# Johns Hopkins AS 110 202 - Past Exam Problems in Integrals (2 pages)

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## Past Exam Problems in Integrals

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## Past Exam Problems in Integrals

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Johns Hopkins University
Course:
As 110 202 - Calculus III
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Past Exam Problems in Integrals Prof Qiao Zhang Course 110 202 December 6 2004 The following is a list of the problems concerning integrals that appeared in the midterm and final exams of Calc III 110 202 within the last several years You may use them to check your understanding of the relevant material Some other exam problems may be found at http reserves library jhu edu access reserves findit exams 110 110202 php Note These problems do not imply in any sense my taste or preference for our own exam Some of the problems here may be more or less challenging than what will appear in our exam 1 Show that there is no vector field G such that curl G 2xi 3yzj xz 2 k Hint Recall that curl G is the same as G 2 a State Green s Theorem b Use Green s Theorem to evaluate the contour integral Z 1 y 8 dx x2 ey dy C where C denotes the boundary of the region enclosed by the curve y x and the lines x 1 and y 0 3 a State the Divergence Theorem Explain briefly what each symbol in the theorem stands for You may assume all the differentiability you want 1 b Use the Divergence Theorem to evaluate RR S F n dS where 2 F x y z xyi y 2 exz j sin xy k and S is the boundary surface of the region E bounded by the parabolic cylinder z 1 x2 and the planes z 0 y 0 and y 5 4 Let F x y z 2x y 2 i 2xy 3y 2 j a Show that curl F 0 b Find a function f x y z such that f F R c Use b to evaluate the integral C F ds where C is the arc of the curve y sin3 x from 0 0 to 2 1 in the xy plane 5 Use Gauss s Divergence Theorem to calculate the total flux out of the cube given by 1 x 1 1 y 1 1 z 1 of the vector field v r 2xyi y y 2 j x2 y z k 6 Consider the vector field G r y 2 zi x2 y z 2 3z j 2yz ez k R a Use Stokes theorem to express the line integral C G ds as a surface integral where C denotes the piecewise linear square contour that goes from the origin to 0 2 0 then to 0 2 2 then to 0 0 2 and back to the origin b Hence evaluate the line integral HINT Do not evaluate the line integral directly unless you

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