#$&* course Mth277 Question: Use Green's Theorem to evaluate Int[ y^3 dx - x^3 dy, C] where C is parameterized by x = cos(theta), y = sin(theta) and 0 <= theta <= 2pi. Check your answer by computing the line integral without Green's Theorem.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
.............................................
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): ????When applying Green’s theorem I used polar coord., which gave me 3`pi/2, instead of the -3`pi/2. Is there an obvious reason for this???? ********************************************* 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: Green’s Theorem states Int[2x arctan(y) dx - x^2*y^2 (1 + y^2) dy, C] = Int(Int((dN/dx - dM/dy) dA), where N = x^2*y^2 (1 + y^2) and M =2x arctan(y) dN/dx = (-2x)(y^2)(1+y^2), dM/dy = (2x)(1/(y^2 + 1)), plugging into our equation Int(Int(([(-2x)(y^2)(1+y^2)] - [(2x)(1/(y^2 + 1))] dA) We paratermize with 0<=x<=3 and 0<=y<=(3-x) So we have Int(Int(([(-2x)(y^2)(1+y^2)] - [(2x)(1/(y^2 + 1))] dy,0,(3-x)) dx,0,3) = Int(Int([(-2x)(y^2)(1+y^2)] dy,0,(3-x)) dx,0,3) - Int(Int([(2x)(1/(y^2 + 1))] dy,0,(3-x)) dx,0,3) Int(Int([(-2x)(y^2)(1+y^2)] dy,0,(3-x)) dx,0,3) Int([(-2x)(y^2)(1+y^2)] dy) =-2x(Int(y^2 dy) + Int(y^4 dy) =-2x(y^3/3 + y^5/5), evaluated -2x((3-x)^3/3 + (3-x)^5/5) Int(-2x((3-x)^3/3 + (3-x)^5/5) dx) = Int(-2x((3-x)^3/3) dx) + Int(-2x((3-x)^5/5) dx) Int(-2x((3-x)^3/3) dx) = to save time and space will just provide solution, evaluated -81/10 - 0 = -8.1 Int(-2x((3-x)^5/5) dx) = to save time and space will just provide solution, evaluated -729/35 - 0 = -approx -20.8 So that Int(Int([(-2x)(y^2)(1+y^2)] dy,0,(3-x)) dx,0,3)= approx. -29 Now for second part of integral Int(Int([(2x)(1/(y^2 + 1))] dy,0,(3-x)) dx,0,3) Int((2x)(1/(y^2 + 1)) dy) = 2x*Int(1/(y^2 + 1) dy) = 2x*tan^-1(y), evaluated 2x*tan^-1(3-x) - 0 Int(2x*tan^-1(3-x) dx) using integration by parts u= tan^-1(3-x), u’ = 1/(x^2 - 6x +10),v’ = 2x, v = x^2 Int(2x*tan^-1(3-x) dx) = tan^-1(3-x)*x^2 - Int( x^2/(x^2-6x +10) dx), using partial fractions we get Int( x^2/(x^2-6x +10) dx) = 3*ln|x^2 - 6x +10|+ 8*tan^-1(3-x) + x, so that Int(2x*tan^-1(3-x) dx) = tan^-1(3-x)*x^2 - [3*ln|x^2 - 6x +10|+ 8*tan^-1(3-x) + x], evaluated we get 3 - approx. 16.9 = -13.9 So finally we have Int(Int(([(-2x)(y^2)(1+y^2)] - [(2x)(1/(y^2 + 1))] dy,0,(3-x)) dx,0,3) = approx. -29 - (-13.9) = -15.1 or = approx. -15 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I’m not sure where I went wrong, there is so much work that goes into calculating these integrals its hard for me to see where my error was but clearly there was an error my solution was (1/35) the solution of yours. ********************************************* 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: Force =
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Rushed through and only accounted for right side of circle but apart from that I believe solution should have been close wrt logic. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course Mth277 Question: Use Green's Theorem to evaluate Int[ y^3 dx - x^3 dy, C] where C is parameterized by x = cos(theta), y = sin(theta) and 0 <= theta <= 2pi. Check your answer by computing the line integral without Green's Theorem.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
.............................................
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): ????When applying Green’s theorem I used polar coord., which gave me 3`pi/2, instead of the -3`pi/2. Is there an obvious reason for this????
.............................................
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.
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Rushed through and only accounted for right side of circle but apart from that I believe solution should have been close wrt logic.