EXAM03-solutions-1 - Version 012 – EXAM03 – gilbert...

Info iconThis preview shows pages 1–3. 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 Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 012 – EXAM03 – gilbert – (55485) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The solid E shown in x y z is bounded by the circular cylinders x 2 + y 2 = 1 , y 2 + z 2 = 1 , and the coordinate planes. Write the triple integral I = integraldisplay integraldisplay integraldisplay E f ( x, y, z ) dV as a repeated integral, integrating first with respect to z , then x , and finally y . 1. integraldisplay 1 integraldisplay 1 − y 2 integraldisplay 1 − x 2 f ( x, y, z ) dz dx dy 2. integraldisplay 1 integraldisplay √ 1 − y 2 integraldisplay √ 1 − y 2 f ( x, y, z ) dz dx dy correct 3. integraldisplay 1 integraldisplay √ 1 − y 2 integraldisplay √ 1 − x 2 f ( x, y, z ) dz dx dy 4. integraldisplay 1 integraldisplay 1 − y 2 integraldisplay 1 − y 2 f ( x, y, z ) dz dx dy Explanation: Because E is bounded by the coordinate planes x, y, z = 0, the graph of the solid shows that E consists of all points ( x, y, z ) in 3-space satisfying ≤ z ≤ radicalbig 1 − y 2 , ≤ x ≤ radicalbig 1 − y 2 , and 0 ≤ y ≤ 1 . Consequently, as a repeated integral, I be- comes integraldisplay 1 parenleftBig integraldisplay √ 1 − y 2 parenleftBig integraldisplay √ 1 − y 2 f ( x, y, z ) dz parenrightBig dx parenrightBig dy . keywords: clicker 002 10.0 points Evaluate the triple integral I = integraldisplay integraldisplay integraldisplay E 6 xy 2 dxdydz when E is the set of points ( x, y, z ) such that ≤ z ≤ y ≤ (1 − x 2 ) 1 / 4 , and 0 ≤ x ≤ 1. 1. I = 1 8 2. I = 9 8 3. I = 3 8 correct 4. I = 5 8 5. I = 7 8 Explanation: As a repeated integral I = integraldisplay 1 parenleftBig integraldisplay (1 − x 2 ) 1 / 4 parenleftBig integraldisplay y 6 xy 2 dz parenrightBig dy parenrightBig dx . Version 012 – EXAM03 – gilbert – (55485) 2 Now integraldisplay y 6 xy 2 dz = bracketleftBig 6 xy 2 z bracketrightBig y = 6 xy 3 , while integraldisplay (1 − x 2 ) 1 / 4 6 xy 3 dy = 3 2 bracketleftBig xy 4 bracketrightBig (1 − x 2 ) 1 / 4 = 3 2 x ( − x 2 ) . Thus I = 3 2 integraldisplay 1 x ( − x 2 ) dx = bracketleftBig 3 4 x 2 − 3 8 x 4 bracketrightBig 1 . Consequently, I = 3 8 . keywords: integral, triple integral, repeated integral, linear function, limits of integration, general region, polynomial integrand, mono- mial integrand, evaluation of triple integral 003 10.0 points Evaluate the double integral I = integraldisplay integraldisplay A 4 xye y dxdy when A = braceleftBig ( x, y ) : 0 ≤ x ≤ 2 , ≤ y ≤ 1 bracerightBig . 1. I = 9 2. I = 10 3. I = 6 4. I = 8 correct 5. I = 7 Explanation: The double integral can be rewritten as the repeated integral I = integraldisplay 1 parenleftBig integraldisplay 2 xye y dx parenrightBig dy , integrating first with respect to x . Now integraldisplay 2 4 xye y dx = 2 bracketleftBig x 2 ye y bracketrightBig 2 = 8 ye y ....
View Full Document

{[ snackBarMessage ]}

Page1 / 9

EXAM03-solutions-1 - Version 012 – EXAM03 – gilbert...

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

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