exam2sample1

exam2sample1 - MATH 32B SECOND MIDTERM MAY 26, 2004 LAST...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 32B SECOND MIDTERM MAY 26, 2004 LAST NAME FIRST NAME ID NO. Please write clearly and legibly. There are four 20-point problems and four multiple choice problems worth 5 points each. To receive credit, you must circle your answer . DO NOT WRITE BELOW THIS LINE 1 5 2 6 3 7 4 8 TOTAL 2 PROBLEM 1 (20 Points) Use the mapping T ( u, v ) = ( u- 2 v, v ) and the change of variables formula to compute integraldisplay integraldisplay R ( x + 4 y ) dx dy where R is the shaded region in the Figure. CIRCLE YOUR ANSWER x+2y = 10 6 x y 10 3 5 1 x+2y = 6 Solution: We have x = u- 2 v and y = v , so u = x + 2 y . The region R is defined by the inequalities 6 u 10 , 1 v 3 Furthermore, vextendsingle vextendsingle vextendsingle ( x, y ) ( u, v ) vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 1- 2 1 vextendsingle vextendsingle vextendsingle vextendsingle = 1 so the Jacobian factor is 1. Since x + 4 y = u + 2 v , the change of variables formula gives us: integraldisplay integraldisplay R ( x + 3 y ) dx dy = integraldisplay 10 u =6 integraldisplay 3 v =1 ( u + 2 v ) dudv = 1 2 (10 2- 6 2 )(2) + 4(3 2- 1 2 ) = 64 + 32 = 96 3 PROBLEM 2 (20 Points) Compute...
View Full Document

Page1 / 10

exam2sample1 - MATH 32B SECOND MIDTERM MAY 26, 2004 LAST...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online