog{(y)})\widehat{j}+(y^xlog{(y)} - z^ylog{(z)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xy^{(x - 1)})\widehat{i}+(yz^{(y - 1)})\widehat{j}+(x^{(z - 1)}z)\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(xy^{(x - 1)})\widehat{i}+(yz^{(y - 1)})\widehat{j}+(x^{(z - 1)}z)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-yz^{(y - 1)})\widehat{i}+(-x^{(z - 1)}z)\widehat{j}+(-xy^{(x - 1)})\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-yz^{(y - 1)})\widehat{i}+(-x^{(z - 1)}z)\widehat{j}+(-xy^{(x - 1)})\widehat{k}]\cdot\overrightarrow{A}}$

Correct answer:

Explanation:

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

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[yz^{(y - 1)}]=-yz^{(y - 1)} \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[x^{(z - 1)}z]=-x^{(z - 1)}z \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[xy^{(x - 1)}]=-xy^{(x - 1)}\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-yz^{(y - 1)})\widehat{i}+(-x^{(z - 1)}z)\widehat{j}+(-xy^{(x - 1)})\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$ $\displaystyle \begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[yz^{(y - 1)}]=-yz^{(y - 1)} \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[x^{(z - 1)}z]=-x^{(z - 1)}z \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[xy^{(x - 1)}]=-xy^{(x - 1)}\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-yz^{(y - 1)})\widehat{i}+(-x^{(z - 1)}z)\widehat{j}+(-xy^{(x - 1)})\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

Report an Error

Example Question #42 : Stokes' Theorem

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3sin{(x)})}dx+{(y^2 - cos{(y)}sin{(y)})}dy+{(x + y + z)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$ $\displaystyle \begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3sin{(x)})}dx+{(y^2 - cos{(y)}sin{(y)})}dy+{(x + y + z)}dz\\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

Possible Answers:

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(1)\widehat{k}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(1)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(-1)\widehat{j}]\cdot\overrightarrow{A}}$ $\displaystyle {\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(-1)\widehat{j}]\cdot\overrightarrow{A}}$

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

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

Correct answer:

Explanation:

$\begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3sin{(x)})}dx+{(y^2 - cos{(y)}sin{(y)})}dy+{(x + y + z)}dz\\&\text{Meaning that}\\&F_x=x^3sin{(x)};F_y=y^2 - cos{(y)}sin{(y)};F_z=x + y + z\\&\text{From this we can derive our curl vectors:}\end{align*}$ $\displaystyle \begin{align*}&\text{From what we are told}\\&\overrightarrow{F}\cdot d\overrightarrow{s}={(x^3sin{(x)})}dx+{(y^2 - cos{(y)}sin{(y)})}dy+{(x + y + z)}dz\\&\text{Meaning that}\\&F_x=x^3sin{(x)};F_y=y^2 - cos{(y)}sin{(y)};F_z=x + y + z\\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=1-[0]=1 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[1]=-1 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(-1)\widehat{j}]\cdot\overrightarrow{A}}\end{align*}$ $\displaystyle \begin{align*}\\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=1-[0]=1 \\ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[1]=-1 \\ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\\&\text{And in turn our surface integral}:\\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(1)\widehat{i}+(-1)\widehat{j}]\cdot\overrightarrow{A}}\end{align*}$

Report an Error

View Calculus 3 Tutors

Reuben
Certified Tutor

University of Alberta, Bachelor of Science, Mathematics.

View Calculus 3 Tutors

Scott
Certified Tutor

Florida State University, Bachelor of Science, Applied Mathematics. Florida State University, Bachelor of Science, Astrophysi...

View Calculus 3 Tutors

Katherine
Certified Tutor

Northeastern University, Bachelor of Engineering, Electrical Engineering.

All Calculus 3 Resources

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

Popular Subjects

SAT Tutors in Houston, Algebra Tutors in Atlanta, French Tutors in Washington DC, GRE Tutors in Washington DC, French Tutors in Houston, Physics Tutors in Houston, Chemistry Tutors in Philadelphia, Statistics Tutors in New York City, Statistics Tutors in San Diego, ISEE Tutors in Boston

Popular Courses & Classes

LSAT Courses & Classes in Washington DC, SAT Courses & Classes in Boston, SAT Courses & Classes in Seattle, LSAT Courses & Classes in New York City, ISEE Courses & Classes in Los Angeles, GMAT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Seattle, ISEE Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Houston, GRE Courses & Classes in New York City

Popular Test Prep

MCAT Test Prep in Miami, SSAT Test Prep in New York City, ACT Test Prep in Atlanta, ISEE Test Prep in Phoenix, LSAT Test Prep in Denver, GMAT Test Prep in San Francisco-Bay Area, ISEE Test Prep in Houston, GMAT Test Prep in San Diego, GMAT Test Prep in Dallas Fort Worth, MCAT Test Prep in San Francisco-Bay Area

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 3 : Surface Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #41 : Stokes' Theorem

Example Question #41 : Surface Integrals

Example Question #43 : Stokes' Theorem

Example Question #3872 : Calculus 3

Example Question #45 : Stokes' Theorem

Example Question #46 : Stokes' Theorem

Example Question #47 : Stokes' Theorem

Example Question #41 : Stokes' Theorem

Example Question #49 : Stokes' Theorem

Example Question #42 : Stokes' Theorem

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: