HW13S - Version 100 – Homework 13 – Gilbert – (59825)...

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: Version 100 – Homework 13 – Gilbert – (59825) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 10.0 points Evaluate the iterated integral I = integraldisplay 3 π/ 2 integraldisplay cos θ 2 e sin θ drdθ . 1. I = e- 2 2. I = 2( e- 1) 3. I = 2 parenleftBig 1 e- 1 parenrightBig correct 4. I = 1 e- 2 5. I = 2 e 6. I = 0 Explanation: After simple integration integraldisplay cos θ 2 e sin θ dr = bracketleftBig 2 r e sin θ bracketrightBig cos θ = 2 cos θ e sin θ . In this case, I = integraldisplay 3 π/ 2 2 cos θ e sin θ dθ = bracketleftBig 2 e sin θ bracketrightBig 3 π/ 2 . Consequently, I = 2 parenleftBig 1 e- 1 parenrightBig . 002 10.0 points Find the value of the integral I = integraldisplay integraldisplay A (5 x 2- 3 y 2 ) dxdy when A = braceleftBig ( x, y ) : 0 ≤ y ≤ 2 x, ≤ x ≤ 1 bracerightBig . 1. I = 7 6 2. I = 5 6 3. I = 1 4. I = 2 3 5. I = 1 2 correct Explanation: As an iterated integral, I = integraldisplay 1 bracketleftbiggintegraldisplay 2 x (5 x 2- 3 y 2 ) dy bracketrightbigg dx = integraldisplay 1 bracketleftbig 5 x 2 y- y 3 bracketrightbig 2 x dx = integraldisplay 1 2 x 3 dx . Consequently, I = 1 2 . 003 10.0 points The graph of f ( x, y ) = 4 xy over the bounded region A in the first quad- rant enclosed by y = radicalbig 9- x 2 and the x, y-axes is the surface Version 100 – Homework 13 – Gilbert – (59825) 2 Find the volume of the solid under this graph over the region A . 1. Volume = 81 cu. units 2. Volume = 27 cu. units 3. Volume = 81 4 cu. units 4. Volume = 81 2 cu. units correct 5. Volume = 81 8 cu. units Explanation: The volume of the solid under the graph of f is given by the double integral V = integraldisplay integraldisplay A f ( x, y ) dxdy, which in turn can be written as the repeated integral integraldisplay 3 parenleftBig integraldisplay √ 9- x 2 4 xy dy parenrightBig dx. Now the inner integral is equal to bracketleftBig 2 xy 2 bracketrightBig √ 9- x 2 = 2 x (9- x 2 ) . Thus V = 2 integraldisplay 3 x (9- x 2 ) dx = bracketleftBig- 1 2 (9- x 2 ) 2 bracketrightBig 3 . Consequently, Volume = 81 2 cu. units. 004 10.0 points Evaluate the double integral I = integraldisplay integraldisplay A ( x + 2 y ) dxdy when A is the region enclosed by the graphs of x = 1 , x- y = 1 , y = 1 . 1. I = 4 3 correct 2. I = 7 3 3. I = 2 4. I = 1 5. I = 5 3 Explanation: The graphs of x = 1 , x- y = 1 , y = 1 are straight lines intersecting at the points (1 , 0) , (1 , 1) , (2 , 1) . Thus the region of integration is the shaded triangular region (1 , 0) (2 , 1) (1 , 1) ( x, x- 1) ( x, 1) ( x, 0) Version 100 – Homework 13 – Gilbert – (59825) 3 so I can be written as the repeated integral I = integraldisplay 2 1 parenleftbiggintegraldisplay 1 x- 1 ( x + 2 y ) dy parenrightbigg dx , integrating first with respect to y from y = x- 1 to y = 1. Now the inner integral is equal to bracketleftBig...
View Full Document

This homework help was uploaded on 03/19/2008 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas.

Page1 / 10

HW13S - Version 100 – Homework 13 – Gilbert – (59825)...

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