Stokes’ Theorem
For a positively oriented boundary curve \(C\) of a surface \(S\),
\[\oint_C\vec F\cdot d\vec r=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
Stokes’ Theorem
Line Integrals Converted into Surface Integrals
Stokes’ theorem is developed through two IOE past-paper questions followed by some important past-paper questions. Every verification is worked out from the line integral to the surface integral.
UnitVector Calculus
TopicStokes’ Theorem
Question Type4 Mark Problems
Theory Required for These Problems
Curl
If \(\vec F=F_1\hat i+F_2\hat j+F_3\hat k\), then
\[{\rm curl}\,\vec F=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\F_1&F_2&F_3\end{array}\right|.\]
Stokes’ Theorem in the \(xy\)-Plane
Note: While applying or verifying Stokes’ theorem in two dimensions, take
\[\hat n=\hat k,\qquad dS=dx\,dy.\]Normal and Surface Element
For a surface \(\phi(x,y,z)=0\),
\[\vec n={\rm grad}\,\phi,\qquad\hat n=\frac{{\rm grad}\,\phi}{\left|{\rm grad}\,\phi\right|}.\]
\[\vec n={\rm grad}\,\phi\]\[\hat n=\frac{{\rm grad}\,\phi}{\left|{\rm grad}\,\phi\right|}.\]
For projection on the \(xy\)-plane,
\[dS=\frac{dx\,dy}{\left|\hat n\cdot\hat k\right|}.\]Verification: First evaluate \(\displaystyle\oint_C\vec F\cdot d\vec r\) along the complete boundary. Then evaluate \(\displaystyle\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS\). Stokes’ theorem is verified when both values are equal.
Continue through the IOE Engineering Mathematics II Digital Workbook. This stokes’ theorem page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
37
4 Marks
Line Integral Around a Square
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 11 OR • 4 Marks
Apply Stokes’ theorem to evaluate
\[\oint_C(xy\,dx+xy^2\,dy),\]
where \(C\) is the boundary of the square in the \(xy\)-plane with vertices
\[(-1,0),\qquad(1,0),\qquad(0,1),\qquad(0,-1).\]
Solution
Stokes’ theorem in the (xy)-plane
Take
\[\vec F=xy\hat i+xy^2\hat j+0\hat k.\]Therefore,
\[{\rm curl}\,\vec F=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\xy&xy^2&0\end{array}\right|.\]
Expanding the determinant,
\[{\rm curl}\,\vec F=\hat i(0-0)-\hat j(0-0)+\hat k(y^2-x).\]
\[{\rm curl}\,\vec F=\hat i(0-0)-\hat j(0-0)\]\[{}+\hat k(y^2-x).\]
Hence,
\[{\rm curl}\,\vec F=0\hat i+0\hat j+(y^2-x)\hat k.\]The square lies in the \(xy\)-plane. Therefore,
\[\hat n=\hat k,\qquad dS=dx\,dy.\]By Stokes’ theorem,
\[\oint_C\vec F\cdot d\vec r=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
Now,
\[{\rm curl}\,\vec F\cdot\hat n=y^2-x.\]Therefore,
\[\oint_C\vec F\cdot d\vec r=\iint_S(y^2-x)\,dx\,dy.\]The region is symmetric about the \(y\)-axis. Hence,
\[\iint_Sx\,dx\,dy=0.\]Thus,
\[\oint_C\vec F\cdot d\vec r=\iint_Sy^2\,dx\,dy.\]The square and the integrand \(y^2\) are symmetric in all four quadrants. In the first quadrant,
\[0\le y\le1,\qquad0\le x\le1-y.\]Therefore,
\[\oint_C\vec F\cdot d\vec r=4\int_0^1\int_0^{1-y}y^2\,dx\,dy.\]Integrating with respect to \(x\),
\[=4\int_0^1\left[xy^2\right]_{x=0}^{x=1-y}dy.\]Hence,
\[=4\int_0^1y^2(1-y)\,dy\]\[=4\int_0^1(y^2-y^3)\,dy.\]Therefore,
\[=4\left[\frac{y^3}{3}-\frac{y^4}{4}\right]_0^1\]\[=4\left(\frac13-\frac14\right)\]\[=\frac13.\]\[\oint_C(xy\,dx+xy^2\,dy)=\frac13.\]
\[\oint_C(xy\,dx+xy^2\,dy)\]\[=\frac13.\]
38
4 Marks
Line Integral Around a Triangle
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 11 OR • 4 Marks
Apply Stokes’ theorem to evaluate
\[\oint_C\left\{(x+y)\,dx+(2x-z)\,dy+(y+z)\,dz\right\},\]
\[\oint_C\Big\{(x+y)\,dx+(2x-z)\,dy\]\[{}+(y+z)\,dz\Big\},\]
where \(C\) is the boundary of the triangle with vertices
\[(2,0,0),\qquad(0,3,0),\qquad(0,0,6).\]
Solution
Stokes’ theorem and \(xy\)-projection
Take
\[\vec F=(x+y)\hat i+(2x-z)\hat j+(y+z)\hat k.\]
\[\vec F=(x+y)\hat i+(2x-z)\hat j\]\[{}+(y+z)\hat k.\]
Therefore,
\[{\rm curl}\,\vec F=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x+y&2x-z&y+z\end{array}\right|.\]
Expanding the determinant,
\[{\rm curl}\,\vec F=\hat i\{1-(-1)\}-\hat j(0-0)+\hat k(2-1).\]
\[{\rm curl}\,\vec F=\hat i\{1-(-1)\}\]\[{}-\hat j(0-0)+\hat k(2-1).\]
Hence,
\[{\rm curl}\,\vec F=2\hat i+0\hat j+\hat k.\]The equation of the plane through \((2,0,0)\), \((0,3,0)\) and \((0,0,6)\) is
\[\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1.\]Therefore,
\[\phi=3x+2y+z-6=0.\]Now,
\[{\rm grad}\,\phi=3\hat i+2\hat j+\hat k.\]Hence, the unit normal in the direction fixed by \(A\to B\to C\to A\) is
\[\hat n=\frac{3\hat i+2\hat j+\hat k}{\sqrt{3^2+2^2+1^2}}=\frac{3\hat i+2\hat j+\hat k}{\sqrt{14}}.\]
\[\hat n=\frac{3\hat i+2\hat j+\hat k}{\sqrt{3^2+2^2+1^2}}\]\[=\frac{3\hat i+2\hat j+\hat k}{\sqrt{14}}.\]
Therefore,
\[{\rm curl}\,\vec F\cdot\hat n=(2\hat i+\hat k)\cdot\frac{3\hat i+2\hat j+\hat k}{\sqrt{14}}=\frac{7}{\sqrt{14}}.\]
\[{\rm curl}\,\vec F\cdot\hat n\]\[=(2\hat i+\hat k)\cdot\frac{3\hat i+2\hat j+\hat k}{\sqrt{14}}\]\[=\frac{7}{\sqrt{14}}.\]
For projection on the \(xy\)-plane,
\[dS=\frac{dx\,dy}{\left|\hat n\cdot\hat k\right|}.\]Since
\[\hat n\cdot\hat k=\frac{1}{\sqrt{14}},\]it follows that
\[dS=\sqrt{14}\,dx\,dy.\]The projection of the triangular surface on the \(xy\)-plane is
\[3x+2y\le6,\qquad x\ge0,\qquad y\ge0.\]Therefore, the limits are
\[0\le x\le2,\qquad0\le y\le3-\frac{3x}{2}.\]By Stokes’ theorem,
\[\oint_C\vec F\cdot d\vec r=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
Hence,
\[\oint_C\vec F\cdot d\vec r=\int_0^2\int_0^{\,3-\frac{3x}{2}}\frac{7}{\sqrt{14}}\left(\sqrt{14}\right)dy\,dx.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\int_0^2\int_0^{\,3-\frac{3x}{2}}\frac{7}{\sqrt{14}}\left(\sqrt{14}\right)dy\,dx.\]
Therefore,
\[=7\int_0^2\int_0^{\,3-\frac{3x}{2}}dy\,dx\]\[=7\int_0^2\left(3-\frac{3x}{2}\right)dx.\]Hence,
\[=7\left[3x-\frac{3x^2}{4}\right]_0^2\]\[=7(6-3)\]\[=21.\]\[\oint_C\vec F\cdot d\vec r=21\]Orientation: \(A\to B\to C\to A\).
Some Important Past-Paper Questions
Complete verification: Evaluate the line integral around \(C\), evaluate the surface integral over \(S\), and show that both values are equal.
Stokes’ Theorem on a First-Octant Plane
SOME IMPORTANT PAST-PAPER QUESTION • COMPLETE VERIFICATION
Verify Stokes’ theorem for
\[\vec F=x\hat i+y^2\hat k,\]
where \(S\) is the part of the plane
\[x+y+z=1\]
in the first octant and \(C\) is its positively oriented boundary.
Solution
Line integral and surface integralBy Stokes’ theorem,
\[\oint_C\vec F\cdot d\vec r=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
Here,
\[\vec F=x\hat i+0\hat j+y^2\hat k.\]Therefore,
\[\vec F\cdot d\vec r=x\,dx+y^2\,dz.\]The boundary is
\[C=C_1+C_2+C_3.\]For \(C_1\), the line \(AB\), the direction ratios are \(\langle-1,1,0\rangle\). Hence,
\[\frac{x-1}{-1}=\frac{y-0}{1}=\frac{z-0}{0}=t.\]
\[\frac{x-1}{-1}=\frac{y}{1}=\frac{z}{0}=t.\]
Therefore,
\[x=1-t,\qquad y=t,\qquad z=0,\qquad0\le t\le1.\]
\[x=1-t,\qquad y=t,\qquad z=0\]\[0\le t\le1.\]
Differentiating,
\[dx=-dt,\qquad dy=dt,\qquad dz=0.\]
\[dx=-dt,\qquad dy=dt,\qquad dz=0.\]
Thus,
\[\int_{C_1}\vec F\cdot d\vec r=\int_0^1(1-t)(-dt).\]Hence,
\[\int_{C_1}\vec F\cdot d\vec r=-\int_0^1(1-t)\,dt\]\[=-\left[t-\frac{t^2}{2}\right]_0^1=-\frac12.\]For \(C_2\), the line \(BC\), the direction ratios are \(\langle0,-1,1\rangle\). Therefore,
\[x=0,\qquad y=1-t,\qquad z=t,\qquad0\le t\le1.\]
\[x=0,\qquad y=1-t,\qquad z=t\]\[0\le t\le1.\]
Differentiating,
\[dx=0,\qquad dy=-dt,\qquad dz=dt.\]Thus,
\[\int_{C_2}\vec F\cdot d\vec r=\int_0^1(1-t)^2\,dt.\]Hence,
\[\int_{C_2}\vec F\cdot d\vec r=\left[-\frac{(1-t)^3}{3}\right]_0^1=\frac13.\]For \(C_3\), the line \(CA\), the direction ratios are \(\langle1,0,-1\rangle\). Therefore,
\[x=t,\qquad y=0,\qquad z=1-t,\qquad0\le t\le1.\]
\[x=t,\qquad y=0,\qquad z=1-t\]\[0\le t\le1.\]
Differentiating,
\[dx=dt,\qquad dy=0,\qquad dz=-dt.\]Thus,
\[\int_{C_3}\vec F\cdot d\vec r=\int_0^1t\,dt=\left[\frac{t^2}{2}\right]_0^1=\frac12.\]Therefore,
\[\oint_C\vec F\cdot d\vec r=\int_{C_1}\vec F\cdot d\vec r+\int_{C_2}\vec F\cdot d\vec r+\int_{C_3}\vec F\cdot d\vec r.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\int_{C_1}\vec F\cdot d\vec r+\int_{C_2}\vec F\cdot d\vec r\]\[{}+\int_{C_3}\vec F\cdot d\vec r.\]
Hence,
\[\oint_C\vec F\cdot d\vec r=-\frac12+\frac13+\frac12=\frac13.\]VERIFICATION
\[\oint_C\vec F\cdot d\vec r=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
Now,
\[{\rm curl}\,\vec F=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x&0&y^2\end{array}\right|.\]
Expanding the determinant,
\[{\rm curl}\,\vec F=\hat i(2y-0)-\hat j(0-0)+\hat k(0-0).\]
\[{\rm curl}\,\vec F=\hat i(2y-0)-\hat j(0-0)\]\[{}+\hat k(0-0).\]
Hence,
\[{\rm curl}\,\vec F=2y\hat i+0\hat j+0\hat k.\]The surface is
\[\phi=x+y+z-1=0.\]Therefore,
\[{\rm grad}\,\phi=\hat i+\hat j+\hat k.\]Hence,
\[\hat n=\frac{\hat i+\hat j+\hat k}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt3}(\hat i+\hat j+\hat k).\]
\[\hat n=\frac{\hat i+\hat j+\hat k}{\sqrt{1^2+1^2+1^2}}\]\[=\frac{1}{\sqrt3}(\hat i+\hat j+\hat k).\]
Therefore,
\[{\rm curl}\,\vec F\cdot\hat n=\frac{2y}{\sqrt3}.\]For projection on the \(xy\)-plane,
\[dS=\frac{dx\,dy}{\left|\hat n\cdot\hat k\right|}.\]Since
\[\hat n\cdot\hat k=\frac{1}{\sqrt3},\]it follows that
\[dS=\sqrt3\,dx\,dy.\]The projection of \(S\) on the \(xy\)-plane is
\[x+y\le1,\qquad x\ge0,\qquad y\ge0.\]Therefore, the limits are
\[0\le x\le1,\qquad0\le y\le1-x.\]Thus,
\[\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS=\int_0^1\int_0^{1-x}\frac{2y}{\sqrt3}\left(\sqrt3\right)dy\,dx.\]
\[\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS\]\[=\int_0^1\int_0^{1-x}\frac{2y}{\sqrt3}\left(\sqrt3\right)dy\,dx.\]
Therefore,
\[=2\int_0^1\int_0^{1-x}y\,dy\,dx\]\[=2\int_0^1\left[\frac{y^2}{2}\right]_0^{1-x}dx.\]Hence,
\[=\int_0^1(1-x)^2\,dx\]\[=\left[-\frac{(1-x)^3}{3}\right]_0^1\]\[=\frac13.\]\[\oint_C\vec F\cdot d\vec r=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS=\frac13.\]
\[\oint_C\vec F\cdot d\vec r=\frac13\]\[\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS=\frac13.\]
Hence, Stokes’ theorem is verified.Stokes’ Theorem on an Upper Hemisphere
SOME IMPORTANT PAST-PAPER QUESTION • COMPLETE VERIFICATION
Verify Stokes’ theorem for
\[\vec F=2y\hat i+3x\hat j-z^2\hat k,\]
where \(S\) is the upper half of the sphere
\[x^2+y^2+z^2=9\]
and \(C\) is its positively oriented boundary.
Solution
Circular parametrisation and projectionBy Stokes’ theorem,
\[\oint_C\vec F\cdot d\vec r=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
The boundary \(C\) is
\[x^2+y^2=9,\qquad z=0.\]Using the parametric form,
\[x=3\cos t,\qquad y=3\sin t,\qquad z=0,\qquad0\le t\le2\pi.\]
\[x=3\cos t,\qquad y=3\sin t,\qquad z=0\]\[0\le t\le2\pi.\]
Differentiating,
\[dx=-3\sin t\,dt,\qquad dy=3\cos t\,dt,\qquad dz=0.\]
\[dx=-3\sin t\,dt\]\[dy=3\cos t\,dt,\qquad dz=0.\]
Now,
\[\vec F\cdot d\vec r=2y\,dx+3x\,dy-z^2\,dz.\]Substituting \(x\), \(y\), \(z\), \(dx\), \(dy\) and \(dz\),
\[\vec F\cdot d\vec r=2(3\sin t)(-3\sin t)\,dt+3(3\cos t)(3\cos t)\,dt-0.\]
\[\vec F\cdot d\vec r=2(3\sin t)(-3\sin t)\,dt\]\[{}+3(3\cos t)(3\cos t)\,dt-0.\]
Therefore,
\[\vec F\cdot d\vec r=(-18\sin^2t+27\cos^2t)\,dt.\]Hence,
\[\oint_C\vec F\cdot d\vec r=\int_0^{2\pi}(-18\sin^2t+27\cos^2t)\,dt.\]Using
\[\sin^2t=\frac{1-\cos2t}{2},\qquad\cos^2t=\frac{1+\cos2t}{2},\]
\[\sin^2t=\frac{1-\cos2t}{2}\]\[\cos^2t=\frac{1+\cos2t}{2},\]
it follows that
\[\oint_C\vec F\cdot d\vec r=\frac12\int_0^{2\pi}\{-18(1-\cos2t)+27(1+\cos2t)\}\,dt.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\frac12\int_0^{2\pi}\{-18(1-\cos2t)\]\[{}+27(1+\cos2t)\}\,dt.\]
Therefore,
\[=\frac12\int_0^{2\pi}(9+45\cos2t)\,dt.\]Hence,
\[=\frac12\left[9t+\frac{45}{2}\sin2t\right]_0^{2\pi}\]\[=\frac12(18\pi+0)\]\[=9\pi.\]VERIFICATION
\[\oint_C\vec F\cdot d\vec r=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
Here,
\[\vec F=2y\hat i+3x\hat j-z^2\hat k.\]Therefore,
\[{\rm curl}\,\vec F=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\2y&3x&-z^2\end{array}\right|.\]
Expanding the determinant,
\[{\rm curl}\,\vec F=\hat i(0-0)-\hat j(0-0)+\hat k(3-2).\]
\[{\rm curl}\,\vec F=\hat i(0-0)-\hat j(0-0)\]\[{}+\hat k(3-2).\]
Hence,
\[{\rm curl}\,\vec F=0\hat i+0\hat j+\hat k.\]The surface is
\[\phi=x^2+y^2+z^2-9=0.\]Therefore,
\[{\rm grad}\,\phi=2x\hat i+2y\hat j+2z\hat k.\]Hence, the outward unit normal is
\[\hat n=\frac{2x\hat i+2y\hat j+2z\hat k}{\sqrt{4x^2+4y^2+4z^2}}.\]
\[\hat n=\frac{2x\hat i+2y\hat j+2z\hat k}{\sqrt{4x^2+4y^2+4z^2}}.\]
Since \(x^2+y^2+z^2=9\),
\[\hat n=\frac{2(x\hat i+y\hat j+z\hat k)}{\sqrt{4(9)}}=\frac{x\hat i+y\hat j+z\hat k}{3}.\]
\[\hat n=\frac{2(x\hat i+y\hat j+z\hat k)}{\sqrt{4(9)}}\]\[=\frac{x\hat i+y\hat j+z\hat k}{3}.\]
Therefore,
\[{\rm curl}\,\vec F\cdot\hat n=\frac{z}{3}.\]For projection on the \(xy\)-plane,
\[dS=\frac{dx\,dy}{\left|\hat n\cdot\hat k\right|}.\]On the upper hemisphere,
\[\hat n\cdot\hat k=\frac{z}{3}.\]Therefore,
\[dS=\frac{3}{z}\,dx\,dy.\]Thus,
\[\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS=\iint_D\frac{z}{3}\left(\frac{3}{z}\right)dx\,dy.\]
\[\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS\]\[=\iint_D\frac{z}{3}\left(\frac{3}{z}\right)dx\,dy.\]
Hence,
\[\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS=\iint_Ddx\,dy,\]where \(D\) is the circular region
\[x^2+y^2\le9.\]Therefore,
\[\iint_Ddx\,dy=\pi(3)^2=9\pi.\]\[\oint_C\vec F\cdot d\vec r=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS=9\pi.\]
\[\oint_C\vec F\cdot d\vec r=9\pi\]\[\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS=9\pi.\]
Hence, Stokes’ theorem is verified.Stokes’ Theorem Around Four Lines
SOME IMPORTANT PAST-PAPER QUESTION • COMPLETE VERIFICATION
Verify Stokes’ theorem for
\[\vec F=(x^2+y^2)\hat i-2xy\hat j,\]
taken around the lines
\[x=-a,\qquad x=a,\qquad y=0,\qquad y=b.\]
Solution
Four boundary lines and \(xy\)-planeNote: The curve lies in the \(xy\)-plane. Therefore,
\[\hat n=\hat k,\qquad dS=dx\,dy.\]By Stokes’ theorem,
\[\oint_C\vec F\cdot d\vec r=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
Here,
\[\vec F=(x^2+y^2)\hat i-2xy\hat j+0\hat k.\]Therefore,
\[\vec F\cdot d\vec r=(x^2+y^2)\,dx-2xy\,dy.\]The boundary is
\[C=C_1+C_2+C_3+C_4.\]For \(C_1\), the line \(AB\),
\[y=0,\qquad dy=0,\qquad -a\le x\le a.\]Therefore,
\[\int_{C_1}\vec F\cdot d\vec r=\int_{-a}^{a}x^2\,dx.\]Hence,
\[\int_{C_1}\vec F\cdot d\vec r=\left[\frac{x^3}{3}\right]_{-a}^{a}=\frac{2a^3}{3}.\]For \(C_2\), the line \(BC\),
\[x=a,\qquad dx=0,\qquad0\le y\le b.\]Therefore,
\[\int_{C_2}\vec F\cdot d\vec r=\int_0^b-2ay\,dy.\]Hence,
\[\int_{C_2}\vec F\cdot d\vec r=-2a\left[\frac{y^2}{2}\right]_0^b=-ab^2.\]For \(C_3\), the line \(CD\),
\[y=b,\qquad dy=0,\qquad a\ge x\ge-a.\]Therefore,
\[\int_{C_3}\vec F\cdot d\vec r=\int_a^{-a}(x^2+b^2)\,dx.\]Hence,
\[\int_{C_3}\vec F\cdot d\vec r=\left[\frac{x^3}{3}+b^2x\right]_a^{-a}=-2\left(\frac{a^3}{3}+ab^2\right).\]
\[\int_{C_3}\vec F\cdot d\vec r\]\[=\left[\frac{x^3}{3}+b^2x\right]_a^{-a}\]\[=-2\left(\frac{a^3}{3}+ab^2\right).\]
For \(C_4\), the line \(DA\),
\[x=-a,\qquad dx=0,\qquad b\ge y\ge0.\]Therefore,
\[\int_{C_4}\vec F\cdot d\vec r=\int_b^0 2ay\,dy.\]Hence,
\[\int_{C_4}\vec F\cdot d\vec r=2a\left[\frac{y^2}{2}\right]_b^0=-ab^2.\]Therefore,
\[\oint_C\vec F\cdot d\vec r=\int_{C_1}\vec F\cdot d\vec r+\int_{C_2}\vec F\cdot d\vec r+\int_{C_3}\vec F\cdot d\vec r+\int_{C_4}\vec F\cdot d\vec r.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\int_{C_1}\vec F\cdot d\vec r+\int_{C_2}\vec F\cdot d\vec r\]\[{}+\int_{C_3}\vec F\cdot d\vec r+\int_{C_4}\vec F\cdot d\vec r.\]
Hence,
\[\oint_C\vec F\cdot d\vec r=\frac{2a^3}{3}-ab^2-2\left(\frac{a^3}{3}+ab^2\right)-ab^2.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\frac{2a^3}{3}-ab^2-2\left(\frac{a^3}{3}+ab^2\right)-ab^2.\]
Therefore,
\[\oint_C\vec F\cdot d\vec r=-4ab^2.\]VERIFICATION
\[\oint_C\vec F\cdot d\vec r=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
\[\oint_C\vec F\cdot d\vec r\]\[=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS.\]
Now,
\[{\rm curl}\,\vec F=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x^2+y^2&-2xy&0\end{array}\right|.\]
Expanding the determinant,
\[{\rm curl}\,\vec F=\hat i(0-0)-\hat j(0-0)+\hat k(-2y-2y).\]
\[{\rm curl}\,\vec F=\hat i(0-0)-\hat j(0-0)\]\[{}+\hat k(-2y-2y).\]
Hence,
\[{\rm curl}\,\vec F=0\hat i+0\hat j-4y\hat k.\]Since \(\hat n=\hat k\),
\[{\rm curl}\,\vec F\cdot\hat n=-4y.\]The limits of the rectangular region are
\[-a\le x\le a,\qquad0\le y\le b.\]Therefore,
\[\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS=\int_{-a}^{a}\int_0^b-4y\,dy\,dx.\]Integrating with respect to \(y\),
\[=-4\int_{-a}^{a}\left[\frac{y^2}{2}\right]_0^b dx\]\[=-2b^2\int_{-a}^{a}dx.\]Hence,
\[=-2b^2[x]_{-a}^{a}\]\[=-2b^2(2a)\]\[=-4ab^2.\]\[\oint_C\vec F\cdot d\vec r=\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS=-4ab^2.\]
\[\oint_C\vec F\cdot d\vec r=-4ab^2\]\[\iint_S{\rm curl}\,\vec F\cdot\hat n\,dS=-4ab^2.\]
Hence, Stokes’ theorem is verified.