Stokes' Theorem

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AP Calculus BC › Stokes' Theorem

Questions 1 - 10

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:

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

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

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

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

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

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:

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

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

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

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

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

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

Explanation

In order to utilize Stokes' theorem, note its form

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

The curl of a vector 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

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

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

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

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4 - 10x)\widehat{j}+(-x^yln{(x)})\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(10x - 4)\widehat{i}+(x^ylog{(x)})\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(x^yln{(x)})\widehat{i}+(10x - 4)\widehat{k}]\cdot\overrightarrow{A}}$

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

Explanation

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

$\begin{align*}\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-[0]=0 \ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-[10x - 4]=4 - 10x \ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[x^yln{(x)}]=-x^yln{(x)}\&\text{And in turn our surface integral}:\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4 - 10x)\widehat{j}+(-x^yln{(x)})\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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

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

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5z)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(4y^3 - 5x)\widehat{i}+(5x - x^{(y + z - 1)}{(y + z)})\widehat{j}+(x^{(y + z - 1)}{(y + z)} - 4y^3)\widehat{k}]\cdot\overrightarrow{A}}$

Explanation

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

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

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

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

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(tan{(x + y)}^2 + 1)\widehat{i}+(cos{(z + 5)} - tan{(x + y)}^2 - 1)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(tan{(x + y)}^2 - 1)\widehat{i}+(cos{(z + 5)} - tan{(x + y)}^2 - 1)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(tan{(x + y)}^2 - 1)\widehat{i}+(cos{(z + 5)} + tan{(x + y)}^2 - 1)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(tan{(x + y)}^2 + 1)\widehat{i}+(cos{(z + 5)} + tan{(x + y)}^2 - 1)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(tan{(x + y)}^2 + 1)\widehat{j}+(cos{(z + 5)} + tan{(x + y)}^2 - 1)\widehat{k}]\cdot\overrightarrow{A}}$

Explanation

In order to utilize Stokes' theorem, note its form

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

The curl of a vector 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

$\begin{align*}&\text{From what we are told}\&\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(z + 5)})}dx+{(tan{(x + y)})}dz\&\text{Meaning that}\&F_x=sin{(z + 5)};F_y=0;F_z=tan{(x + y)}\&\text{From this we can derive our curl vectors:}\end{align*}$

$\begin{align*}\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=tan{(x + y)}^2 + 1-[0]=tan{(x + y)}^2 + 1 \ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=cos{(z + 5)}-[tan{(x + y)}^2 + 1] \ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-[0]=0\&\text{And in turn our surface integral}:\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(tan{(x + y)}^2 + 1)\widehat{i}+(cos{(z + 5)} - tan{(x + y)}^2 - 1)\widehat{j}]\cdot\overrightarrow{A}}\end{align*}$

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

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(3x)\widehat{j}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(3x)\widehat{i}]\cdot\overrightarrow{A}}$

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

Explanation

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

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

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

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

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

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

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

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

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

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

Explanation

In order to utilize Stokes' theorem, note its form

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

The curl of a vector 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

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

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

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5y)\widehat{k}]\cdot\overrightarrow{A}}$

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

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

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

Explanation

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

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

$\begin{align*}&\text{Let }\textbf{S}\text{ be a known surface with a boundary curve, }\textbf{C}\text{.}\&\text{Considering the integral }\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(yz)})}dx+{(x^3)}dy+{(z + 5)}dz\&\text{Utilize Stokes' Theorem to determine an equivalent integral of the form:}\&\iint_{S} curl(\overrightarrow{F})\cdot d\overrightarrow{A}\end{align*}$

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

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

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

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

Explanation

$\begin{align*}&\text{From what we are told}\&\overrightarrow{F}\cdot d\overrightarrow{s}={(sin{(yz)})}dx+{(x^3)}dy+{(z + 5)}dz\&\text{Meaning that}\&F_x=sin{(yz)};F_y=x^3;F_z=z + 5\&\text{From this we can derive our curl vectors:}\end{align*}$

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

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

Considering the integral $\oint_{C} \overrightarrow{F}\cdot d\overrightarrow{s}\oint_{C}\overrightarrow{F}\cdot d\overrightarrow{s}={(30y^2)}dx+{(5z)}dy+{(8z)}dz$ , utilize Stokes' Theorem to determine an equivalent integral of the form:

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

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-5)\widehat{i}+(-60y)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5)\widehat{i}+(-60y)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(5)\widehat{i}+(60y)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(60y+5)\widehat{k}]\cdot\overrightarrow{A}}$

${\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(60y-5)\widehat{k}]\cdot\overrightarrow{A}}$

Explanation

In order to utilize Stokes' theorem, note its form

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

The curl of a vector 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

$\begin{align*}&\text{From what we are told}\&\overrightarrow{F}\cdot d\overrightarrow{s}={(30y^2)}dx+{(5z)}dy+{(8z)}dz\&\text{Meaning that}\&F_x=30y^2;F_y=5z;F_z=8z\&\text{From this we can derive our curl vectors:}\&\frac{\delta F_z}{\delta y}-\frac{\delta F_y}{\delta z}=0-5=-5 \ &\frac{\delta F_x}{\delta z}-\frac{\delta F_z}{\delta x}=0-0=0 \ &\frac{\delta F_y}{\delta x}-\frac{\delta F_x}{\delta y}=0-60y=-60y\&\text{This in turn allows us to set up our surface integral}:\&{\iint_{S}curl(\overrightarrow{F})\cdot d\overrightarrow{A} =[(-5)\widehat{i}+(-60y)\widehat{k}]\cdot\overrightarrow{A}}\end{align*}$

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