# MSU MTH 234 - mth234-009-E3-f2010 (4 pages)

Previewing page*1*of 4 page document

**View the full content.**## mth234-009-E3-f2010

Previewing page *1*
of
actual document.

**View the full content.**View Full Document

## mth234-009-E3-f2010

0 0 76 views

Lecture Notes

- Pages:
- 4
- School:
- Michigan State University
- Course:
- Mth 234 - Multivariable Calculus

**Unformatted text preview: **

Math 234 Practice Test 3 Show your work in all the problems 1 Find the volume of the region bounded above by the paraboloid z 9 x2 y 2 below by the xy plane and lying outside the cylinder x2 y 2 1 2 Evaluate the integral by changing to polar coordinates Z Z 1 y 2 1 1 1 y 2 ln x2 y 2 1 dx dy 3 Describe the region of integration Convert the integral to spherical coordinates and evaluate it Z 1 1 Z 1 x2 1 x2 Z 1 x2 y 2 dz dy dx 4 Sketch the region of integration and write an integral with the order of integration reversed Do not evaluate the integral Z 0 4 Z y 4 2 4 y dx dy 5 Find the centroid of the triangular region cut from the first quadrant by the line x y 3 1 Solutions 1 The most convenient coordinates for this problem are cylindrical coordinates x r cos y r sin We have to figure out when the paraboloid intersects the xy plane This is the case for 0 z 9 x2 y 2 i e x2 y 2 9 or r 3 The integral is then given by Z 0 2 Z 1 3 Z 9 r 2 0 r dz dr d Z 2 Z 2 0 0 32 Z 3 1 9 r 2 r dr d 3 9r 2 r 4 d 2 4 1 2 The equations x 1 y 2 are equivalent to x2 y 2 1 which describes a circle with radius 1 centered at the origin In polar coordinates x r cos y r sin this is given by r 1 Hence Z Z Z Z 1 y 2 1 1 1 y 2 ln x2 y 2 1 dx dy 2 0 1 0 ln r 2 1 r dr d substitute u r 2 1 du 2rdr Z Z 1 2 2 ln u du 2 0 1 Z 2 2 1 u ln u u d 1 2 0 2 ln 2 1 Recall from Calculus II that the integral of the logarithm function is calculated as follows Z ln x dx Z 1 ln x dx integration by parts du 1 dx v ln x dv dx x u x x ln x Z dx x ln x x C 3 The equation z x2 y 2 describes a cone in upper half space z 0 with tip at the origin and opening angle of 4 The region of integration is bounded below by the cone and above by the horizontal plane 2 z 1 In spherical coordinates the plane z 1 corresponds to 1 z cos i e sec The converted integral is then Z 0 2 Z 0 4 Z sec 0 2 sin d d d 1 2 4 3 sec sin d d 3 0 0 1 Z 2 Z 4 sec sec tan d

View Full Document