Exam 17: Integrals and Vector Fields

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Solve the problem. -The radial flow field of an incompressible fluid is shown below. Which of the closed paths would exhibit a non-zero flux? Solve the problem. -The radial flow field of an incompressible fluid is shown below. Which of the closed paths would exhibit a non-zero flux?

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Using Green's Theorem, calculate the area of the indicated region. -The circle r(t)=(11cost)i+(11sint)j,0t2π\mathbf { r } ( \mathrm { t } ) = ( 11 \cos \mathrm { t } ) \mathbf { i } + ( 11 \sin \mathrm { t } ) \mathbf { j } , 0 \leq \mathrm { t } \leq 2 \pi

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Find the equation for the plane tangent to the parametrized surface S at the point P. - S is the cone r(r,θ)=rcosθi+rsinθj+4rk;P=(52,52,40)\mathrm { S } \text { is the cone } \mathbf { r } ( \mathrm { r } , \theta ) = \mathrm { r } \cos \theta \mathbf { i } + \mathrm { r } \sin \theta \mathbf { j } + 4 \mathrm { rk } ; \mathrm { P } = ( - 5 \sqrt { 2 } , 5 \sqrt { 2 } , 40 ) \text {. }

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Evaluate the surface integral of the function g over the surface S. - G(x,y,z)=x2y2z2;SG ( x , y , z ) = x ^ { 2 } y ^ { 2 } z ^ { 2 } ; S is the surface of the rectangular prism formed from the coordinate planes and the planes xx =1,y=1= 1 , y = 1 , and z=3z = 3

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Test the vector field F to determine if it is conservative. -F = xyi + yj + zk

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Find the flux of the vector field F across the surface S in the indicated direction. - F=x5yizk\mathbf { F } = \mathrm { x } ^ { 5 } \mathrm { yi } - \mathrm { zk } ; S\mathrm { S } is portion of the cone z=4x2+y2\mathrm { z } = 4 \sqrt { \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } } between z=0\mathrm { z } = 0 and z=3\mathrm { z } = 3 ; direction is outward

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Solve the problem. -The velocity field FF of a fluid is the spin field F=kyx2+y2i+kxx2+y2jF = - \frac { k y } { \sqrt { x ^ { 2 } + y ^ { 2 } } } \mathrm { i } + \frac { k x } { \sqrt { x ^ { 2 } + y ^ { 2 } } } \mathrm { j } . Following the smooth curve y=f(x)y = f ( x ) from (a,f(a))( a , f ( a ) ) to (b,f(b))( b , f ( b ) ) , show that the flux across the curve is CFnds=k[(a2+(f(a))2)1/2(b2+(f(b))2)1/2]\int _ { C } F \cdot n d s = k \left[ \left( a ^ { 2 } + ( f ( a ) ) ^ { 2 } \right) ^ { 1 / 2 } - \left( b ^ { 2 } + ( f ( b ) ) ^ { 2 } \right) ^ { 1 / 2 } \right]

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Find the limit. - r(t)=32ti+(2t)k;0t2r ( t ) = \frac { 3 } { 2 } t i + ( 2 - t ) k ; 0 \leq t \leq 2

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Evaluate. The differential is exact. - (0,0,0)(π/4,π/20,π/16)4sin4xdx+5sec25ydy8cos8zdz\int _ { ( 0,0,0 ) } ^ { ( \pi / 4 , \pi / 20 , \pi / 16 ) } 4 \sin 4 x d x + 5 \sec ^ { 2 } 5 y d y - 8 \cos 8 z d z

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Find the work done by F over the curve in the direction of increasing t. - F=zi+3xj+6yk;C:r(t)=ti+tj+tk,0t1\mathbf { F } = \mathrm { zi } + 3 \mathrm { x } \mathbf { j } + 6 y \mathbf { k } ; \mathrm { C } : \mathbf { r } ( \mathrm { t } ) = \mathrm { ti } + \mathrm { tj } + \mathrm { t } \mathbf { k } , 0 \leq \mathrm { t } \leq 1

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Evaluate. The differential is exact. - (1,1,1)(5,3,4)1xdx+1ydy+1zdz\int _ { ( 1,1,1 ) } ^ { ( 5,3,4 ) } \frac { 1 } { x } d x + \frac { 1 } { y } d y + \frac { 1 } { z } d z

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Using the Divergence Theorem, find the outward flux of F across the boundary of the region D. - F=sinyi+xzj+5zkF = \sin y i + x z j + 5 z k ; D: the thick sphere 4x2+y2+z294 \leq x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \leq 9

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Find the flux of the curl of field F through the shell S. - F=(zy)i+(xz)j+(zx)k;S:r(r,θ)=rcosθi+rsinθj+(9r2)k,0r3 and 0θ2π\mathbf { F } = ( \mathrm { z } - \mathrm { y } ) \mathbf { i } + ( \mathrm { x } - \mathrm { z } ) \mathbf { j } + ( \mathrm { z } - \mathrm { x } ) \mathbf { k } ; \mathrm { S } : \mathrm { r } ( \mathrm { r } , \theta ) = \mathrm { r } \cos \theta \mathbf { i } + \mathrm { r } \sin \theta \mathbf { j } + \left( 9 - \mathrm { r } ^ { 2 } \right) \mathbf { k } , 0 \leq \mathrm { r } \leq 3 \text { and } 0 \leq \theta \leq 2 \pi

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Find the flux of the curl of field F through the shell S. - F=x2i+6xj+9k;S\mathbf { F } = x ^ { 2 } \mathbf { i } + 6 x \mathbf { j } + 9 \mathbf { k } ; \mathrm { S } is the upper hemisphere of x2+y2+z2=16\mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } + \mathrm { z } ^ { 2 } = 16

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Using Green's Theorem, find the outward flux of F across the closed curve C. -F = (x - y)i + (x + y)j; C is the triangle with vertices at (0, 0), (9, 0), and (0, 9)

(Multiple Choice)
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Find the potential function f for the field F. - F=xz2x2+y2i+yz2x2+y2j2x2+y2z3kF = \frac { x } { z ^ { 2 } \sqrt { x ^ { 2 } + y ^ { 2 } } } \mathbf { i } + \frac { y } { z ^ { 2 } \sqrt { x ^ { 2 } + y ^ { 2 } } } j - \frac { 2 \sqrt { x ^ { 2 } + y ^ { 2 } } } { z ^ { 3 } } \mathbf { k }

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Evaluate. The differential is exact. - (0,0,0)(3,7,9)(3x+1)e3xdx+zdy+ydz\int _ { ( 0,0,0 ) } ^ { ( 3,7,9 ) } ( 3 x + 1 ) e ^ { 3 x } d x + z d y + y d z

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Evaluate the surface integral of the function g over the surface S. -G(x, y, z) = x + z; S is the surface of the wedge formed from the coordinate planes and the planes x + z = 3 and y = 5

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Find the potential function f for the field F. - F=(x(x2+y2+z2)3/2)i(y(x2+y2+z2)3/2)j(z(x2+y2+z2)3/2)k\mathbf { F } = - \left( \frac { \mathrm { x } } { \left( \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } + \mathrm { z } ^ { 2 } \right) ^ { 3 / 2 } } \right) \mathbf { i } - \left( \frac { \mathrm { y } } { \left( \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } + \mathrm { z } ^ { 2 } \right) ^ { 3 / 2 } } \right) \mathbf { j } - \left( \frac { \mathrm { z } } { \left( \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } + \mathrm { z } ^ { 2 } \right) ^ { 3 / 2 } } \right) \mathbf { k }

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Find the surface area of the surface S. -S is the area cut from the plane z=3yz = 3 y by the cylinder x2+y2=36x ^ { 2 } + y ^ { 2 } = 36 .

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