F = (0.x^2 + y^2); R = {(x.y):x^2+ y^2 lessthanorequalto 1} Area of regions Use
ID: 2862412 • Letter: F
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F = (0.x^2 + y^2); R = {(x.y):x^2+ y^2 lessthanorequalto 1} Area of regions Use a line integrl on the boundary to find the area of the following regions. A disk of radius 5 A region bounded by an ellipse with major and mnor axes of length 12 and H. respectively.{ (x, y):x^2 + y^2 lessthanoreuqalto 16} The region show n in the figure The region hounded by the parabolas r(t) = ( t, 2r^2) and r(t) = (r, 12 minus r^2), for minus2 lessthanorequalto r lessthanorequalto 2 The region hounded by the curve r(t) = (r(I minus r^2), 1 minus r^2), for minus1 lessthanorequalto t lessthanorequalto 1 Green's Theorem, flux form Consider the following regions R and vector fields F Compute the two-dimensional dnergcnce of the sector field. Evaluate hath integrals in Green's Theorem and check for consistency. State whether the vector field is source free. F = ; R = {(x, y):x^2 + y^2 lessthanorequalto 4} F = (y,minus x): R is the square with vertices (0,0). (1,0). (1, 1), and (0,1). F = (y, minus3x); R is the region hounded by y = 4 minus x^2 and v = 0. F = (minus3y, 3x); R is lire triangle with vertices (0.0). (3.0). and (0.1). F = {2xy, x^2 minus y^2 ); R is the region hounded by y = x(2 minus x) and y = 0. F = (x^2 + y^2, 0); R = {(x, y): x^2 + y^2 lessthanorequalto 1} Line Integrals Use Green 's Theorem to evaluate the following line integrals. Unless staled otherwise, assume all cun es are oriented terclockwise. Integral (2x + x^y^2)dy minus (4y^2 + e^x^2)dx. Where c is the boundary of the square with vertices (0,0) (1,0) (1,1) and (0,1) The circulation line integral of F = (a^2 + y^2,4x + y^2), where C is the boundary of {(x, y):0 lessthanorequalto y lessthanorequalto sinx, 0 lessthanorequalto x lessthanorequalto pi} The flux line integral of F = (e^x mmminus y,e^y minu x). where C is the boundary of {(x, y):0 lessthanorequalto y lessthanorequalto x,0 lessthanorequalto x lessthanorequalto 1} General regions For the following sector fields, compute the circulation on and the outward flux across the boundary of the given region. Assume boundars curves are oriented counterclockwise. F = (x.y): R is the half-annulus {(r,theta): 1 lessthanorequalto r lessthanorequalto 2,0 lessthanorequalto theta lessthanorequalto pi}. F = (minusy, x); R is t lie annulus {(r,theta): 1 lessthanorequalto r lessthanorequalto 3,0 lessthanorequalto theta lessthanorequalto 2pi}. F = (2x + y,x minus 4y); R is die quarter annulus {(r,theta): 1 lessthanorequalto r lessthanorequalto 4,0 lessthanorequalto theta lessthanorequalto pi/2}. F = (x minus y, minusx + 2y); R is the parallelogram {(x, y): 1 minus x lessthanorequalto y lessthanorequalto 3 minus x,0 lessthanorequalto x lessthanorequalto 1}. Further Explanations Explain why or why mil Determine whether tlic following statements;ire tnic and give an explanation or counterexample The work required to move an object around a closed curve C in the presence of a vector force field is the circulation of the force field on the curve, If a vector field has zero divergence throughout a region (on which the conditions of (irecn's Theorem are met), then the circulation on the boundary of that region is zero. If the two-dimensional curl of a vector field is positive throughout a region (on which the conditions of Green's Theorem are met), then the circulation on the boundary of that region is positive (assuming counterclockwise orientation). Circulation and flux For the following sector fields, compute (ai the circulation on and tb) the outward flux across the boundary of the given region. Assume boundary curves have counterclockwise orientation. F = ,where R is the annamulus {(r,theta); 1 lessthanorequalto r lessthanorequalto 2, 0 lessthanorequalto theta lessthanorequalto 2pi} F = V(squarerootx^2 + y^2). where R is the half annulus {(r,theta): 1 lessthanorequalto r lessthanorequalto 3,0 lessthanorequalto theta lessthanorequalto pi} F = (y cosx. minussin x). where R is the square {(x, y): 0 lessthanorequalto x lessthanorequalto pi/2, 0 lessthanorequalto y lessthanorequalto pi/2) F = (x + y^2, x^2 minus y), where R = (x, y):3y^2 lessthanorequalto x lessthanorequalto 36 minus y2}Explanation / Answer
ans -33
By Green's Theorem, this equals
[(/x)(4x + y^3) - (/y)(x^2 + y^2)] dA
= (x = 0 to ) (y = 0 to sin x) (4 – 2y) dy dx
= (x = 0 to ) (4 y - y^2) {for y = 0 to sin x} dx
= (x = 0 to ) (4 sin x - sin^2(x)) dx
= (x = 0 to ) [-4cos x - (1 - cos(2x))] dx
= (x = 0 to ) [-4cos x - x + sin(2x)/2 ] dx
= [ ( -4cos - + sin(2 )/2 ) - ( -4cos 0 - 0 + sin(20)/2) ] {for x = 0 to }
= (-4 * -1) – + 0+ 4 - 0+ 0
= 4+4 - = 8- ------------ans.
= 8- ------------ans.
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