Question

I hope to be given problems on solids of revolution and surfaces of revolution as a review for my exam =)

Answer 1

i hope that your answers will be in the form of word problems. =)

Answer 2

if not, then it's ok. =)

Answer 3

heres an easy one to start

Find volume generated when the area between x = 0 and x = 2 under the curve y=x^2 is
rotated through 360 degrees about the x-axis.

Answer 4

ok. i'll draw first. which do you prefer, shells or disks? or both? =)))

Answer 5

hmmmm - disks

Answer 6

via disks:

\[A =\pi \int\limits_{0}^{2} (x^{2})^{2} dx = \pi \int\limits_{0}^{2} x^{4} dx\]

Answer 7

to be honest , maybe its because i'm from UK i only know the disk method

Answer 8

shell?

Answer 9

32/5 square units. =))) . yes, cylindrical shells. =)))

Answer 10

this problem will get too complicated when done with cylindrical shells. XD =)))

Answer 11

ok

Answer 12

hold on - i've got another for you

Answer 13

answer via cylindrical shells:

\[A = 2\pi \int\limits_{0}^{4} y ( 2- \sqrt{y} ) dy\]

Answer 14

see, needlessly complicated. =))) for the same answer =)))

Answer 15

yes

Answer 16

try this one
find volume generated when the area external to the hyperbola xy = 3, in the first quadrant, bounded by the lines x=2, y = 3 is rotated about the line x=2 through 2 pi radians.
a diagram would obviously help here.

Answer 17

2pi radians means one whole rotation? aye?

Answer 18

yes

Answer 19

excuse the drawing!!!

Answer 20

it's ok. good thing.; i misinterpreted it... external to parabola means over the graph or under da graph? =)))

Answer 21

to the hyperbola, i mean... =))