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Given Solution: This function is of the form M dx + N dy, with M = y^3 and N = x^3. This function is integrated over a positively oriented closed curve, so by Green's Theorem the given line integral is equal to the integral of N_x - M_y over the region enclosed by the curve. The curve is the unit circle, traversed counterclockwise. The region can be described by -1 <= x <= 1, -sqrt(1 - x^2) <= y <= sqrt(1 - x^2), so our integral is integral ( integral( N_x - M_y) dy, -sqrt(1 - x^2), sqrt(1 - x^2)), dx, -1, 1) = integral ( integral( -3 x^2 - 3 y^2 ) dy, -sqrt(1 - x^2), sqrt(1 - x^2)), dx, -1, 1) An antiderivative of -3 x^2 - 3 y^2 with respect to y is -3 x^2 y - y^3, which between the given limits changes from 3 x^2 sqrt(1 - x^2) + (1 - x^2)^(3/2) to -3 x^2 sqrt(1 - x^2) - (1 - x^2)^(3/2) , a change of -6 x^2 sqrt(1 - x^2) - 2 (1 - x^2)^(3/2). We are thus left with integral ( -6 x^2 sqrt(1 - x^2) - 2 (1 - x^2)^(3/2), x, -1, 1). Evaluating this integral we get -3 pi / 2. The line integral around the curve is easily written down. We have x = cos(t), y = sin(t), dx = - sin(t) dt, dy = cos(t) dt so that Int[ y^3 dx - x^3 dy, C] = int( sin^3(t) * (-sin(t) dt) - cos^3(t) * cos(t) dt , t from 0 to 2 pi) = int(-sin^4(t) - cos^4(t) dt, 0, 2 pi). This integral comes out to -3 pi / 2, in agreement with the area integral. Notes on details of integrations: x^2 sqrt(1 - x^2) contains the expression x sqrt( 1 - x^2), which can be evaluated with substitution x^2 sqrt(1 - x^2) can then be written as x ( x sqrt(1 - x^2)), and integrated by parts sin^4(t) can be written sin^2(t) ( 1 - cos^2(t)) = sin^2(t) - sin^2(t) cos^2(t). sin^2(t) can be integrated by parts, letting u = sin(t) and dv = sin(t) dt. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: Use Green's Theorem to evaluate the integral Int[2x arctan(y) dx - x^2*y^2 (1 + y^2) dy, C] where C is the square with vertices (0,0), (3,0), (3,3), (0,3) traversed clockwise. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int[2x arctan(y) dx - x^2*y^2 (1 + y^2) dy, C] M=2x arctan(y) N=- x^2*y^2 (1 + y^2) Int[ Int[ N_x - M_y]dy 0, 3, 3, 0] dx 0, 0, 3, 3 Int[ Int[ 2x*y^2(1+y^2) - 1/(1+y^2)] dy (0,0), (3,0),(3,3), (0,3) Int[ (2xy^3) / 3 + (2xy^5)/5 - ln(y^2 + y + 1) Int[ 54x/3 + 566x/5 - ln(13)dx 9x^2 + (283/5)x^2 - ln(13)x 81+(2547/5) - 9ln(13) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This function is of the form M dx + N dy, with M = 2x arcTan(y) and N = -x^2 y^2 ( 1 + y^2). This function is integrated over a positively oriented closed curve, so by Green's Theorem the given line integral is equal to the integral of N_x - M_y over the region enclosed by the curve. N_x = 2 x y^2 ( 1 + y^2) and M_y = 2 x / (1 + y^2), so N_x - M_y = -2 x y^2 ( 1 + y^2) - 2 x / (1 + y^2). The region is easily described, so the integrand of our area integral will be -2 x y^2 ( 1 + y^2) - 2 x / (1 + y^2). We can choose to integrate first with respect to x or y. Integrating first with respect to x we get int(int( -2 x y^2 ( 1 + y^2) - 2 x / (1 + y^2) dx, 0, 3) dy, 0, 3) = - 2 int (int x dx,0,3) (y^2 ( 1 + y^2) + 1 / (1 + y^2)) dy, 0, 3) = -2 int ( 9/2 ( y^2 + y^4 + 1 / (1 + y^2) ) ) dy, 0, 3) = - 9 int ( (y^2 + y^4 + 1 / ( 1 + y^2) dy, 0, 3) An antiderivative is y^3 / 3 + y^5 / 5 + arcTan(y). This antiderivative changes from 0 at y = 0 to 288 / 5 + arcTan(3) at y = 3, a change of 288 / 5 + arcTan(3). Our integral is therefore -9 ( 288/5 + arcTan(3) ) = -9 arcTan(3) - 2592 / 5. The line integral around the square can be broken into four integrals. The four paths can be parameterized as x = t, y = 0 x = 3, y = t x = (3 - t), y = 0 x = 0, y = 3 - t all for 0 <= t <= 4. On the first and third paths y is constant so the dy term will be zero. On the second and fourth paths x is constant so the dx term will be zero. On the fourth path both terms have x as a factor. Since x = 0 on this path, the integrand is therefore zero. On the third path x = 3 - t so dx = - dt; on the fourth y = 3 - t so dy = - dt. Our integral around the boundary of the region is therefore int( 2 t arcTan(0) dt, 0, 3) + int ( -(3^2 t^2 ( 1 + t^2) dt, 0, 3) + int( 2 ( 3 - t) arcTan(3) * (-dt), 0, 3) + int ( 0 (- dt), 0, 3). arcTan(0) = 0 so the first integral is just zero. This leaves us with - int( 9 t^2 ( 1 + t^2 ) dt, 0, 3) - 2 arcTan(3) int( (3 - t) dt, 0, 3). Both integrands are polynomials. You should easily be able to confirm that the result is -2592 / 5 - 9 arcTan(3)., in agreement with the previous area integral. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: Find the work done when an object moves in the force field F(x,y) = 2y^2i + 3x^2j counterclockwise around the circular path x^2 + y^2 = 4. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(x,y) = 2y^2i + 3x^2j x^2 + y^2 = 4 M=2y^2 N=2x^2 Nx-My=6x-4y Int((6x - 4y)dy)=6xy-2y^2 6x*`sqrt(4-x^2)-2(4-x^2) Int(6x*`sqrt(4-x^2)-2(4-x^2)dx) Int(6x*`sqrt(4-x^2)-2(4-x^2)dx) Int(6x*`sqrt(4-x^2) dx) - Int(8 dx) + Int(2x^2 dx) Int(6x*`sqrt(4-x^2) dx) x^2, u’ = 2x xd Int((1/3)*`sqrt(4-u) du) (1/3)((2/3)(4-u)^(3/2) (2/9)(4-x^2)^(3/2) Int(8 dx) 8x Int(2x^2 dx) (2/3)*x^3 Int(6x*`sqrt(4-x^2) - 2(4-x^2)dx) = [(2/9)(4-x^2)^(3/2)] - [8x] + [(2/3)*x^3] -16 + (15/3) - (16/9) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ The work would be the line integral int ( F dot ds ) around the curve. If ds = dx i + dy j the integral would be int(2 y^2 dx + 3 x^2 dy, over C), which is of the form int(M dx + N dy, over C), with M = 2 y^2 and N = 3 x^2. By Green's Theorem this is the integral of N_x - M_y = 6 x - 4 y the enclosed region. In terms of x and y the region is -2 <= x <= 2, -sqrt(4 - x^2) <= y <= sqrt(4 - x^2). The integral would therefore be int(int ( (6 x - 4 y) dy), -sqrt(4 - x^2), sqrt(4 - x^2))dx, -2, 2) The antiderivative for the inner integral is 6 x y - 2 y^2. The antiderivative changes by 12 x sqrt( 4 - x^2). The outer integral is therefore int( 12 x sqrt(4 - x^2) dx, -2, 2). An antiderivative is 4 * 2/3 (4 - x^2)^(3/2). The antiderivative has the same value at x = -2 and at x = 2, so the integral is zero. The path could be parameterized by x = 4 cos(t), y = 4 sin(t). The integral would be int( 3 * 16 cos^2(t) * (-sin t) dt + 2 * 16 sin^2(t) * cos(t) dt, 0, 2 pi). The antiderivative is a multiple of cos^3(t) added to a multiple of sin^3(t), and hence will not change between t = 0 and t = 2 pi. The resulting integral is therefore zero.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK"