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March 25, Mathematics 206a Mr. Haines 2005 Multivariable Calculus Examination #
(15) I. Consider the path f : [0, ∞ ) ⊂ → ^2 with f ( ) t =( e −^ t^ cos , t e − t sin t ).
A. Sketch a graph of f.
B. Give an integral that computes the total length of this path.
C. Calculate the value of this integral.
(10) II. Let C be the curve along the graph of y = x^3 − x^2 between the points (0, 0) and (1, 0). A. Give a parametrization for C.
B. Set up but do not evaluate the line integral ∫ C ( x + y dL ).
(15) IV. Evaluate:
1
0 0 0
z^ y
∫ ∫ ∫ z dx dy dz.
(15) V. Let M be the triangular surface in the plane 1 2
x
- y + z = that is cut off by the three coordinate planes. (M lies in the first octant, where x ≥ 0 , y ≥ 0 ,and z ≥ 0 .)
A. Give a parametrization for the surface M.
B. Set up, but do not evaluate, an iterated integral that gives the area of M.
C. Set up, but do not evaluate, an iterated integral that gives the surface integral of the vector field F (x, y, z) = (0, x, 0) over the surface M.
(15) VIII. Let f : R ⊂ℜ^2 →ℜ^3 be defined by f^ ( , ) s t^ =( , s t^^2 ,^ s t^2 )with R = [0,1] x [0, 2]. Let M be the surface parametrized by f. A. Set up, but do not evaluate, an iterated integral that gives σ ( M ), the surface area of M.
B. Let g : 3 → be defined by (^2 )
x y z x y
g
Set up, but do not evaluate, an iterated integral that gives M
∫∫ g d σ
C. Let F : 3 → ^3 be defined by F ( , x y z , ) = ( , x − y , 6) Set up, but do not evaluate, an iterated integral that gives M
∫∫ F n d σ
