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(math204)[2009](s)final~PPSpider^_10481.pdf
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Math 204 Final Exam
May 25, 2009
Your Name
Student Number
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Total
1. (20 points) Suppose f(x) is Riemann integrable on [0, 1]. Prove that the function
g(x, y)= f(x2 + y2) is Riemann integrable on x2 + y2 1. Moreover,
.. 1
g(x, y)d = f( )d.
x2+y210
2.
(20 points) Use the Lagrange multipliers to .nd the shortest distance from the straight line 3x +4y = 10 to the unit circle x2 + y2 = 1.
3.
(20 points) Suppose a change of variable x = x(u, v),y = y(u, v) maps a Jordan measurable region D . R2 to . R2 . If the transform satis.es
.x .y .x .y
= , = .,
.u .v .v .u
prove:
... .2 ..2 . ... .2 ..2 .
.f .f .f .f
+d(x,y) = +d(u,v)
D .x .y .u .v
4. (20 points) Show that the curve integral
I =[f(x)+ ay + bz]dx +[ax + g(y)+ cz]dy +[bx + cy + h(z)] dz
C
is independent of path. Evaluate the integral from (0, 0, 0) to (X, Y, Z). Verify your answer by using a potential function ..
5. (20 points) Show that the area enclosed by the curve
Ax2 +2Bxy + Cy2 =1, A> 0, AC . B2 > 0 is given by the curve integral
1 x dy . y dx
I = .
2 2+y=R2 Ax2 +2Bxy + Cy2
2
x
What is the area of ?
Answer to Math 204 Final, Spring 2010
1. (20 points) Suppose f(x) is Riemann integrable on [0, 1]. Prove that the function
g(x, y)= f(x2 + y2) is Riemann integrable on x2 + y2 1. Moreover,
.. 1
g(x, y)d = f( )d.
x2+y210
Solution: We can partition the unit disk by a family of eccentric rings {Ri}, char-acterizing by their radii:
P :0= r0 <r1 < .... < rn =1.
In this partition, the Riemann sum of the oscillation is
g(Ri)(Ii)
that is the same as
f ([ri.1,ri]) (ri . ri.1).
By the condition, the later goes to zero as maxi(ri . ri.1) 0. Thus, the function g is Riemann integrable on the unit disk. By Fubini theorem, we have
.. 2 . 1 . 1 g(x, y)d = f(r 2)rdrd = f( )d.
x2+y2100 0
2. (20 points) Use the Lagrange multipliers to .nd the shortest distance from the straight line 3x +4y = 10 to the unit circle x2 + y2 = 1.
Solution:
Let (x, y) be on the unit circle x2 + y2 =1 and (X, Y ) be the point on the straight line 3X +4Y = 10 such that the distance between them is the shortest. This is equivalent to minimize the function f(x, y, X, Y )=(X . x)2 +(Y . y)2 . By the Lagrange Multipliers, we have
(.2(X . x), .2(Y . y), 2(X . x), 2(Y . y)) = (2x, 2y, 0, 0) + (0, 0, 3, 4).
Together with the two equations, we have the following system
.
.
.2(X . x)= 2x .2(Y . y)= 2y
.
2(X . x)=3 .
2(Y . y)=4 x2 + y2 =1
.
3X +4Y = 10.
Since |x| 1 and |y| 1, we know that the straight line has no intersection with the circle. Hence, X ..These imply = 0 and = 0. The .rst four
= x and Y = y. ..
equations give
3
x = . ,y =(.2) ,
2
The unit circle gives
..2
4
= .
25
Thus,
3
X = (1 . )x = (1 . ) . ,Y = (1 . )y =