Theorems
We will cover following topics
Green’s Theorem
Green’s theorem gives a relation between a double integral over a region \(R\) in the \(x y\) plane and the line integral over a closed curve \(C\) enclosing the region \(R\). It helps to evaluate line integral easily.
Statement of Green’s theorem: If \(P(x, y)\) and \(Q(x, y)\) are continuous functions with continuous partial derivatives in a region \(R\) in the \(x y\) plane and on its boundary \(C\) which is a simple closed curve then \(\oint_{C}(P d x+Q d y)=\iint_{R}\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right) d x d y\) where \(C\) is described in the anticlockwise sense (which is the positive sense).
The vector form of Green’s theorem is given by \(\oint_{C} \vec{F} \cdot d \vec{r}=\iint_{R} \nabla \times \vec{F} \cdot \vec{k} dR\), where \(dR=dx dy\).
Gauss’s Divergence Theorem
Gauss’s Divergence Theorem is used to convert a volume integral to a surface integral and vice-versa.
Let \(V\) be a region in space with piecewise smooth boundary \(S\) and let \(F\) is a continuously differentiable vector field defined on a neighbourhood of \(V\), then, according to the divergence theorem:
\[\iiint_{V}(\nabla \cdot \mathbf{F}) d V=\iint_{S} (\mathbf{n} \times \mathbf{F} )d S\]Stokes’ Theorem
Stokes’ Theorem is used to convert a line integral to a surface integral or vice-versa.
According to Stokes’ theorem, if \(\mathbf{F}\) be a vector-field, then:
\[\iint_{S}(\operatorname{curl} \mathbf{F} \cdot \hat{\mathbf{n}}) dS=\int_{C} (\mathbf{F} \cdot d \mathbf{r})\]where \(C\) is the boundary of the surface \(S\), and \(C\) and \(S\) have same orientations.
Green’s Identities
Green’s Identities are a set of three vector identities.
The first of these can be derived from the vector derivative identitities and Gauss’ divergence theorem: The two vector derivative equations are given by:
\[\nabla \cdot(\psi \nabla \phi)=\psi \nabla^{2} \phi+(\nabla \psi) \cdot(\nabla \phi) \qquad (1)\] \[\nabla \cdot(\phi \nabla \psi)=\phi \nabla^{2} \psi+(\nabla \phi) \cdot(\nabla \psi) \qquad (2)\]From the Gauss’s divergence theorem,
\[\int_{V}(\nabla \cdot \mathbf{F}) d V=\int_{S} \mathbf{F} \cdot d \mathbf{a} \qquad (3)\]From (2) and (3), we get:
\[\int_{S} \phi(\nabla \psi) \cdot d \mathbf{a}=\int_{V}\left[\phi \nabla^{2} \psi+(\nabla \phi) \cdot(\nabla \psi)\right] d V\]Subtracting (2) from (1), we get Green’s First Identity:
\[\nabla \cdot(\phi \nabla \psi-\psi \nabla \phi)=\phi \nabla^{2} \psi-\psi \nabla^{2} \phi (4)\]From (3) and (4), we get Green’s Second Identity:
\[\int_{V}\left(\phi \nabla^{2} \psi-\psi \nabla^{2} \phi\right) d V=\int_{5}(\phi \nabla \psi-\psi \nabla \phi) \cdot d \mathbf{a}\]Let \(u\) have continuous first partial derivatives and be harmonic inside the region of integration.
Then, according to Green’s Third Identity:
\[u(x, y)=\frac{1}{2 \pi} \oint_{C}\left[\ln \left(\frac{1}{r}\right) \frac{\partial u}{\partial n}-u \frac{\partial}{\partial n} \ln \left(\frac{1}{r}\right)\right] d s\]PYQs
Gauss’s Divergence Theorem
1) Let \(\vec{F}=xy^2\hat{i}+(y+x)\hat{j}\). Integrate \((\nabla\times\vec{F})\cdot\vec{k}\) over the region in the first quadrant bounded by the curve \(y=x^2\) and \(y=x\) using Green’s theorem.
[2018, 12M]
2) Using Green theorem, evaluate the \(\int_{C} F(\vec{r}) . d \vec{r}\) counterclockwise where \(F(r)=\left(x^{2}+y^{2}\right) \hat{i}+\left(x^{2}-y^{2}\right) \hat{j}\) and \(d \vec{r}=x \hat{i}+d y \hat{j}\) and the curve \(C\) is the boundary of the region \(R=\left\{(x, y) \vert 1 \leq y \leq 2-x^{2}\right\}\).
[2017, 8M]
3) Evaluate \(\int_{C} e^{-x}(\sin y d x+\cos y d y)\), where \(C\) is the rectangle with vertices \((0,0)\), \((\pi, 0)\), \(\left(\pi, \dfrac{\pi}{2}\right)\), \(\left(0, \dfrac{\pi}{2}\right)\).
[2015, 12M]
4) Verify Green’s theorem in the plane for \(\oint_{C}\left[(x y+y^{2}) d x+x^{2} d y\right]\) where \(C\) is the closed curve of the region bounded by \(y=x\) and \(y=x^{2}\).
[2012, 20M]
5) Verify Green’s theorem for \(e^{-x} \sin y d x+e^{-x} \cos y\) by the path of integration being the boundary of the square whose vertices are \((0,0)\), \(\left(\dfrac{\pi}{2},0\right)\), \(\left(\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\) and \(\left(0, \dfrac{\pi}{2}\right)\).
[2010, 20M]
Gauss’s Divergence Theorem
1) State Gauss divergence theorem. Verify this theorem for \(\vec{F}=4x\hat{i}-2y^2\hat{j}+z^2\hat{k}\), taken over the region bounded by \(x^2+y^2=4\), \(z=0\) and \(z=3\).
[2019, 15M]
2) If \(S\) is the surface of the sphere \(x^2+y^2+z^2=a^2\), then evaluate
\[\iint_S [(x+y) dydz+(y+z) dzdx+(x+y) dxdy]\]using Gauss’s divergence theorem.
[2018, 12M]
3) Evaluate the integral \(\iint_{S} F nds\), where \(\overline{F}=3 x y^{2} \hat{i}+\left(y x^{2}-y^{3}\right) j+3 z x^{2} K\) and \(S\) is a surface of the cylinder \(y^{2}+z^{2} \leq 4,-3 \leq x \leq 3\) using divergence theorem.
[2017, 9M]
4) By using Divergence Theorem of Gauss, evaluate the surface integral \(\iint\left(a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}\right)^{-\dfrac{1}{2}} d S\), where \(S\) is the surface \(e\) of the ellipsoid \(a x^{2}+b y^{2}+c z^{2}=1\), \(a\), \(b\) and \(c\) being all positive constants.
[2013, 15M]
5) Verify Gauss’ Divergence Theorem for the vector \(\vec{v}=x^{2} \hat{i}+y^{2} \hat{j}+z^{2} \hat{k}\) taken over the cube \(0 \leq x, y, z \leq 1\).
[2011, 15M]
6) Use the divergence theorem to evaluate \(\iint_{s} \vec{V}ndA\), where \(\vec{V}=x^{2} z \vec{i} y \vec{j}-x z^{2} \vec{k}\) and \(S\) is he boundary of the region bounded by the paraboloid \(z=x^{2}+y^{2}\) and the plane \(z=4y\).
[2010, 20M]
7) Using divergence theorem, evaluate \(\iint_{s} \overline{A} \cdot d \overline{S}\) where \(\overline{A}=x^{3} \hat{i}+y^{3} \hat{j}+z^{3} \hat{k}\) and \(S\) is the surface of the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\).
[2009, 20M]
8) Evaluate
\[\iint_S ( x^3 dy dz + x^2 y dz dx + x^2z dx dy )\]by Gauss divergence theorem, where \(S\) is the surface of the cylinder \(x^2+y^2=a^2\) bounded by \(z=0\) and \(z=b\).
[2005, 15M]
9) Verify Gauss’ divergence theorem of \(A=\left(4 x,-2 y^{2}, z^{2}\right)\) taken over the region bounded by \(x^{2}+y^{2}=4, z=0\) and \(z=3\).
[2001, 15M]
Stokes’ Theorem
1) Evalute by Stokes’ theorem \(\oint_C e^xdx+2ydy-dz\), where \(C\) is the curve \(x^2+y^2=4\), \(z=2\).
[2019, 5M]
2) Evaluate the line integral \(\int_C -y^3dx +x^3dy +z^3dz\) using Stokes’ theorem. Here \(C\) is the intersection of the cylinder \(x^2+y^2=1\) and the plane \(x+y+z=1\). The orientation on \(C\) corresponds to counterclockwise motion in the \(xy-plane\).
[2018, 13M]
3) Evaluate by Stokes’ theorem:
\[\int (ydx+zdy+xdz)\]where \(\) is the curve given by \(x^2+y^2+z^2-2ax-2ay=0\), \(x+2y=2a\), starting from \((2a,0,0)\) and then going below the z-plane.
[2014, 20M]
4) Use Stokes’ theorem to evaluate the line integral \(\int_{C}\left(-y^{3} d x+x^{3} d y-z^{3} d z\right)\), where \(C\) is the intersection of the cylinder \(x^{2}+y^{2}=1\) and the plane \(x+y+z=1\).
[2013, 15M]
5) If \(\vec{F}=y \vec{i}+(x-2 x z) \vec{j}-x y \vec{k}\), evaluate \(\iint_{S}(\vec{\nabla} \times \vec{F}) . \vec{n} d \vec{s}\), where \(S\) is the surface of the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) above the \(xy-plane\).
[2012, 20M]
6) If \(\vec{u}=4 y \hat{i}+x \hat{j}+2 z \hat{k}\), calculate the double integral \(\iint(\nabla \times \vec{u}) d \vec{s}\) over the hemisphere given by \(x^{2}+y^{2}+z^{2}=a^{2}\), \(z \geq 0\).
[2011, 15M]
7) Find the work done in moving the particle once round the ellipse \(\dfrac{x^{2}}{25}+\dfrac{y^{2}}{16}=1\), \(z=0\) under the field of force of given by \(\overline{F}=(2 x-y+z) \hat{i}+\left(x+y-z^{2}\right) \hat{j}+(3 x-2 y+4 z) \hat{k}\).
[2009, 20M]
8) Find the value of \(\iint_{s}(\vec{\nabla} \times \vec{f}) . d s\) taken over the upper portion of the surface \(x^{2}+y^{2}-2 a x+a z=0\) and the bounding curve lies in the plane \(z=0\), when \(\vec{F}=\left(y^{2}+z^{2}-x^{2}\right) \hat{i}+\left(z^{2}+x^{2}-y^{2}\right) \hat{j}+\left(x^{2}+y^{2}-z^{2}\right) \hat{k}\).
[2009, 20M]
9) Determine \(\int_{c}(y d x+z d y+x d z)\) by using Stoke’s theorem, where \(C\) is the curve defined by \((x-a)^{2}+(y-a)^{2}+z^{2}=2 a^{2}\), \(x+y=2 a\) that starts from the point \((2 a, 0,0)\) goes at first below the \(z-plane\).
[2007, 15M]
10) Verify Stokes’ theorem for the function \(\overline{F}=x^{2} \hat{i}-x y \hat{j}\) integrated round the square in the plane \(z=0\) and bounded by the lines \(x=0\), \(y=0\), \(x=a\) and \(y=a, a>0\).
[2006, 15M]
11) Verify Stokes’ theorem for \(\hat{f}=(2 x-y) \hat{i}-y z^{2} \hat{j} z \hat{k}\) where \(S\) is the upper half surface of the sphere \(x^{2}+y^{2}+z^{2}=1\) and \(C\) is its boundary.
[2004, 15M]
12) Evaluate \(\iint_S curl A. dS\), where \(S\) is the open surface
\[x^2+y^2-4x+4z=0, z \geq 0\]and
\(A=(y^2+z^2-x^2) \hat{i}\) + \((2z^2+x^2-y^2) \hat{j}\) + \((x^2+y^2-3z^2) \hat{k}\)
[2003, 15M]