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 xy 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 xy plane and on its boundary C which is a simple closed curve then ∮C(Pdx+Qdy)=∬R(∂Q∂x−∂P∂y)dxdy where C is described in the anticlockwise sense (which is the positive sense).
The vector form of Green’s theorem is given by ∮C→F⋅d→r=∬R∇×→F⋅→kdR, where dR=dxdy.
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:
∭V(∇⋅F)dV=∬S(n×F)dSStokes’ Theorem
Stokes’ Theorem is used to convert a line integral to a surface integral or vice-versa.
According to Stokes’ theorem, if F be a vector-field, then:
∬S(curlF⋅ˆn)dS=∫C(F⋅dr)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:
∇⋅(ψ∇ϕ)=ψ∇2ϕ+(∇ψ)⋅(∇ϕ)(1)From the Gauss’s divergence theorem,
∫V(∇⋅F)dV=∫SF⋅da(3)From (2) and (3), we get:
∫Sϕ(∇ψ)⋅da=∫V[ϕ∇2ψ+(∇ϕ)⋅(∇ψ)]dVSubtracting (2) from (1), we get Green’s First Identity:
∇⋅(ϕ∇ψ−ψ∇ϕ)=ϕ∇2ψ−ψ∇2ϕ(4)From (3) and (4), we get Green’s Second Identity:
∫V(ϕ∇2ψ−ψ∇2ϕ)dV=∫5(ϕ∇ψ−ψ∇ϕ)⋅daLet u have continuous first partial derivatives and be harmonic inside the region of integration.
Then, according to Green’s Third Identity:
u(x,y)=12π∮C[ln(1r)∂u∂n−u∂∂nln(1r)]dsPYQs
Gauss’s Divergence Theorem
1) Let →F=xy2ˆi+(y+x)ˆj. Integrate (∇×→F)⋅→k over the region in the first quadrant bounded by the curve y=x2 and y=x using Green’s theorem.
[2018, 12M]
2) Using Green theorem, evaluate the ∫CF(→r).d→r counterclockwise where F(r)=(x2+y2)ˆi+(x2−y2)ˆj and d→r=xˆi+dyˆj and the curve C is the boundary of the region R={(x,y)|1≤y≤2−x2}.
[2017, 8M]
3) Evaluate ∫Ce−x(sinydx+cosydy), where C is the rectangle with vertices (0,0), (π,0), (π,π2), (0,π2).
[2015, 12M]
4) Verify Green’s theorem in the plane for ∮C[(xy+y2)dx+x2dy] where C is the closed curve of the region bounded by y=x and y=x2.
[2012, 20M]
5) Verify Green’s theorem for e−xsinydx+e−xcosy by the path of integration being the boundary of the square whose vertices are (0,0), (π2,0), (π2,π2) and (0,π2).
[2010, 20M]
Gauss’s Divergence Theorem
1) State Gauss divergence theorem. Verify this theorem for →F=4xˆi−2y2ˆj+z2ˆk, taken over the region bounded by x2+y2=4, z=0 and z=3.
[2019, 15M]
2) If S is the surface of the sphere x2+y2+z2=a2, then evaluate
∬S[(x+y)dydz+(y+z)dzdx+(x+y)dxdy]using Gauss’s divergence theorem.
[2018, 12M]
3) Evaluate the integral ∬SFnds, where ¯F=3xy2ˆi+(yx2−y3)j+3zx2K and S is a surface of the cylinder y2+z2≤4,−3≤x≤3 using divergence theorem.
[2017, 9M]
4) By using Divergence Theorem of Gauss, evaluate the surface integral ∬(a2x2+b2y2+c2z2)−12dS, where S is the surface e of the ellipsoid ax2+by2+cz2=1, a, b and c being all positive constants.
[2013, 15M]
5) Verify Gauss’ Divergence Theorem for the vector →v=x2ˆi+y2ˆj+z2ˆk taken over the cube 0≤x,y,z≤1.
[2011, 15M]
6) Use the divergence theorem to evaluate ∬s→VndA, where →V=x2z→iy→j−xz2→k and S is he boundary of the region bounded by the paraboloid z=x2+y2 and the plane z=4y.
[2010, 20M]
7) Using divergence theorem, evaluate ∬s¯A⋅d¯S where ¯A=x3ˆi+y3ˆj+z3ˆk and S is the surface of the sphere x2+y2+z2=a2.
[2009, 20M]
8) Evaluate
∬S(x3dydz+x2ydzdx+x2zdxdy)by Gauss divergence theorem, where S is the surface of the cylinder x2+y2=a2 bounded by z=0 and z=b.
[2005, 15M]
9) Verify Gauss’ divergence theorem of A=(4x,−2y2,z2) taken over the region bounded by x2+y2=4,z=0 and z=3.
[2001, 15M]
Stokes’ Theorem
1) Evalute by Stokes’ theorem ∮Cexdx+2ydy−dz, where C is the curve x2+y2=4, z=2.
[2019, 5M]
2) Evaluate the line integral ∫C−y3dx+x3dy+z3dz using Stokes’ theorem. Here C is the intersection of the cylinder x2+y2=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:
∫(ydx+zdy+xdz)where is the curve given by x2+y2+z2−2ax−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 ∫C(−y3dx+x3dy−z3dz), where C is the intersection of the cylinder x2+y2=1 and the plane x+y+z=1.
[2013, 15M]
5) If →F=y→i+(x−2xz)→j−xy→k, evaluate ∬S(→∇×→F).→nd→s, where S is the surface of the sphere x2+y2+z2=a2 above the xy−plane.
[2012, 20M]
6) If →u=4yˆi+xˆj+2zˆk, calculate the double integral ∬(∇×→u)d→s over the hemisphere given by x2+y2+z2=a2, z≥0.
[2011, 15M]
7) Find the work done in moving the particle once round the ellipse x225+y216=1, z=0 under the field of force of given by ¯F=(2x−y+z)ˆi+(x+y−z2)ˆj+(3x−2y+4z)ˆk.
[2009, 20M]
8) Find the value of ∬s(→∇×→f).ds taken over the upper portion of the surface x2+y2−2ax+az=0 and the bounding curve lies in the plane z=0, when →F=(y2+z2−x2)ˆi+(z2+x2−y2)ˆj+(x2+y2−z2)ˆk.
[2009, 20M]
9) Determine ∫c(ydx+zdy+xdz) by using Stoke’s theorem, where C is the curve defined by (x−a)2+(y−a)2+z2=2a2, x+y=2a that starts from the point (2a,0,0) goes at first below the z−plane.
[2007, 15M]
10) Verify Stokes’ theorem for the function ¯F=x2ˆi−xyˆj 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 ˆf=(2x−y)ˆi−yz2ˆjzˆk where S is the upper half surface of the sphere x2+y2+z2=1 and C is its boundary.
[2004, 15M]
12) Evaluate ∬ScurlA.dS, where S is the open surface
x2+y2−4x+4z=0,z≥0and
A=(y2+z2−x2)ˆi + (2z2+x2−y2)ˆj + (x2+y2−3z2)ˆk
[2003, 15M]