Hw online 19 sol - Webwork con respuesta PDF

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Mei Qin Chen

citadel-math231

WeBWorK assignment number Homework19 is due : 11/11/2011 at 09:00am EST. The (* replace with url for the course home page *) for the course contains the syllabus, grading policy and other information. This file is /conf/snippets/setHeader.pg you can use it as a model for creating files which introduce each problem set. The primary purpose of WeBWorK is to let you know that you are getting the correct answer or to alert you if you are making some kind of mistake. Usually you can attempt a problem as many times as you want before the due date. However, if you are having trouble figuring out your error, you should consult the book, or ask a fellow student, one of the TA’s or your professor for help. Don’t spend a lot of time guessing – it’s not very efficient or effective. Give 4 or 5 significant digits for (floating point) numerical answers. For most problems when entering numerical answers, you can if you wish enter elementary expressions such as 2 ∧ 3 instead of 8, sin(3 ∗ pi/2)instead of -1, e ∧ (ln(2)) instead of 2, (2 + tan(3)) ∗ (4 − sin(5)) ∧ 6 − 7/8 instead of 27620.3413, etc. Here’s the list of the functions which WeBWorK understands. You can use the Feedback button on each problem page to send e-mail to the professors. √ √ Z 2π 1 3 3 63 π 63 π )π. ) = ( − dθ = ( − 1. (1 pt) Library/Dartmouth/setMTWCh1S6/urvc 3 1a.pg 4 4 2 6 2 6 2 0 Correct Answers: • (pi/6)*(6ˆ3 - 3ˆ3)*(2*pi/6 - sin(2*pi/6))

What are the rectangular coordinates of the point whose cylindrical coordinates are (r = 6, θ = 3π 9 , z = 5) ?

4. (1 pt) Library/Dartmouth/setMTWCh5S5/ur vc 10 10.pg Use spherical coordinates to evaluate the triple integral RR R 2 x + y 2 + z2 dV , where E is the ball: x2 + y 2 + z2 ≤ 49. E

x= y= z=

Correct Answers: • 42240.5981831069

5. (1 pt) Library/272/setStewart15 8/problem 8.pg Use spherical coordinates to evaluate the triple integral

Correct Answers: • 3 • 5.19615242270663 • 5

ZZZ

E

2. (1 pt) Library/Michigan/Chap16Sec5/Q03.pg Find an equation for the paraboloid z = x2 + y 2 in spherical coordinates. (Enter rho, phi and theta for ρ, φ and θ, respectively.) equation: SOLUTION cos(phi) . The paraboloid has equation rho = 2

2

2

2

e−(x +y +z ) p dV, x2 + y 2 + z2

where E is the region bounded by the spheres x2 +y 2 +z2 = 4 and x2 + y 2 + z2 = 25. Answer = Correct Answers: • 2 pi (exp(- 4) - exp(- 25))

sin (phi)

Correct Answers:

6. (3 pts) Library/Michigan/Chap16Sec5/Q32.pg The region W is the cone shown below.

• rho = [cos(phi)]/([sin(phi)]ˆ2)

3. (1 pt) Library/Michigan/Chap16Sec5/Q11.pg Evaluate, in spherical coordinates, the triple integral of f (ρ, θ, φ) = sin φ, over the region 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/6, 3 ≤ ρ ≤ 6. integral = SOLUTION The angle at the vertex is π/2, and the top is flat and at a height of 3.  Z Z 2π Z π/6 Z 2π Z π/6 Z 6 R 1 3 ρ=6 2 2 Write limits ( sin φ )( (sin φ)ρ sin φ dρd φ dθ = f dV = )(ρ )  thedφ dθ of integration for W dV in the following co3 0 3 W 0 0 0 ordinates (do not reduce the domain of integration by taking ρ=3 advantage of symmetry):  Z 2π Z π/6 Z 2π  φ=π/6 1 1 1 (a) Cartesian:  = 63 sin 2φ) (φ− (1−cos2φ) dφ dθ = 63 dθ  2 2 With a = ,b= , 0 0 2 0 φ=0 1

c= ,d= , e= , and f = , R R R d d d Volume = ab cd ef (b) Cylindrical: With a = ,b= , c= ,d= , e= , and f = , R R R Volume = ab cd ef d d d (c) Spherical: With a = ,b= , c= ,d= , e= , and f = , R R R Volume = ab cd ef d d d SOLUTION (a) Since the cone has an angle of π/2 at its vertex, it has equation p z = x2 + y 2 . The top of the cone, the plane with equation z = 3, intersects the cone in the circle x2 + y 2 = 3. Thus, in cartesian coordinates we have √ Z Z

dV =

W

Z 3

−3 −

r2−x2 Z 3

√

0−x2

√

x2 +y2

Correct Answers: • -3 • 3 • -sqrt(3ˆ2-xˆ2) • sqrt(3ˆ2-xˆ2) • 1*sqrt(xˆ2+yˆ2) • 3 • 1 • z • y • x • 0 • 2*pi • 0 • 3 • 1*r • 3 • r • z • r • theta • 0 • 2*pi • 0 • pi/4 • 0 • 3/[cos(phi)] • (rhoˆ2)*sin(phi) • rho • phi • theta

1 dz dy dx.

(b) In cylindrical coordinates the cone has equation z = r, so the integral becomes Z

dV =

W

Z 2π Z 0

03

Z 3

r dz dr dθ.

r

(c) In spherical coordinates, the cone has equation φ = π/4 and the plane z = 3 is ρcosφ = 3. Thus we have Z

W

dV =

Z 2π Z π/4 Z 3/ cos φ 0

0

ρ2 sin φ dρdφ dθ.

0

Generated by the WeBWorK system WeBWorK c Team, Department of Mathematics, University of Rochester

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