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Engineering Mathematics Test 5
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Engineering Mathematics Test 5
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  • Question 1/10
    1 / -0

    Let C be the positively oriented square with vertices (0, 0), (1, 0), (1, 1), (0, 1). The value of line integral C(4y2xdx+8x2ydy) is ______
    Solutions

    Concept:

    Green’s Theorem:

    If M(x, y), N(x, y), My and Nx be continuous in a region E of the xy-plane bounded by a closed curve C, then

    C(Mdx+Ndy)=E(NxMy)dxdy

    Calculation:

    The given function is in the form of

    (M dx + N dy)

    M = 4y2x, N = 8x2y

    My=8xy,Nx=16xy

    C(4y2xdx+8x2ydy)=S(NxMy)dxdy

    =S(16xy8xy)dxdy

    =S8xydxdy

    =01018xydxdy

    =01[4xy2]01dx

    =014xdx

    =[2x2]01=2

  • Question 2/10
    1 / -0

    If C is the path along the curve y = x2 – 4x + 4 from (0, 4) to (2, 0), then C(yi^3xj^)dr is
    Solutions

    y = x2 – 4x + 4

    dy = 2x dx – 4 dx = (2x - 4) dx

    dr=dxi^+dyj^ 

    C(yi^3xj^)(dxi^+dyj^) 

    =Cydx3xdy 

    =C(x24x+4)dx3x(2x4)dx 

    =C(5x2+8x+4)dx 

    x=02(5x2+8x+4)dx 

    =[5x33+4x2+4x]02 

    =53(8)+4(4)+4(2) 

    403+24=323

  • Question 3/10
    1 / -0

    The value of the integral

    sr.nds

    over the closed surface S bounding a volume V, where r=xi+yj+zk is the position vector and n̂ is normal to the surface S, is
    Solutions

    Concept:

    According to Gauss Divergence theorem:

    A.ds=(.A)dv

    F.n^ds=(.F)dv

    Calculation:

    Given that S is a closed surface:

    r.n^ds=(.r)dv

    r=xi^+yj^+zk^

    .r=x(x)+y(y)+z(z)=3

    r.n^ds=(.r)dv=3dv=3V

  • Question 4/10
    1 / -0

    The value of C(2x+3y)dx(3x4y)dy where c is the circle with radius as 1 and centre at origin.

    Solutions

    C ≡ x2 + y2 = 1

    Let x = r cos θ, y = r sin θ

    ⇒ dx = -sin θ dθ, dy = cos θ dθ

    θ=02π[(2cosθ+3sinθ)(sinθ)dθ(3cosθ4sinθ)(cosθ)dθ]=02π[2sinθcosθ3sin2θ3cos2θ+4sinθcosθ]dθ=02π(2sinθcosθ3)dθ=02π(sin2θ3)dθ 

    = -3(2π) = -6π

  • Question 5/10
    1 / -0

     Evaluate CFdr where F(x,y,z)=xi^+yj^+3(x2+y2)k^ and C is the boundary of the part of the paraboid where z2 = 64 – x2 – y2 which lies above the xy-plane and C is oriented counter clockwise when viewed from above.
    Solutions

    Concept:

    By stokes theorem,

    CF.dr=SCurlF.Nds

    Calculation:

    F(x,y,z)=xi^+yj^+3(x2+y2)k^

     

    ×F=|ijkxyzxy3(x2+y2)|

    = i (6y) – j(6x) = 6y î - 6x ĵ

    S: x2 + y2 + z2 – 64 = 0

    ds=+2xi^+2yj^+2zk^

    (×F)ds=(6yi^6xj^)(2xi^+2yj^+2zk^)

    = 12xy – 12xy + 0 = 0

    CFdr=0

  • Question 6/10
    1 / -0

    If S is the surface of the sphere x2 + y2 + z2 = a2, then the value of

    S(x+z)dydz+(y+z)dzdx+(x+y)dxdy is
    Solutions

    From Gauss divergence theorem,

    SF.ds=V.FdV

    F=(x+z)i+(y+z)j+(x+y)k

    .F=1+1+0=2

    V.FdV=2dV

    2dV

    The volume of the sphere =43πa3

    V.FdV=2×43πa3=83πa3

    SF.ds=83πa3
  • Question 7/10
    1 / -0

    If S be any closed surface, evaluate SCurlF.ds

    Solutions

    Explanation:

    Cut open the surface S by any plane and Let S1, S2 denotes its upper and lower portions.

    Let C be the common curve bounding both these portions.

    SCurlF.ds=S1CurlF.ds+S2CurlF.ds

    By stokes theorem,

    SCurlF.ds=CF.drSCurlF.ds=CF.drCF.dr=0

    The second integral is negative because it is traversed in a direction opposite to that of the first.

  • Question 8/10
    1 / -0

    The numerical value of work done (rounded off to the nearest integer) by a position-dependent force F=xi^+5xyj^ (where i^ and j^ are unit vectors) along the path, y=x22, from (0, 0) to (2, 2) in the xy plane is ________.

    Solutions

    F=xi^+5xyj^

    Work done is given by:

    W=CF.dr

    r=xi^+yj^

    W=CF.dr=C(xi^+5xyj^).d(xi^+yj^)=C(xdx+5xydy)

    Substituting y=x22 and x is varying from 0 to 2

    dy = x dx

    CF.dr=C(xdx+5x(x22)xdx)

    =02(x+52x4)dx

    =[x22]02+[x52]02

    =42+322=18
  • Question 9/10
    1 / -0

    For vectors F=3xyi^y2j^ and R=xi^+yj^, the value of CF.dR on the curve C (y = 2x2) in the x-y plane from (0, 0) to (1, 2) is
    Solutions

    F=3xyi^y2j^ and R=xi^+yj^

    CF.dR=C(3xyi^y2j^).d(xi^+yj^)=C(3xydxy2dy)

    Substituting y = 2x2 and x is varying from 0 to 1

    ⇒ dy = 4x dx

    CF.dR=C(3x(2x2)dx(2x2)2(4xdx)

    =01(6x316x5)dx

    =[64x4]01[166x6]01

    =6486=183212=76=1.17

    Alternate method:

    F=3xyi^y2j^ and R=xi^+yj^

    CF.dR=C(3xyi^y2j^).d(xi^+yj^)=C(3xydxy2dy)

    Let x = t

    ⇒ dx = dt

    y = 2x2 = 2t2

    ⇒ dy = 4tdt

    t varies from 0 to 1.

    CF.dR=C(3(t)(2t2)dt(2t2)2(4tdt))

    =01(6t316t5)dt

    =[64t4]01[166t6]01

    =6486=183212=76=1.17
  • Question 10/10
    1 / -0

    The value of the line integral 2πγ(y3dx+x3dy), where γ is the circle x2 + y2 = 1 oriented counter clockwise, is ________.

    Solutions

    Concept:

    Green’s Theorem:

    If M(x, y), N(x, y), My and Nx be continuous in a region E of the xy-plane bounded by a closed curve C, then

    C(Mdx+Ndy)=E(NxMy)dxdy

    Calculation:

    2πγ(y3dx+x3dy)

    It is in the form of C(Mdx+Ndy)

    M = -y3 and N = x3

    Nx=3x2

    My=3y2

    2πγ(y3dx+x3dy)=2πE(3x2+3y2)dxdy

    Changing to polar coordinates (r, θ), r varies from 0 to 1 and θ varies from 0 to 2π.

    =2π0102π(3(rcosθ)2+3(rsinθ)2)rdθdr

    =2π0102π3r3dθdr

    =2π01[3r3θ]02πdr

    =6π012πr3dr

    =6π[2πr44]01=3
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