Problem Points S C O R E

Transcription

1 MATH 34F Final Exam March 19, 13 Name Student I # Your exam should consist of this cover sheet, followed by 7 problems. Check that you have a complete exam. Unless otherwise indicated, show all your work and justify your answers. Unless otherwise indicated, your answers should be exact values rather than decimal approximations. For example, π is an exact answer and is preferable to You may use a scientific calculator and one double-sided inch sheet of handwritten notes. All other electronic devices, including graphing or programmable calculators, and calculators which can do calculus, are forbidden. The use of headphones or earbuds during the exam is not permitted. Show your work, unless instructed otherwise. If you need more space, write on the back and indicate this. If you still need more space, raise your hand and I ll give you some extra paper to staple onto the back of your test. Academic misconduct will guarantee a score of zero on this exam. O NOT CHEAT. Turn your cell phone OFF and put it AWAY for the duration of the exam. Problem Points S C O R E Total: 7

2 1. (1 points) Consider the function f(x, y, z) = x + y + z. Take n =< 1/3, /3, /3 > and calculate f n (, 1, ). Solution: We have: at the given point x =, y = 1, z =, so f =< x, y, z >=<,, 4 > f n (, 1, ) = f(, 1, ) n = ( 4) 3 = 4 3

3 . (1 points) Write parametric equations of the plane that passes through the points P (1,, ), Q(, 1, ), R(,, 1). Solution: This plane contains vectors a = P Q =< 1, 1, > and b = P R =< 1,, 1 >. It passes through the point given by the position vector r =< 1,, >. Therefore, its parametric equations are r = r + ua + vb. In coordinate form: x = 1 + u v, y = u, z = v

4 3. (1 points) Calculate the integral z ds, where C is the curve r(t) = cos ti + sin tj + tk, t π. C Solution: We have: r (t) = sin ti + cos tj + k, r (t) = ( sin t) + cos t + 1 =. Therefore, ds = r (t) dt = dt. And z = t. Thus, z ds = π C t dt = t 3 3 t=π t= = 8 3 π3

5 4. (1 points) Calculate the integral xy dv, where E is the solid region between the planes y = x, y = x, x =, x = 1, z = 1, z = 1. E Solution: We have: 1 x E = { x 1, 1 z 1, x y x}. Therefore, this triple integral can be expressed as an iterated integral: x xy xy 3 y=x dydz x 4 3 dz 3 dzdx. This integral can be split into a product of two: 3 x 4 dx 1 1 dz = 3 x5 5 x=1 x= y= x 1 = = 4 15

6 5. (1 points) Calculate the integral x + y + z dv, E where E is the part of the unit ball that lies in the first octant above the half-cone z = x + y. Solution: Use spherical coordinates. The unit ball is given by ρ 1. The first octant: θ π/, ϕ π/. The half-cone is given by ρ cos ϕ = ρ sin ϕ cos ϕ = sin ϕ ϕ = π/4, so the part above it is given by ϕ π/4. Therefore, E = { θ π/, ϕ π/4, ρ 1}, x + y + z = ρ, dv = ρ sin ϕ dϕdθdρ. This integral is written as π/ π/4 It can be split into a product of three integrals: ρ 3 dρ π/ dθ π/4 sin ϕ dϕ = ρ4 4 ρ=1 ρ= ρ ρ sin ϕdϕdθdρ. π ( cos ϕ) ϕ=π/4 ϕ= = 1 4 π ( ) 1, which can be simplified up to π( 1)/16

7 6. (1 points) Using the ivergence Theorem, calculate F ds, where and Σ is the boundary of the region with the standard outward orientation. Solution: We have: Calculate div F: Σ F = xyz(i + j + k), E = { x y z x + y, x, y, x + y 1} div F = (xyz) x Σ F ds = + (xyz) y E + (xyz) z div FdV. = yz + xz + xy. Therefore, the triple integral over E can be written as an iterated integral: [ x+y ] (yz + xz + xy)dz da. x y Here, = {x, y, x + y 1} is the projection of E onto the xy-plane. The inner integral is equal to ( ) yz + xz z=x+y + xyz z= x y = xy(x + y) xy( x y) = xy(x + y) = x y + xy. Therefore, the triple integral is equal to (x y + xy )da = x (x (1 x) + 3 x(1 x)3 ) ) (x y + xy3 y=1 x 3 y= ( x ( 1 x + x ) + 3 x ( 1 3x + 3x x 3)) (x y + xy )dy (x x 3 + x x x + x 3 3 x4 ) ( x 3 x3 3 + x5 15 ( 3 x x + 1 ) 3 x4 ) x=1 x= = = 1 15

8 7. (1 points) Using Stokes Theorem, calculate F dr, where C F = yj + yzk, and C is the triangular contour between P (1,, ), Q(, 1, ), R(,, 1) (traversed counterclockwise, if you look from above). Solution: Consider the part Σ of the plane that passes through the points P, Q, R which is bounded by the contour C. Orient it so that the normal vector points upwards, then its orientation is adjusted to the orientation of C. Then, by Stokes Theorem, we have: F dr = curl FdS. Calculate the curl of the field F: i j k curl F = x y z y yz = zi. C Let us express Σ in coordinates. Find this plane: it contains the point P and the vectors P Q =< 1, 1, > and P R =< 1,, 1 >, so its normal vector is n = P Q P R =< 1, 1, 1 >. Its equation is 1(x 1) + 1(y ) + 1(z ) = x + y + z = 1. The part Σ of this plane has projection which is the triangle with vertices (, ), (1, ), (, 1). Therefore, = {x, y, x + y 1} = { x 1, y 1 x}. Thus, Σ is the graph of the function z = g(x, y) = 1 x y above the region on the xy-plane. And F = P i + Qj + Rk, P = z, Q =, R =. Therefore, curl FdS = Σ ( P g x Qg y + R) da = but we should plug in z = g(x, y) = 1 x y, so (1 x y)da = x Σ (1 x y)dy ((1 x) x(1 x) 1 (1 x) ) ( z( 1) ( 1) + ) da = zda, ) (y xy y y=1 x y= ((1 x)(1 x) 1 ) (1 x) 1 (1 x) 1 (1 x) 3 x=1 3 = 1 1 x= 3 = 1 6