Question

Question: Verify that Stokes' Theorem is true for the given vector field F and surface S.

\({\rm{F(x,y,z) = - 2yzi + yj + 3xk}}\), S is the part of the paraboloid \({\rm{z = 5 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}\)that lies above the plane, oriented upward

Short answer

The Stokes' theorem for \({\rm{F(x,y,z) = - 2yzi + yj + 3xk}}\) is True.

Step-by-step solution

Definition of Concept

Integrals: An integral is a mathematical concept that assigns numbers to functions in order to describe displacement, area, volume, and other concepts that arise from combining infinitesimal data. The process of determining integrals is known as integration.

Verify that Stokes' Theorem is true for the given vector field F and surface S

Considering the given information:

The field is \({\rm{F(x,y,z) = - 2yzi + yj + 3xk}}\) and

Consider the expression for surface that is part parabolid above the plane is,\({\rm{z = 5 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}{\rm{(1)}}\)

Apply the formula:

Write the expression for curl of\({\rm{F(x,y,z) = Pi + Qj + Rk}}\).

\(\begin{array}{l}{\mathop{\rm curl}\nolimits} {\bf{F}} = \left| {\begin{array}{*{20}{l}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\P&Q&R\end{array}} \right|\\{\mathop{\rm curl}\nolimits} {\bf{F}} = \left( {\frac{{\partial R}}{{\partial y}} - \frac{{\partial Q}}{{\partial z}}} \right){\bf{i}} - \left( {\frac{{\partial R}}{{\partial x}} - \frac{{\partial P}}{{\partial z}}} \right){\bf{j}} + \left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right){\bf{k}}(2)\end{array}\)

Write the Stokes' theorem expression.

Here,

Surface is denoted by S.

Consider the surface\({\rm{S,z = g(x,y)}}\), which is oriented upward. Create an expression for the surface integral of F over the surface S.

Here,

A stands for area.

A plane\({\rm{z = 1}}\)intersects the parabolic.

In the equation, replace z with 1. (1),

\(\begin{array}{c}{\rm{1 = 5 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}\\{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 5 - 1}}\\{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 4}}\end{array}\)

Because the surface is oriented upward, consider the parametric equations.

\(\begin{array}{l}x = 2\cos t\\y = 2\sin t\\z = 1\quad ,0 \le t \le 2\pi \end{array}\)

Create an expression for the vector function\({\rm{r(t)}}\).

\({\rm{r(t) = xi + yj + zk}}\)

Substitute\({\rm{2cost}}\)for\({\rm{x,2sint}}\)for\({\rm{y,1}}\)for z

\(\begin{array}{c}{\rm{r(t) = 2costi + 2sintj + (1)k}}\\{\rm{ = 2costi + 2sintj + k}}\end{array}\)

Differentiate the expression in terms of t.

\(\begin{array}{c}{{\rm{r}}^{\rm{\cent}}}{\rm{(t) = }}\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{(2costi + 2sintj + k)}}\\{\rm{ = }}\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{(2cost)i + }}\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{(2sint)j + }}\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{(1)k}}\\{\rm{ = 2( - sint)i + (2cost)j + (0)k}}\\{\rm{ = - 2sinti + 2costj}}\end{array}\)

Determine the value of\({\rm{F(r(t))}}\).

\(\begin{array}{c}{\rm{F(r(t)) = - 2(2sint)(1)i + (2sint)j + 3(2sint)k}}\\{\rm{ = - 4sinti + 2sintj + 6sintk}}\end{array}\)

Determine the value of\({\bf{F}}({\bf{r}}(t)) \cdot {{\bf{r}}^\prime }(t)\).

\(\begin{array}{c}{\rm{F(r(t)) \times }}{{\rm{r}}^{\rm{\cent}}}{\rm{(t) = ( - 4sinti + 2sintj + 6sintk) \times ( - 2sinti + 2costj)}}\\{\rm{ = ( - 4sint)( - 2sint) + (2sint)(2cost) + (6sint)(0)}}\\{\rm{ = 8si}}{{\rm{n}}^{\rm{2}}}{\rm{t + 4sintcost + 0}}\\{\rm{ = 8si}}{{\rm{n}}^{\rm{2}}}{\rm{t + 4sintcost}}\end{array}\)

Create an expression for\(\oint_C F \cdot dr\).

\(\oint_C F \cdot dr = \int_0^{2\pi } {\bf{F}} ({\bf{r}}(t)) \cdot {{\bf{r}}^\prime }(t)dt\)

Putting\({\rm{8si}}{{\rm{n}}^{\rm{2}}}{\rm{t + 4sintcost}}\)for\({\bf{F}}({\bf{r}}(t)) \cdot {{\bf{r}}^\prime }(t)\),

\(\begin{array}{l} = \int_0^{2\pi } {(4(} 1 - \cos 2t) + 2\sin 2t)dt\\ = \left[ {4\left( {t - \frac{{\sin 2t}}{2}} \right) + 2\left( {\frac{{ - \cos 2t}}{2}} \right)} \right)_0^{2\pi }\\\oint_C F \cdot dr = \left( {\begin{array}{*{20}{c}}{\left. {4\left( {2\pi - \frac{{\sin 2(2\pi )}}{2}} \right) + 2\left( {\frac{{ - \cos 2(2\pi )}}{2}} \right) - } \right)}\\{4\left( {0 - \frac{{\sin 2(0)}}{2}} \right) + 2\left( {\frac{{ - \cos 2(0)}}{2}} \right)}\end{array}} \right)\\ = 8\pi - 0 + 2\left( { - \frac{1}{2}} \right) - 0 - 2\left( { - \frac{1}{2}} \right)\\ = 8\pi - 1 + 1\\ = 8\pi \end{array}\)

Using equation, calculate the value of \({\rm{curlF}}\). (2).

\(\begin{array}{c}{\mathop{\rm curl}\nolimits} F = \left( {\frac{{\partial (3x)}}{{\partial y}} - \frac{{\partial (y)}}{{\partial z}}} \right){\bf{i}} - \left( {\frac{{\partial (3x)}}{{\partial x}} - \frac{{\partial ( - 2yz)}}{{\partial z}}} \right){\bf{j}} + \left( {\frac{{\partial (y)}}{{\partial x}} - \frac{{\partial ( - 2yz)}}{{\partial y}}} \right){\bf{k}}\\ = (0 - 0)\;i - (3(1) - ( - 2y(1)))\;j + (0 - ( - 2z(1)))\;k\\ = ( - 3 - 2y){\bf{j}} + 2z{\bf{k}}\end{array}\)

The expression for surface\(S,D = \{ (r,\theta )\mid 0 \le r \le 2,0 \le \theta \le 2\pi \} \)in polar coordinate system is,\({\rm{z = 5 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}\)

Here,

\({\rm{x = rcos\theta and y = rsin\theta }}\)

As a result, the equation is\({\rm{z = g(x,y)}}\).

The surface S has an upward orientation.

Compare the equations \({\rm{curlF = Pi + Qj + Rk}}\)and \({\rm{curlF = ( - 3 - 2y)j + 2zk}}\).

\(\begin{array}{l}{\rm{P = 0}}\\{\rm{Q = ( - 3 - 2y)}}\\{\rm{R = 2z}}\end{array}\)

Using equation, calculate the value of

Putting \({\rm{rcos\theta }}\) for \({\rm{x,rsin\theta }}\) for y, and \(rdrd\theta \)for\({\rm{dA}}\),

To make the equation easier to understand, simplify it as follows.

\(\begin{array}{l} = \int_0^{2\pi } {\int_0^2 {\left( { - 4{r^3} + 2{r^3}\cos 2\theta + 6{r^2}\cos \theta + 10r} \right)} } drd\theta \\ = \int_0^{2\pi } {\left( { - 4\left( {\frac{{{r^4}}}{4}} \right) + 2\left( {\frac{{{r^4}}}{4}} \right)\cos 2\theta + 6\left( {\frac{{{r^3}}}{3}} \right)\cos \theta + 10\left( {\frac{{{r^2}}}{2}} \right)} \right)_0^2} d\theta \\ = \int_0^{2\pi } {\left( { - {r^4} + \frac{{{r^4}}}{2}\cos 2\theta + 2{r^3}\cos \theta + 5{r^2}} \right)_0^2} d\theta \end{array}\)

Limit values should be used.

Limit values should be used to simplify the equation.

Putting \({\rm{8\pi }}\) for in equation (3),

\(\int_{\rm{C}} {\rm{F}} {\rm{ \times dr = 8\pi }}\)

The value of \(\int_C F \cdot dr\) is the same when Stokes' theorem is used and when it is not used, and Stokes' theorem is verified.

Therefore, the Stokes' theorem for \({\rm{F(x,y,z) = - 2yzi + yj + 3xk}}\) is True.