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Thomas Calculus 13th [Solutions]

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1232 Chapter 16 Integrals and Vector Fields<br />

i j k<br />

r( x, y) xi yj ( ax by)<br />

k be a parametrization of the surface. Then rx ry 1 0 a ai bj k<br />

0 1 b<br />

2 2<br />

d rx<br />

r y dx dy a b 1 dx dy.<br />

Also,<br />

i j k<br />

F<br />

x y z<br />

i j k and n<br />

z x y<br />

2 2<br />

a b 1<br />

xy xy xy<br />

2 2<br />

ai<br />

bj k<br />

a b<br />

2 2 1<br />

1 2 2<br />

F n d a b a b 1 dx dy ( a b 1) dx dy ( a b 1) dx dy.<br />

Now<br />

S R R R<br />

2 2 2 1 2 1 2<br />

x y ( ax by) 4 a x b y ab xy 1 the region R<br />

4 4 2<br />

xy is the interior of the ellipse<br />

2 2<br />

2 2<br />

Ax Bxy Cy 1 in the xy -plane, where A a 1, B ab ,and C b 1.<br />

The area of the ellipse is<br />

4 2 4<br />

2<br />

2 4<br />

4 ( a b 1)<br />

( a b 1)<br />

F dr h( a, b) . Thus we optimize H ( a, b) :<br />

2 2 2 2 2<br />

4AC B a b 1 C<br />

2 2<br />

a b 1<br />

a b 1<br />

H<br />

a<br />

2( a b 1) b 1 a ab<br />

a<br />

2<br />

2 2<br />

b<br />

1<br />

2<br />

0<br />

and<br />

H<br />

b<br />

2( a b 1) a 1 b ab<br />

a<br />

2<br />

2 2<br />

b<br />

1<br />

2<br />

2<br />

0 a b 1 0, or b 1 a ab 0 and<br />

2 2 2<br />

a 1 b ab 0 a b 1 0, or a b ( b a) 0 a b 1 0, or ( a b)( a b 1) 0<br />

a b 1 0 or a b . The critical values a b 1 0 give a saddle.<br />

2<br />

If a b, then 0 b 1 a ab<br />

2<br />

a 1 a<br />

2<br />

a 0 a 1 b 1. Thus, the point ( a, b ) ( 1, 1) gives a local extremum for<br />

F<br />

C<br />

dr<br />

z x y x y z 0 is the desired plane, if c 0.<br />

Note: Since h ( 1, 1) is negative, the circulation about n is clockwise, so n is the correct pointing normal<br />

for the counterclockwise circulation. Thus F ( n ) d actually gives the maximum circulation.<br />

S<br />

If c 0, one can see that the corresponding problem is equivalent to the calculation above when b 0, which<br />

does not lead to a local extreme.<br />

11. (a) Partition the string into small pieces. Let i s be the length of the i th piece. Let( xi<br />

, y i ) be a point in the<br />

th<br />

i piece. The work done by gravity in moving the i th piece to the x-<br />

axis is approximately<br />

Wi ( gxi yi is) yi where xi yi is is approximately the mass of the i th piece. The total work done by<br />

2 2<br />

gravity in moving the string to the x-axis is Wi gxi yi is Work gxy ds<br />

i i<br />

C<br />

(b) Work<br />

2 /2 2 2 2 /2 2<br />

gxy ds g(2cos t)(4sin t) 4 sin t 4cos tdt 16g cos t sin t dt<br />

C 0 0<br />

16g<br />

/2<br />

3<br />

sin t 16 g<br />

3 3<br />

0<br />

(c) x<br />

x( xy) ds<br />

C<br />

and y<br />

y( xy)<br />

ds<br />

C<br />

; the mass of the string is<br />

xy ds<br />

xy ds<br />

xy ds and the weight of the string is<br />

C<br />

C<br />

C<br />

g xy ds . Therefore, the work done in moving the point mass at ( x, y) to the x-axis is<br />

C<br />

W g 2 16 .<br />

C<br />

xy ds y g C<br />

xy ds 3<br />

g<br />

Copyright<br />

2014 Pearson Education, Inc.

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