Sample Midterm Test - II 1. Let F~ = grad f , where f (x, y) = ex y + ey x. (a) Find the formula for the vector field F~ . R (b) Calculate the line integral C F~ · d~r by direct calculation, where C is the line segment from P = (0, 0) to Q = (1, 1). (c) Calculate the same line integral using the Fundamental Theorem of Calculus for line integrals. Solution (a) The vector field F~ is given by ∂f ∂f F~ (x, y) = ~i + ~j = (ex y + ey )~i + (ex + ey x)~j. ∂x ∂y (b) The line segment from P to Q is parameterized by the equation ~r(t) = t~i + t~j, where t changes from 0 to 1. The velocity vector is ~r0 (t) = ~i + ~j. The vector field in terms of the parameter t is given by F~ (t) = (tet + et )~i + (et + tet )~j. Thus, we can calculate the line integral Z F~ · d~r = C Z 1 F~ (t) · ~r0 (t) dt 0 = Z 1 ((tet + et )~i + (et + tet )~j) · (~i + ~j) dt 0 = Z 1 2et (1 + t) dt. 0 We can calculate this integral by integration by parts: 2 Z 1 t (1 + t) de = 2 0 Z 1 (1 + t) det 0 t = 2(1 + t)e − 2 Z 1 et d(1 + t) 0 = 1 t 2(1 + t)e 0 1 t − 2e = 2e. 0 R (c) Let us calculate the integral C F~ · d~r using the Fundamental Theorem of Calculus. By the theorem, we have Z grad f · d~r = f (Q) − f (P ), C thus, to find the line integral we just have to calculate f (P ) and f (Q) and substitute these values into the formula. We have f (P ) = f (0, 0) = e0 · 0 + e0 · 0 = 0, f (Q) = e1 · 1 + e1 · 1 = 2e. It follows that Z (e y + e )~i + (ex + ey x)~j · d~r = x y C Z grad (ex y + ey x) · d~r C = f (Q) − f (P ) = 2e. 2. Decide whether the vector field could be a gradient field. Justify your answer. (a) F~ = 3y 2~i − 6xy~j; (b) F~ = x1~i + 2y~j. Solution (a) The curl of F~ = F1~i + F2~j is ∂F2 ∂F1 curl F~ = − = −6y − 6y = −12y ∂x ∂y and it is not equal to zero. Thus, F~ is not a gradient field. (b) The curl of F~ is equal to zero, ∂F2 ∂F1 − = 0 − 0 = 0. ∂x ∂y However, we cannot apply the curl test, because the domain of the vector field F~ = x1~i + 2y~j has holes in it. Indeed, the field is not defined for x = 0. Thus, we have to try to find a potential function f (x, y) of F~ such that grad f = F~ . We have the system of equations ∂f 1 = ; ∂x x ∂f = 2y. ∂y The first equation implies f (x, y) = log x + C(y), the second equation implies f (x, y) = y 2 + C(x), where C(x) and C(y) are constants that depend on x and y, respectively. Combining these two results, we obtain f (x, y) = log x + y 2 + C, where C is a constant. We found the potential function for F~ , thus, F~ is a gradient field. 3. Let F~ = ~i + x2~j. (a) Sketch the vector field F~ . The sketch must include at least four arrows in each quadrant and overall enough arrows to give a reasonable representation of the field. (b) Prove that this vector field is not path-independent by using of the curl test. R (c) Find and sketch a closed curve C such that C F~ · d~r 6= 0. Explain why this integral is not equal to zero. Solution (a) The sketch of F~ is shown in the figure 1 2 (b) curl F~ = ∂F − ∂F = 2x − 0 = 2x. The curl is not equal to zero, thus, ∂x ∂y the vector field is not path-independent. (c) If C is a closed curve joining the points (0, 0), (1, 0), (1, 1), and (0, 1) R oriented counterclockwise, then the line integral C F~ · d~r is positive. The line integral can be represented as the sum of line integrals along the sides of the square ABCD oriented as shown in the figure, Z F~ · d~r = C Z F~ · d~r + AB Z F~ · d~r + BC Z F~ · d~r + Z CD F~ · d~r. DA The line integral along DA is equal to zero, because F~ is perpendicular to the path DA. The line integrals along AB and CD cancel out, because the vector field along these paths has the same magnitude, but opposite orientation. The integral along the path BC is positive, because the angle between the vectors of F~ and the velocity vector of the path BC is less than π/2. Thus, the dot product of F~ and the velocity vector ~j of path BC is positive and the integral along BC is positive. You can also show that the integral is not equal to zero by direct calculation or by using of Green’s theorem. 4. Use Green’s theorem to calculate the circulation of F~ around the unit circle oriented counterclockwise, where F~ = (sin(x2 ) + 4y)~i + (2x + cos(y 2 ))~j. Solution By Green’s theorem, Z F~ · d~r = Z curl F~ dx dy, R C where R is the region inside the unit circle C. The curl of F~ is curl F = ∂ ∂ (2x + cos(y 2 )) − (sin(x2 ) + 4y) = 2 − 4 = −2. ∂x ∂y Thus, we have Z C F~ · d~r = Z (−2) dx dy = −2Area(R) = −2π, R because the area of R is equal to π · 12 = π. 5. Find the flux of F~ (x, y, z) = 145~i + 28y~j + ~k through the rectangle S in the plane y = −1 oriented in the direction of positive y-axis with vertices at (0, −1, 0), (0, −1, 1), (2, −1, 1), (2, −1, 0). Solution. The value of the vector field on the rectangle S is F~ (x, −1, z) = 145~i − 28~j + ~k. We just substitute y = −1 in the formula for the vector field, because the ~ = Area(S)~n. The vector field depends just on y. Next, the area vector A ~ = 2~j. normal vector is ~n = ~j. The area of the rectangle is equal to 2. Thus, A Finally, the flux is ~ = F~ (x, −1, z) · 2~j = (145~i − 28~j + ~k) · ~j = −28. F~ (x, −1, z) · A 6. ∗ Compute the flux of the vector field F~ = x~i + y~j + 3~k through the surface S, where S is the part of the surface z = 25 − x2 − y 2 above the disk of radius 5 centered at the origin, oriented upward. Solution. The graph of function z = f (x, y) = 25 − x2 − y 2 is parameterized by x = x, y = y, z = f (x, y), where (x, y) ∈ R, R is the disk of radius 5 centered at the origin in the xy-plane. Thus, the area vector is given by ~ = − ∂f ~i − ∂f ~j + ~k = 2x~i + 2y~j + ~k. dA ∂x ∂y Thus, we can write the flux integral, Z Z ∂f ∂f ~ ~ F~ (x, y, f (x, y)) · (− ~i − ~j + ~k) dx dy F · dA = ∂x ∂y S R Z (x~i + y~j + 3~k) · (2x~i + 2y~j + ~k) dx dy = = ZR (2x2 + 2y 2 + 3) dx dy R Converting the integral to polar coordinates, x = r cos θ, y = r sin θ, where 0 ≤ r ≤ 5, 0 ≤ θ ≤ π, we obtain Z (2x2 + 2y 2 + 3) dx dy = R Z 2π Z 5 0 (2r2 + 3)r dr dθ = 0 0 Thus, we have Z S 5 4 2 2π(r /2 + 3r /2) ~ = 700π. F~ · dA = 700π.