微积分代考-MTH 252代写-Final Exam代写
微积分代考

微积分代考-MTH 252代写-Final Exam代写

Final Exam (MTH 252)-Spring 2023

微积分代考 Evaluate ∫∫ xz dS, where S is the boundary of the region enclosed by the cylinder y2 + z2 = 1 and the planes y + x = 5 and x = 0.

Final Exam of MTH 252 will be on May 10, from 10:15 am to 12:15 pm and covers

  • Sections 12.4 and 12.5;
  • Sections 14.1-14.7;
  • Sections 15.1-15.3, 15.5-15.8;
  • Sections 16.1-16.9.

I will hold offiffiffice hours on Tuesday (May 09) from 3:00-4:30 pm over zoom. You can reserve a time using the link of offiffiffice hours in the syllabus.

Below are some problems to practice for the fifinal exam. For practice problems for chapter 15 and sections 16.1-16.3, you can use the study guide for Exam 3. Below are some problems to practice for sections 16.4-16.9 and Chapters 12 and 14. The solutions of Questions 10-20 can be found in the study guides for Exams 1-2.

1.Evaluate ∫∫S curl  · dS, where (x, y, z) = 〈 xz, yz, xy〉 and S is the part of the sphere x2 + y2+ z2= 4 that lies inside the cylinder x2 + y2 = 1 and above the xy-plane.

2.Evaluate C · , where (x, y, z) = 〈 2y, xz, x + y〉 and C is the curve of intersection of the plane z = y + 2 and the cylinder x 2 + y 2 = 1.

3.Evaluate ∫∫ · dS for the vector fifield (x, y, z) = 〈 z, y, zx〉 and S is the surface of the tetrahedron enclosed by the coordinate planes and the plane x/a + y/b + z/c = 1, where a, b, and c are positive numbers.

4.Evaluate ∫∫ · dS for the vector fifield (x, y, z) = 〈 x4 , x3 z2 , 4xy2 z〉 and S is the surface of the solid bounded by the cylinder x2 + y2 = 9 and the planes z + x = 2 and z = 0.  微积分代考

5.Evaluate ∫∫  · dS for the vector fifield (x, y, z) = 〈 y, z y, x〉 and the oriented surface S the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1) in two ways: 1) using the defifinition of the flflux integral and 2) the divergence theorem.

6.Evaluate ∫∫ xz dS, where S is the boundary of the region enclosed by the cylinder y2 + z2 = 1 and the planes y + x = 5 and x = 0.

7.Evaluate C  · d~r, where (x, y) = 〈 e x + y2 , ey + x2 〉 and C consists of of the arc of the curve y = cos x from (π/2, 0) to (π/2, 0) and the line segment from (π/2, 0) to (π/2, 0).

8.Suppose that C is the oriented curve pictured in the fifigure below.

(a) If (x, y) = (x4 y5 2y)~i + (3x + x5 y4 )~ j, evaluate C ·

(b) If (x, y) = (xy2 2y)~i + (3x + y)~ j, evaluate C ·  ds.

微积分代考
微积分代考

9.Evaluate C · , where (x, y, z) = 〈 xy, yz, zx〉 and C is the boundary of the part of the paraboloidz = 1x2y2 in the fifirst ontant.

10.Evaluate ∫∫s curl dS, where F~(x, y, z) = 〈 tan1 (x2 yz2 ), x2 y, x2 z 2and S is the cone

x =0 x 2, oriented in the the direction of positive x-axis.

11.Find the equation of the plane that contains the line

x = 1 + t, y = 2 t, z = 4 3t

and is parallel to the plane 5x + 2y + z = 1.

12.Find the equation of the plane that contains the line

x = 4 t, y = 2t 1, z = 3t

and passes through P(3, 5, 1).

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微积分代考

13.Determine whether the lines

微积分代考
微积分代考

are parallel, skew, or intersecting.

14.Find the limit if it exists or show that it does not exist:

(a) lim(x,y)(0,0)(x2 + y2 ) ln(x2 + y2 ).

(b) lim(x,y)(0,0)

15.Find the fifirst partial derivatives of the following functions:

(a) f(x, y) =

(b) p(u, v, w) = u arctan(v )

16.Find the equations of the tangent plane and normal line to the following surfaces at the given point:  微积分代考

(a) xy + yz + zx = 3 at (1, 1, 1).

(b) z = 3x2 y2 + 2x at (1, 2, 1).

17.Find the points on

x2 + 4y2 z2 = 4

where the tangent plane is parallel to the plane 2x + 2y + z = 5.

18.Find the maximum and minimum rates of change of f at the given point and the direction in which they occur:

(a) f(x, y) = sin(xy) at P(1, 0).

(b) f(x, y, z) = x ln(yz) at (1, 2, 1).

19.Find critical points of

f(x, y) = (x2 + y2 )e−x.

Then classify them as local maximums, minimums, or saddle points.

20.Suppose f is a difffferentiable function of x and y and

g(u, v) = f(eu + sin v, eu + cos v).

Use the table below to calculate gu(0, 0) and gv(0, 0).

21.(6 points) If z = sin(x + sin t) show that

zxzxt = ztzxx.

 

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