(c) The first thing to do for this problem is to create a parametric curve for C. One choice would be. (cos(t), sin(t)) with 0 t 2π

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1 1. Let g(x, y) = (y, x) ompute gds for a circle with radius 1 centered at the origin using the line integral. (Hint: use polar coordinates for your parametrization). (a) Write out f((t)) so that f is a function of only t? (b) ompute (t) (c) Evaluate the line integral Solution: (a) ompute (t) (b) Evaluate the line integral (c) The first thing to do for this problem is to create a parametric curve for. One choice would be (cos(t), sin(t)) with t 2π (As a side remark. Whenever anyone asks you to find a parametric equation for anything involving a circle, this should be the first equation to try. Also note that a parametric equation is not really complete without bounds on t) With this as our parametric equation we know (d) Knowing (t) is is straightforward to compute (t) (e) Finally, we can setup the integral. c f(x, y)ds = (t) = ( sin t, cos t) 2π And computing this integral we end up with c (sin t, cos t).( sin t, cos t)dt f(x, y)ds = 2π 2. Evaluate ey sin(x)dx + e y cos(x)dy + dz over the curve where is the straight line from the point (,, ) to (π, π, 1). Solution: It is not difficult to write a parametric equation for.

2 P (t) = (πt, πt, t) Where t 1. This gives rise to the line integral e y sin(x)dx + e y cos(x)dy + dz = e πt sin(πt)dx + e πt cos(πt) (π, π, 1)dy + dz 1 = πe πt sin(πt) + πe πt cos(πt) + 1dt Which gives the final answer of e π A better way to approach this problem would be to notice that the vector field is conservative. We will notice this by simply finding a potential function. f(x, y, z) = e y cos(x) + z We can check that is really is a potential function since f = ( e y sin(x), e y cos(x), 1) Then we can use the fundamental theorem of line integrals to evaluate the integral. e y sin(x)dx + e y cos(x)dy + dz = f(π, π, 1) f(,, ) = 1 e π 1 = e π 3. ompute F ds where F (x, y, z) = (9x, 4y, z 9x 2 + 4y 2 ) and is parametrized by (t) = (2 cos(t), 3 sin(t), 4t 2 ) for 1 t 2. Solution: There are no obvious tricks to use in the evaluation of this line integral, so we must proceed with our basic definition of line integral. To begin we will compute (t) (t) = ( 2 sin(t), 3 cos(t), 8t)

3 2 F ds = F (t) (t)dt = 1 = (18 cos(t), 12 sin(t), 4t 2 36 cos(t) sin(t) 2 ) ( 2 sin(t), 3 cos(t), 8t) = = 36 cos(t) sin(t) + 36 cos(t) sin(t) + 192t 3 = Evaluate the integral (hint: use Green s Theorem). (x 2 + y 2 )dx + 2xydy Where is the curve which follows the parabola y = x 2 (, ) to (2, 4) then the line from (2, 4) to (2, ) and finally the line (2, ) to (, ). Solution: We will first compute directly, which will be the harder way. The first step is to setup parametric equations. We have three segments, two of them are line segments which are easy to parametrize, the third is a function with 1 dependent variable. So we can easily setup the bounds as x 1 (t) = (2t, 4) ( t 1) x 2 (t) = (t, t 2 ) (2 t ) x 3 (t) = (, 2t) ( t 1) Now we use these in the vector line integral. We setup each line integral the same way 1 F (x 1 (t)) x 1(t)dt = 1 (((2t) 2 + (4) 2 )dx + 2(2t)(4)) (2, ) = = 1 8t dt = 14 3

4 1 F (x 2 (t)) x 2(t)dt = (((t) 2 + (t 2 ) 2 )dx + 2(t)(t 2 )) (1, 2t) = = 2 2 t 3 + 3t 5 dt = F (x 3 (t)) x 3(t)dt = 1 (4t 2, ) (, 2) = So we sum these results to get a final result of. Now if were were to use Green s theorem to do this we would replace this with a double integral. The process goes like this: First we pull out the F 1 and F 2 from the vector field (or P and Q depending on which notation you prefer). We compute F 1 y F 2 x = 2y = 2y Then 2y 2ydA = = So we get the same answer with less computation.

5 5. Use green s theorem to replace the line integral (y sin(y) cos(y)) dx + 2x sin 2 (y) dy with a double integral, where is the counterclockwise path around the region bounded by x = 1, x = 2, y = 4 x 2, and y = x 2. Solution: The vector field written as a dx+b dy so we setup our F 1 and F 2 functions F 1 (x, y) = y sin(y) cos(y) F 2 (x, y) = 2x sin 2 (y)dy In order to use Green s theorem we will need to compute two partial derivatives: F 1 y F 2 x = 1 cos 2 (y) + sin 2 (x) = 2 sin 2 (y) = 2 sin 2 (y) So we setup the double integral from Green s Theorem. R F 2 x F 1 y = = R 2 4 x 2 = 1 x 2 So we were able to evaluate the integral, just by writing it out.

6 6. Find the area between the ellipse x 2 /9 + y 2 /4 = 1 and the circle x 2 + y 2 = 25 Solution: A picture is a helpful starting point for this problem. A naive way to start the problem would be to write 1 da D But this will be a rather difficult integral to evaluate. By using Green s theorem, we can solve this with less work. 1 ydx + xdx 2 D In order to proceed we must find parametric equations for the two curves bounding the area. Remember that in order to use Green s theorem we have to have clockwise orientation. My choice of parametrization is: x 1 (t) = (5 cos t, 5 sin t) x 2 (t) = (3 cos t, 2 sin t) Note that in x 2 we have had to multiply 2 sin t by -1, in order to preserve consistent orientation, ie that the region is always on the left as we walk along the boundary. So we can compute the integral. 1 ydx + xdx = 1 2 D π 2π (( 5 sin t)( 5 sin t) + (5 cos t)(5 cos t)) dt+ ((2 sin t)( 3 sin t) + (3 cos t)( 2 cos t)) dt

7 When we reduce this out we have = 1 2 2π 25dt π ( 6)dt = 19π 7. Find the area of the region enclosed by the parametric equation p(θ) = ( cos(θ) cos 2 (θ), sin(θ) cos(θ) sin(θ) ) for θ 2π Solution: The region we are going to find the area of looks like this We will use Green s theorem to setup and solve this problem. Recall that the contour integral along a closed path of F (x, y) = ( y/2, x/2) gives the area of the region enclosed by. So we will integral F along p(θ). F (x, y)ds = y/2dx + x/2dyds = p = p 2π ( (sin(θ) cos(θ) sin(θ))/2, (cos(θ) cos 2 (θ))/2 ) ((2 cos(t) 1) sin(t), cos(t) cos( 2π = 2 sin(t/2) 4 dt = 3π 2 So a simple application of Green s Theorem has given us an answer which would have been very hard to compute without line integrals.

8 8. Tessa and Scott are skating on a frictionless pond in the wind. Scott skates on the inside, going along circle of radius 2 in a counterclockwise direction while Tessa skates once around on a circle of radius 3, also in the counterclockwise direction. Suppose the force due to the wind at a point (x, y) is F (x, y) = (x 2 y + 1y, x 3 + 2xy 2 ). Use Green s theorem to determine who does more work. Solution: The setup is key in this problem. What we would like to compute is W ork T essa W ork Scott F (x, y) ds F (x, y) ds T The area between the two paths is a annulus, however, as written, the paths do not specify a consistent orientation of the annulus. In order to obtain a consistent orientation, we can rewrite the integral as F (x, y) ds + F (x, y) ds T in which Scott is moving in the clockwise direction. This line integral now has a consistent orientation, and we can use Green s theorem in order to evaluate the line integral by replacing it with the double integral 2x 2 + 2y 2 1 da We can setup this integral by using polar coordinates 3 2π R S 3 2r 3 1rdθdr = 2π S 2r 3 1r dr 2 2 = 2π r4 2 5r2 dr = 15π 2 3 The fact that the answer is positive means that the outside skater (Tessa) does more work. 9. The barometric pressure P (x, y) is given by the expression p(x, y) = x + y. Wind is caused by air moving from areas of high pressure to areas of low pressure so that the wind vector field W (x, y) = P

9 (a) aptain Jack Sparrow and the Black Pearl are located at the point (, ) and are trying to sail to a treasure hoard located at the point (1, 2). The Black Pearl cannot sail directly into the wind, so aptain Sparrow sets a course which sails along the y axis until the point (, 2) then sails parallel to the x axis until the point (1, 2). If the work done moving against the wind is equal to the line integral W ds ompute how much work is done by the Black Pearl as it moves to the point (1, 2) (b) Will Turner is chasing Jack Sparrow in a row boat, and rows directly to the point (1, 2). How much work does Will Turner do? (c) Explain your results using the Fundamental Theorem of Line Integrals. Solution: The wind vector field is given as the gradient W (x, y) = P = ( 1, 1) (a) In order to compute the work done my aptain Jack we ll need to parametrize both paths independently. p 1 (t) = t ( 2 ) p 2 (t) = (1 t) ( 2 ) + t ( 1 2 ) p 1 is the path along the y axis and p 2 is the path parallel to the x axis. Both paths use the bounds t 1. To work done will be the sum W ds + p 1 W ds. p 2 We will compute each integral separately p 1 W ds = 1 ( 1, 1) (, 2)dt = 2, W ds = p 2 1 ( 1, 1) (1, )dt = 1. So sailing to the point (1, 2) requires doing 3 units of work against the wind. (b) Will s path can be parametrized by p(t) = t(1, 2) t 1. Which so we can setup the line integral p F ds = 1 ( 1, 1) (1, 2)dt = 3. So rowing to the point (1, 2) requires doing 3 units of work against the wind, the same amount of work as sailing.

10 (c) The work done is path independent, which is no great surprise, the wind is the gradient of pressure, so W (x, y) is always a gradient vector field. Thus, by the Fundamental Theorem of Line Integrals, no matter what path Will or Jack follow, as long as they start at (, ) and end at (1, 2) the work done will be the same.