Homework 10

# Homework 10 - Rehman (aar638) – HW10 – sachse –...

This preview shows pages 1–3. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Rehman (aar638) – HW10 – sachse – (56620) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The graph of the function z = f ( x, y ) = 4- x is the plane shown in z 4 x y Determine the value of the double integral I = integraldisplay integraldisplay A f ( x, y ) dxdy over the region A = braceleftBig ( x, y ) : 0 ≤ x ≤ 4 , ≤ y ≤ 2 bracerightBig in the xy-plane by first identifying it as the volume of a solid below the graph of f . 1. I = 16 cu. units correct 2. I = 18 cu. units 3. I = 17 cu. units 4. I = 14 cu. units 5. I = 15 cu. units Explanation: The double integral I = integraldisplay integraldisplay A f ( x, y ) dxdy is the volume of the solid below the graph of f having the rectangle A = braceleftBig ( x, y ) : 0 ≤ x ≤ 4 , ≤ y ≤ 2 bracerightBig for its base. Thus the solid is the wedge z 4 4 x y (4 , 2) and so its volume is the area of triangular face multiplied by the thickness of the wedge. Consequently, I = 16 cu. units . keywords: double integral, linear function, volume under graph, volume, rectangular re- gion, prism, triangle 002 10.0 points Determine the value of the iterated integral I = integraldisplay 2 braceleftBig integraldisplay 2 1 (5 + 2 xy ) dx bracerightBig dy . 1. I = 20 2. I = 14 3. I = 22 4. I = 16 correct 5. I = 18 Explanation: Rehman (aar638) – HW10 – sachse – (56620) 2 Integrating with respect to x and holding y fixed, we see that integraldisplay 2 1 (5 + 2 xy ) dx = bracketleftBig 5 x + x 2 y bracketrightBig x =2 x =1 . Thus I = integraldisplay 2 braceleftBig 5 + 3 y bracerightBig dy = bracketleftBig 5 y + 3 2 y 2 bracketrightBig 2 . Consequently, I = braceleftBig 10 + 6 bracerightBig = 16 . keywords: 003 10.0 points Evaluate the iterated integral I = integraldisplay 3 1 braceleftBig integraldisplay 3 1 ( x + y ) 2 dx bracerightBig dy . 1. I = 1 2 ln 2 2. I = 2 ln2 3. I = ln 2 correct 4. I = 1 2 ln 6 5 5. I = 2 ln 6 5 6. I = ln 6 5 Explanation: Integrating the inner integral with respect to x keeping y fixed, we see that integraldisplay 3 1 ( x + y ) 2 dx = bracketleftBig- 1 x + y bracketrightBig 3 = braceleftBig 1 y- 1 3 + y bracerightBig . In this case I = integraldisplay 3 1 braceleftBig 1 y- 1 3 + y bracerightBig dy = bracketleftBig ln y- ln(3 + y ) bracketrightBig 3 1 . Consequently, I = ln parenleftBig (3)(1 + 3) (3 + 3) parenrightBig = ln 2 . 004 10.0 points Determine the value of the double integral I = integraldisplay integraldisplay A ( 3 x 2 y 3- x 4 ) dxdy when A = braceleftBig ( x, y ) : 0 ≤ x ≤ 1 , ≤ y ≤ 2 bracerightBig . 1. I = 18 5 correct 2. I = 13 5 3. I = 3 5 4. I = 23 5 5. I = 8 5 Explanation: Since A = braceleftBig ( x, y ) : 0 ≤ x ≤ 1 , ≤ y ≤ 2 bracerightBig is a rectangle with sides parallel to the coor- dinate axes, the value of I can be found by interpreting the double integral as the iter- ated integral I = integraldisplay 2 bracketleftbiggintegraldisplay...
View Full Document

## This note was uploaded on 02/01/2011 for the course MATH 408L taught by Professor Gogolev during the Fall '09 term at University of Texas.

### Page1 / 9

Homework 10 - Rehman (aar638) – HW10 – sachse –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online