The line segment can be parametrized as x = 1 + t, y = 1 + 2t, z = 1 + 3t 0 t 1. Then dx = dt, dy = 2dt, dz = 3dt so the integral becomes

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1 MATH rd Midterm Review Solutions. Evaluate xz dx z dy + y dz where is the line segment from (,, ) to (,, ). The line segment can be parametrized as x = + t, y = + t, z = + t where t. Then dx = dt, dy = dt, dz = dt so the integral becomes ( + t)( + t) dt ( + t) dt + ( + t) dt = t + t + dt = t + t + t = 5.. The force field F (x, y) = e x, x e y acts on a particle moving from (, ) to (, ). (a) ompute the work done by the force if the particle moves in a straight line. The work is ex dx + (x e y ) dy. The line can be parametrized as x = t, y = t for t so both dx, dy equal dt and the integral becomes et + t e t dt = t dt = t =. (b) ompute the work done by the force if the particle moves first along the x-axis to (, ) and then in a straight line to (, ). These are not easy to evaluate directly, so we will instead use Green s theorem and the results from part a. Let be the triangle with vertices (, ), (, ), (, ) oriented counterclockwise and T the region enclosed by. Let be the path from part b, a straight line from (, ) to (, ) then a straight line to (, ). Let be the line from (, ) to (, ). Then = so F dr = F dr + F dr. Then F dr = because is the negative of the curve in part a. We can calculate F dr using Green s theorem. We have that P = e x, Q = x e y so Q x =, P y = and by Green s theorem, F dr = da = A(T ) = where A(T ) is the area of the triangle T T. Then = + F dr so the work over is F dr = 5.. Evaluate ( + yz) dx + (y + xz) dy + ( x ) dz along the path parametrized by x = t, y = t, z = e t for t. We have that dx = dt, dy = tdt, dz = e t dt so this becomes ( + t e t ) + (t + te t )t + ( t )e t dt = + t dt = t + t =.

2 . Find the mass of that part of the surface z = xy that lies within one unit of the z-axis if the density at the point (x, y) is given by δ(x, y) = x + y. Note: The mass of an object is equal to the integral over the object of the density function. This is the surface integral of the function x + y over the part of the surface z = xy which is inside the cylinder x + y =. If D is the unit disk x + y on the xy-plane then this integral is D (x + y ) ds. Let f(x, y) = xy. Then ds = + (f x ) + (f y ) da = + y + x da so the integral is D (x + y ) + y + x da. hanging to polar, we get the integral π r + r r dr dθ = π r + r dr dθ = π r + r dr. Use the u-substitution u = + r, du = rdr and r = u to get that this is π (u ) ( u du = π u u du = π 5 u 5 ) u ( ) = π A nonuniform piece of wire if bent into the shape of the curve y = sin(x) between x = and x = π. The density of the wire at the point (x, y) is equal to + y. Set up, but do not evaluate, an integral equal to the mass of the wire. This is the line integral of + y over the curve parametrized by x = t, y = sin(t), t π so it s ( + y) ds. Then dx/dt =, dy/dt = cos(t) so ds = + cos (t)dt and the integral is π ( + sin(t)) + cos (t) dt. 6. Find the work done by the force F (x, y) = x cos(x ) + e y, xe y in moving a particle along a semicircle of radius from (, ) to (, ). This would be difficult to evaluate directly so we will check if F is conservative so we can use the fundamental theorem of line integrals. F is defined on the entire xy-plane and P y = e y = Q x so F is conservative. F has potential function f(x, y) = sin(x ) + xe y so the work is f(, ) f(, ) = (sin() ) (sin() + ) =. 7. A force given by F (x, y) = y, e x acts on a particle moving from the point (, ) to the point (, ) along the following path: first along the curve y = e x from (, ) to (, e ) and then along a line from there to (, ). Find the work done by the force.

3 Let be the first curve. Then can be parametrized by x = t, y = e t, and dx = dt, dy = e t dt so the work is y dx + e x dy = et + e t dt = e t + et = e + e. Let be the second part of the curve, the line from (, e ) to (, ). This can be parametrized by x =, y = t, e t. Then dx =, dy = dt and the work is y dx + e x dy = e e dt = e t e = e. The work over the whole path is e + e e = e + e. 8. Evaluate xey dx + (x + x e y ) dy where is the triangular path from (, ) to (, ) to (, ) and back to (, ). Let T be the triangular region enclosed by. Then by Green s theorem, the line integral over the triangle traversed counterclockwise is + T xey xe y da = da = A(T ) = where A(T ) = is the area of T the triangle. The path traverses the triangle clockwise, so the line integral over will be the negative of the counterclockwise integral so it is. 9. Let F (x, y, z) = e y + ze x, xe y e z, e x ye z. (a) ompute the curl and divergence of F. The curl of F is e z ( e z ), e x e x, e y e y =,, and the divergence is ze x + xe y ye z. (b) Determine if F is conservative. If yes, find f such that F = f, if not explain why. F is conservative as it is defined on all R and has curl equal to. To find f, we integrate the first component of F with respect to x and get that f(x, y, z) = xe y + ze x + g(y, z). Then xe y e z = f y = xe y + g y (y, z) so g y (y, z) = e z. Integrating with respect to y we get g(y, z) = ye z + h(z) so f(x, y, z) = xe y + ze x ye z + h(z). Then e x ye z = f z = e x ye z + h (z) so h (z) = and h(z) = c where c is constant. Thus f is of the form f(x, y, z) = xe y + ze x ye z + c. We only need to find one such f so in particular we can take the one where c = so we have f(x, y, z) = xe y + ze x ye z. (c) Integrate F dr along the line segment from (,, ) to (,, ). F is conservative so we can use the fundamental theorem of line integrals to get that F dr = f(,, ) f(,, ) = (e + e e ) (e + e e) = e e.. ompute cos(y )dx + x(x y sin(y ))dy along each of the following paths. (a) The line segment from (, ) to (, ).

4 Parameterize this as x = t, y = with t and dx = dt, dy =. Then the integral becomes dt =. (b) The line segment from (, ) to (, ). Parameterize this as x =, y = t with t. Then dx =, dy = dt and the integral becomes dt =. (c) The line segment from (, ) to (, ). This integral is hard to compute directly so instead use Green s theorem and the results from the other two parts. Let be the triangle from (, ) to (, ) to (, ) and back to (, ) and T the region enclosed by. Let,, be the three line segments in parts a,b,c respectively. Then = so F dr = F dr F dr + F dr. By Green s theorem, F dr = x y T sin(y ) ( y sin(y )) da = T x da = x (/)x x x dydx = x( x) dx = x x dx = = (/). We combine this result with the results of parts a,b to get (/) = F dr + so the line integral over is /.. Let F be the force field F (x, y) = x, x + y. A particle the feels this force starts at the origin. It is moved along the x-axis to the point (, ) and then it is moved along a quarter circle centered at the origin until it reaches the point (, ). Finally the particle is returned to the origin along the y-axis. ompute the total work done by the force field on the particle during this round trip. Let D be the quarter D enclosed by the path the particle moves around. Then by Green s theorem, the work is D x π/ x + y da = = π/ r cos(θ) drdθ = π/ cos(θ) dθ = sin(θ) π/ =. r cos(θ). A certain force F = P, Q, R is not completely known. It is known however that P = yze xy y + Axz, Q = xze xy + e z + Bxy, and R = e xy + x + ye z where A, B, are constants. A particle is moved from (,, ) to (,, ) many times along different paths and it is found that the work done by the force is the same each time. Determine values of A, B, which might explain this result. ompute the work done by the force using these values of A, B,. This result implies that F is probably conservative. We need to find values of A, B, which make F conservative. We need P y = Q x so xyze xy + ze xy y = xyze xy + ze xy + By and it follows that B =. We also

5 need P z = R x so ye xy + Ax = ye xy + 6x so A = 6. Finally we need Q z = R y so xe xy + e z = xe xy + e z so =. So F is conservative (and thus the line integral is independent of path) if A = 6, B =, =. To find the work we need to find a potential function f for F. Integrating P with respect to x we get that f(x, y, z) = ze xy xy + x z + g(y, z). The derivative is xze xy + e z xy = f y = xze xy xy + g y (y, z) so g y (y, z) = e z. Integrating with respect to y we get that g(y, z) = ye z + h(z) and so f(x, y, z) = ze xy xy + x z + ye z + h(z). Then e xy + x + ye z = f z = e xy + x + ye z + h (z) so = h (z) and we can take h(z) = and f(x, y, z) = ze xy xy + x z + ye z. Then the work done is f(,, ) f(,, ) = (e + + e) = e +.. A thin hollow shell has the shape of the paraboloid z = 9 x y for z. Find the surface area of the shell. The (x, y) values with make z are 9 x y which is 9 x + y, a disk of radius centered at the origin. Let D be this disk and f(x, y) = 9 x y. Then the surface area is D + (fx ) + (f y ) da = D + ( x) + ( y ) da = D + x + y da hanging to polar we get π r + r drdθ = π r + r dr. Using u-substitution with u = + r we get π 7 (/8)u/ du = (π)(/8)(/)u / 7 = π 6 (7 7 ).. A sphere of radius is centered at the origin. Find the area of that part of the sphere that lies above the region on the (x, y)-plane where x, y and x + y. Let D be the quarter disk x + y, x, y. The formula for the sphere is x + y + z = and we re looking at part of the sphere above the (x, y)-plane so we re looking at the top part of the sphere which solved for z is z = x y. As our surface has the form z = f(x, y) we see that the surface area is D + (fx ) + (f y ) da. The partial derivatives are f x = x/( x y ) and f y = y/( x y ) so + (f x ) + (f y ) = + x + y =. Then x y x y x y + (fx ) + (f y ) = D = x y x y da. hanging to polar we get x y π/ u-substitution with u = r gives π r drdθ = π r. So the surface area is r r dr. u / du = πu / = π( ). 5

6 5. Evaluate the line integral xy ds where is the part of the ellipse x + y = in the first quadrant. The ellipse can be parametrized as x = cos(t), y = sin(t) and the part in the first quadrant corresponds to t π/. Then dx/dt = sin(t), dy/dt = cos(t) so ds = sin (t) + cos (t)dt = sin (t) + dt. The line integral is π/ cos(t) sin(t) sin (t) + dt. If u = sin (t) + then du = 6 sin(t) cos(t)dt so this is u/ du = 9 u/ = 6/9 /9 = /9. 6. Evaluate the surface integral xy ds where S is the triangular region with S vertices (,, ), (,, ), and (,, ). The surface is a plane so we will first find the equation of the plane. It contains vectors,, and,, so the cross product,, normal vector to the plane, as is any multiple of this vector including,,. Using the point (,, ) and normal vector,, we get that the plane is (x ) + y + z = or x + y + z =. To evaluate the surface integral, we will parameterize the plane as r(x, y) = x, y, x y. We next need to figure out which values of (x, y) correspond to the triangular region. If we draw the triangular region in dimensions we can see that it s projection down to the xy-plane is the triangle with vertices (, ), (, ), (, ) so the surface integral will be taken over this triangle. If f(x, y) = x y then ds = + (f x ) + (f y ) da = + ( ) + ( ) da = 6dA so the surface integral xy ds = S 6 x xy 6 dydx = 6 xy x dx = x( x) dx = 6 x x + x dx = 6( x x + x ) = Let S be the surface with vector equation r(u, v) = u cos(v), u sin(v), v, u, v π. See Figure IV on p. 5 of the textbook for a picture of this surface. (a) Find the equation of the plane tangent to S at the point (,, π/). First find the values of u, v which correspond to the point (,, π/). These are u =, v = π/. Then r u = cos(v), sin(v), and r v = u sin(v), u cos(v), so the cross product is r u r v = sin(v), cos(v), u. When u =, v = π/ this is,, so we are looking for the formula of a plane with point (,, π/) and normal vector,,. The equation of the plane is x + (z π/) = or x + z = π/. (b) Set up, but do not evaluate, and integral for the surface area of S. The surface area is ds. Then S ds = r u r v da = sin (v) + cos (v) + u da = + u da and we re 6

7 integrating over the region on the uv-plane given by u, v π so we get that the surface area is equal to π + u dudv. (c) Evaluate the surface integral S + x + y ds. This is the same integral as in part b except we are now integrating + x + y instead of. We can the parametric formulas of r(u, v) (x = u cos(v), y = u sin(v), z = v) to convert the thing we re integrating into u s and v s. That is, + x + y = + u cos (v) + u sin (v) = + u. So the surface + u + u dudv = π integral becomes π + u dudv = π. (d) Evaluate the surface integral F ds where F = y, x, S z and S has upward orientation. We found the normal vector for S to be r u r v = sin(v), cos(v), u which has positive k-component as u is positive so this corresponds to the upward orientation. Then S F ds = π sin(v), cos(v), u dudv = π u sin(v), u cos(v), v u sin (v) u cos (v) + v u dudv. We F (r(u, v)) (r u r v ) dudv = π use the trig identity that sin (v) cos (v) = cos(v) to rewrite this as π u cos(v) + v u dudv = π u (cos(v) v ) dv = π v cos(v) dv = v sin(v) π = π. 7