MATH 3 Final Eam (Version 1) Solutions May, 8 S. F. Ellermeyer Name Instructions. Your work on this eam will be graded according to two criteria: mathematical correctness and clarity of presentation. In other words, you must know what you are doing (mathematically) and you must also epress yourself clearly. In particular, write answers to questions using correct notation and using complete sentences where appropriate. Also, you must supply su cient detail in your solutions (relevant calculations, written eplanations of why you are doing these calculations, etc.). It is not su cient to just write down an answer with no eplanation of how you arrived at that answer. As a rule of thumb, the harder that I have to work to interpret what you are trying to say, the less credit you will get. You may use your calculator but you may not use any books or notes. 1. (a) Estimate the volume of the solid that lies below the surface z = 5 and above the rectangle R = f(; y) j 5; y g. Use a Riemann Sum with m = 5 and n = and take the sample points to be the upper right corners of each subrectangle. (Note: The point of this part of the problem is to estimate the volume by using a Riemann sum. o not compute the eact volume using a double integral.) (b) Use a double integral to nd the eact volume of the solid described above. Solution: Note that and thus The desired estimate is = 5 5 y = = 1 = A = (1) () =. [f (1; ) + f (1; ) + f (; ) + f (; ) + f (3; ) + f (3; ) + f (; ) + f (; ) + f (5; ) + f (5; )] = [ + + 3 + 3 + + + 1 + 1 + + ] =. The eact volume is ZZ R (5 ) da = Z 5 Z (5 ) dy d = 5.. Set up and evaluate a double integral that gives the volume of the solid lying under the elliptic paraboloid + y 9 + z = 1 1
and above the rectangle R = [ 1; 1] [ ; ]. Solution: The volume is ZZ 1 R y Z 1 Z da = 1 9 1 y dy d = 166 9 7. 3. Sketch the region of integration for the iterated integral Z 3 Z p 9 y f (; y) d dy and then write an equivalent double integral with the order of integration changed. (Hint: The equivalent integral is actually a sum of two integrals.) Solution: The region,, of integration is pictured below. The integral written above views as a Type II region. 8 y = 9? 6 y = 3 6,3 = 6.5 1. 1.5..5 3. By viewing as a Type I Region, the above integral can also be computed as Z p 6 Z 3 f (; y) dy d + Z 3 p 6 Z 9 f (; y) dy d.. The curve r = cos () (in polar coordinates) is pictured below. This curve is called a rose with eight petals.
y 1..8.6.. -1. -.8 -.6 -. -....6.8 1. -. -. -.6 -.8-1. Set up and evaluate a double integral that gives the area of one petal of the rose. Solution: One half of a petal is traced out as ranges from to =8. Thus the area of one petal is ZZ Z =8 Z cos() 1 da = r dr d = 16. 5. Find the center of mass of a lamina in the shape of an isosceles right triangle with equal sides of length a if the density function at any point is proportional to the square of the distance from the verte opposite the hypotenuse. (Assume that the verte opposite the hypotenuse is at the point (; ).) Solution: We assume that the verte opposite the hypotenuse is located at (; ). This means that the density function is (; y) = K ( + y ) where K is some positive constant. The total mass of the lamina is ZZ m = K + y Z a Z a da = K + y dy d = 1 6 Ka. The coordinate of the center of mass is = 1 ZZ K + y da m = 6 a Z a = 5 a Z a + y dy d 3
and the y coordinate of the center of mass is y = 1 ZZ Ky + y da m = 6 a Z a = 5 a. Z a y + y dy d 6. Match the functions, f, with the plots of their gradient vector elds (which are given on the attached sheet). (a) f (; y) = p + y matches Graph 3 (b) f (; y) = y matches Graph 9 (c) f (; y) = y matches Graph (d) f (; y) = y matches Graph 8 (e) f (; y) = + y matches Graph 1 (f) f (; y) = y matches Graph 1 (g) f (; y) = y matches Graph 6 (h) f (; y) = y matches Graph 7 (i) f (; y) = y matches Graph 5 (j) f (; y) = + y matches Graph 7. Find the work done by the force eld F (; y) = sin (y) i + yj on a particle that moves along the parabola y = from the point ( (; ). Solution: The work done is Z F dr 1; 1) to the point where is the curve = t y = t 1 t. Since d = dt and dy = t dt, the work done is Z Z F dr= P d + Q dy = Z 1 t sin t + t 3 dt = 1 cos (1) 1 15 cos () +.
8. Eplain why the line integral Z 3y 3 d + 9y dy is independent of path and evaluate this integral where is any path beginning at the point (; 1) and ending at the point ( 3; ). Solution: The integral is independent of path because P (; y) = 3y 3 and Q (; y) = 9y are both de ned on all of R (which is a simply connected domain) and because A potential function is @P @y = 9y = @Q @. f (; y) = 3y 3 and thus, by the Fundamental Theorem of Line Integrals, Z 3y 3 d + 9y dy = f ( 3; ) f (; 1) = 7. 9. Use Green s Theorem to nd the work done by the force F (; y) = + y i + y j in moving a particle from the origin along the ais to the point (1; ) ; then along the line segment from (1; ) to (; 1), and then back to the origin along the y ais. Solution: By Green s Theorem, the work done is Z + y ZZ @Q d + y dy = @ = Z 1 Z 1 = 1 1. y @P da @y dy d 1. Find the area of the quadrilateral pictured below. 5
(-8,9) (-,-7) (,-7) (8,-8) Solution: By a result derived in class, the area of this quadrilateral is 1 (( ) ( 7) () ( 7) + () ( 75. 6