Stoke's Theorem
This theorem gives relationship between Line Integral and Surface Integral.
If S be an open surface bounded by closed curve C and \(\vec{F}\) is a continuous dufferentiable vector point function then:
\[ \oint_C \vec{F}.d\vec{R} = \iint_S (curl(\vec{F})).\hat{N}ds \]
Sample Problem 1
Verify Stoke's Theorem for \(\hspace{5pt} \vec{F} = (x^2 + y^2)\hat{i} - 2xy\hat{j} \) taken around the rectangle bounded by the lines \[ x = \pm a, y = 0, y = b \]
Solution
Tracing the curves by using given equations: \[ x = \pm a, y = 0, y = b \]
By Stoke's Theorem:
\[ \oint_{ABCD} \vec{F}.d\vec{R} = \iint_{ABCD} (curl(\vec{F})).\hat{N}ds \]
L.H.S. :
\[ \oint_{ABCD} \vec{F}.d\vec{R} = \int_{AB} \vec{F}.d\vec{R} + \int_{BC} \vec{F}.d\vec{R} + \int_{CD} \vec{F}.d\vec{R} + \int_{DA} \vec{F}.d\vec{R} \]
For curve AB :
The equation is:
\[ x = a \Rightarrow dx = 0 \]
Points are A(a,0) to B(a,b)
\[ \int_{AB} \vec{F}.d\vec{R} = \int_{AB} \left[(x^2 + y^2)\hat{i} - 2xy\hat{j} \right].\left(dx\hat{i} + dy\hat{j} + dz\hat{k} \right) \]
\[ = \int_{AB} (x^2 + y^2)dx - 2xydy \]
\[ = \int^b_0 (a^2 + y^2)0 - 2aydy \]
\[ = \int^b_0 -2aydy \]
\[ = \frac{-2ab^2}{2} = \boxed{-ab^2} \]
Along BC :
\[ y = b \Rightarrow dy = 0 \]
Points : B(a,b) to C(-a,b)
\[ \int_{BC} \vec{F}.d\vec{R} = \int_{BC} (x^2 + y^2)dx - 2xydy \]
\[ = \int^{-a}_a (x^2 + b^2)dx \]
\[ = \left[\frac{x^3}{3} + b^2x \right]^{-a}_a \]
\[ = \left[\frac{-a^3}{3} - b^2a \right] - \left[\frac{a^3}{3} + b^2a \right] \]
\[ = \boxed{\frac{-2a^3}{3} - 2ab^2} \]
Along CD :
\[ x = -a \Rightarrow dx = 0 \]
Points : C(-a,b) to D(-a,0)
\[ \int_{CD} \vec{F}.d\vec{R} = \int_{CD} (x^2 + y^2)dx - 2xydy \]
\[ = \int^0_b (a^2 + y^2)0 + 2aydy \]
\[ = \int^0_b 2aydy \]
\[ = \boxed{-ab^2} \]
Along DA :
\[ y = 0 \Rightarrow dy = 0 \]
Points: D(-a,0) to A(a,0)
\[ \int_{DA} \vec{F}.d\vec{R} = \int_{DA} (x^2 + y^2)dx - 2xydy \]
\[ = \int^{a}_{-a} x^2dx \]
\[ = \left[\frac{x^3}{3} \right]_{-a}^a \]
\[ = \frac{a^3}{3} - \frac{-a^3}{3} \]
\[ = \boxed{\frac{2a^3}{3}} \]
\[ \oint_{ABCD} \vec{F}.d\vec{R} = -ab^2 - \frac{2a^3}{3} - 2ab^2 - ab^2 + \frac{2a^3}{3} \]
\[ = \boxed{-4ab^2} \]
R.H.S. :
\[ curl(\vec{F}) = \nabla \times \vec{F} \]
\[ = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} +
\hat{k}\frac{\partial}{\partial z} \right)\times \left[(x^2 + y^2)\hat{i} - 2xy\hat{j}\right] \]
\[ = \large
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\\\
\Large\frac{\partial}{\partial x} & \Large\frac{\partial}{\partial y} & \Large\frac{\partial}{\partial z}
\\\\
x^2 + y^2 & -2xy & 0
\end{vmatrix} \]
\[ = \hat{i}(0 + 0) - \hat{0 - 0} + \hat{k}(-2y - 2y) \]
\[ \Rightarrow curl\vec{F} = -4y\hat{k} \]
Since surface ABCD lying in X-Y plane so it must be projected on X-Y plane
\[ \Rightarrow \hat{N} = \hat{k} \]
\[ \Rightarrow ds = \frac{dxdy}{|\hat{N}\hat{k}|} = dxdy \]
\[ \iint_{ABCD} (curl(\vec{F})).\hat{N}ds \]
\[ = \iint (-4y\hat{k}).\hat{N}dxdy \]
\[ = \iint -4ydxdy \]
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\[ \Rightarrow \int^b_0 \int^a_{-a} -4ydxdy \] \[ = \int^b_0 \left[4yx \right]^a_{-a} dy \] \[ = \int^b_0 8yady = \left[4y^2a \right]^b_0 \] \[ = \boxed{-4b^2a} \] = R.H.S., Hence Proved :)
