Attempt only one of the following.(a) Evaluate the line integral∫C(y2 − 3z + 2) ds, where C is the line segment from (1, 0, 4) to (2, −1, 1).(b) Let C be the upper half of the circle of radius π centered at the origin with counter-clockwiserotation. Also let F be the vector field given by F(x, y) = (sin x sin y, 5 − cos x cos y). Evaluate the lineintegral∫CF · dr. (Hint: It will be of help to determine if F is conservative first.)6. Evaluate one of the following closed line integrals using Green’s Theorem. The curves have positive(counter-clockwise) orientation.(a)∮C(y4 − 2y)dx + (4xy3 − 6x)dy, where C is the rectangle with corners (0, 0), (6, 0), (6, 4) and (0, 4)(b)∮C(6 + x2)dx − 2xydy, where C is the triangle with corners (0, 0), (3, 0) and (0, 6).Remarks: The duration of the exam is 90 minutes. Attempt only 5 questions. In order to get fullmarks, please make explanations whenever necessary.
Question
Attempt only one of the following.(a) Evaluate the line integral∫C(y2 − 3z + 2) ds, where C is the line segment from (1, 0, 4) to (2, −1, 1).(b) Let C be the upper half of the circle of radius π centered at the origin with counter-clockwiserotation. Also let F be the vector field given by F(x, y) = (sin x sin y, 5 − cos x cos y). Evaluate the lineintegral∫CF · dr. (Hint: It will be of help to determine if F is conservative first.)6. Evaluate one of the following closed line integrals using Green’s Theorem. The curves have positive(counter-clockwise) orientation.(a)∮C(y4 − 2y)dx + (4xy3 − 6x)dy, where C is the rectangle with corners (0, 0), (6, 0), (6, 4) and (0, 4)(b)∮C(6 + x2)dx − 2xydy, where C is the triangle with corners (0, 0), (3, 0) and (0, 6).Remarks: The duration of the exam is 90 minutes. Attempt only 5 questions. In order to get fullmarks, please make explanations whenever necessary.
Solution
Sure, let's solve the first part of the first question:
(a) Evaluate the line integral ∫C(y2 − 3z + 2) ds, where C is the line segment from (1, 0, 4) to (2, −1, 1).
Step 1: Parameterize the line segment C. We can do this by creating a vector function r(t) that traces the line segment from (1, 0, 4) to (2, -1, 1) as t ranges from 0 to 1.
r(t) = (1-t)(1, 0, 4) + t(2, -1, 1) = (1 + t, -t, 4 - 3t)
Step 2: Compute the derivative of r(t) with respect to t.
r'(t) = (1, -1, -3)
Step 3: Substitute the parameterization into the integrand.
y^2 - 3z + 2 = (-t)^2 - 3(4 - 3t) + 2 = t^2 - 12 + 9t + 2 = t^2 + 9t - 10
Step 4: Compute the magnitude of r'(t).
||r'(t)|| = sqrt((1)^2 + (-1)^2 + (-3)^2) = sqrt(1 + 1 + 9) = sqrt(11)
Step 5: Evaluate the line integral.
∫C(y2 − 3z + 2) ds = ∫ from 0 to 1 of (t^2 + 9t - 10) * sqrt(11) dt = sqrt(11) * [t^3/3 + 9t^2/2 - 10t] from 0 to 1 = sqrt(11) * (1/3 + 9/2 - 10) = sqrt(11) * (-19/6)
So, the value of the line integral ∫C(y2 − 3z + 2) ds is -sqrt(11) * 19/6.
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