Question

A region bounded by f(x) = sqrt(x + 4) , y = 1, and x = 0 is shown below. Find the volume of the solid formed by revolving the region about the x-axis.

Answer 1

\[V=\pi\int_{-3}^0\left[\left(\sqrt{x+4}\right)^2-(1)^2\right]~dx\]

Answer 2

okay so i integrate that?

Answer 3

i get 34.87 what do i do next?

Answer 4

@Luigi0210

Answer 5

Luigi0210 is not here.. please leave a message after the beep.. *beep*

Ha I'm just messing.. but I think that's set up correctly.
@Loser66

Answer 6

lol okay i solved and i got that so what do i do next?

Answer 7

i got 6pi for the final answer ? :/

Answer 8

@dan815 I might not be good at volume but I'm great at calling for help ^_^'

Answer 9

hahaha okay

Answer 10

@Jhannybean

Answer 11

@ajprincess

Answer 12

I am really sorry @onegirl and @Luigi0210. i too havnt any idea

Answer 13

D: well thanks for coming and trying ^_^'

Answer 14

and I kind of tried the problem and I got (9/2)pi.. which is probably a 99.99% wrong

Answer 15

hmmm 9/2pi is in one of my multiple choices soo

Answer 16

you probably shouldn't take my word for it.. I have a tendency of getting volume problems wrong :/

Answer 17

okay

Answer 18

what are your other possible choices?

Answer 19

show me your work, @Luigi0210

Answer 20

I think SithsAng....gave out the right equation

Answer 21

=20pi

Answer 22

there is 6pi, 5pi and 4pi

Answer 23

\[\Pi \int\limits_{-3}^{0}[(x-4)-(1)]dx\]
\[\Pi [\frac{ x^2 }{ 2 }+4x-(x)]\]
\[\Pi [0-(\frac{ 9 }{ 2 }-12+3)]\]
\[\Pi[0-(\frac{ 9 }{ 2 }-9)]\]
\[\Pi[0-(\frac{ 9 }{ 2 })\]
\[\frac{ 9 }{ 2 } \Pi\]

Answer 24

but 20pi isn't in my choices..

Answer 25

there's suppose to be a negative on the 9/2.. second last :/

Answer 26

okay

Answer 27

sorry if it's wrong :/

Answer 28

its okay

Answer 29

why (x+4 ) not x -4 and then (x+4) -1 =x+3 and then take integral