2313-practice-mid4-soln

2313-practice-mid4-soln - MAC 2313, Fall 2010 Midterm 4...

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

Unformatted text preview: MAC 2313, Fall 2010 Midterm 4 Review Problems Solutions - 1 We will discuss these problems in class on Monday 11/15. Solutions will be posted on the course webpage over the weekend. 1 A bounded region in the first octant of 3-dimensional Euclidean space has the surface x + y + z 2 = 1 as part of its boundary. The remainder of its boundary is given by portions of the places x = 0, y = 0, and z = 0. Compute the triple integral of z over this region in space. (You should probably begin by sketching this region.) Solution: The surface x + y + z 2 = 1 lies in the first octant for x 0, y 0, and x + y 1. So, we can express the integral as integraldisplay 1 integraldisplay 1 x integraldisplay 1 x y z dz dy dx = integraldisplay 1 integraldisplay 1 x bracketleftbigg z 2 2 bracketrightbigg z = 1 x y z =0 dy dx, = integraldisplay 1 integraldisplay 1 x 1 x y 2 dy dx, = integraldisplay 1 bracketleftbigg (1 x ) y 2 y 2 4 bracketrightbigg y =1 x y =0 dx, = integraldisplay 1 (1 x )(1 x ) 2 (1 x ) 2 4 dx, = integraldisplay 1 (1 x ) 2 4 dx, = integraldisplay 1 (1 2 x + x 2 4 dx, = bracketleftbigg x x 2 4 + x 3 12 bracketrightbigg x =1 x =0 = 1 12 . 2 Let R denote the portion of the unit ball { ( x, y, z ) : x 2 + y 2 + z 2 1 } that lies inside the solid cone { ( x, y, z ) : z radicalbig x 2 + y 2 } . Compute the volume of R with triple integrals in both cylindrical and spherical coordinates. Solution: In cylindrical coordinates, we first integrate z from the cone z = radicalbig x 2 + y 2 = r 2 = r to the top of the ball, z = radicalbig 1 x 2 y 2 = 1 r 2 . Then comes the trickiest bound: with respect to r . We need to determine at which value of r the top of the ball meets the bottom of the cone. This value of r satisfies 1 r 2 = r , so 1 r 2 = r 2 , so 2 r 2 = 1, so MAC 2313, Fall 2010 Midterm 4 Review Problems Solutions - 2 r = radicalbig 1 / 2 . We integrate r from 0 to this value. Finally we integrate from 0 to 2 . This integral is therefore integraldisplay 2 integraldisplay 1 / 2 integraldisplay 1...
View Full Document

This note was uploaded on 12/10/2011 for the course MAC 2313 taught by Professor Keeran during the Fall '08 term at University of Florida.

Page1 / 7

2313-practice-mid4-soln - MAC 2313, Fall 2010 Midterm 4...

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