Question

Could someone help double-check 5 shell integration problems please?

Answer 1

sure

Answer 2

thanks, used washer method on 3 and 4

Answer 3

@israa88 you seen anything wrong?

Answer 4

give me a min plz

Answer 5

I am not 10 percent sure but it seems good to me

Answer 6

thx

Answer 7

no problem :)

Answer 8

first one seems wrong

Answer 9

wait, you doing shells on 1 ... let me look at it again

Answer 10

int 2pir h

r = x, h = x^(1/2)

2pi int[0,4] x^(3/2) dx

setups fine

Answer 11

2nd one, tell me about your limits .... you did a u sub but i dont see you changing back to x, but you are using the limits of x

Answer 12

the interval is defined as

0 < x < sqrt(pi) ; let u = x^2

0^2 < x^2 < sqrt(pi)^2

0 < u < pi

your solution is not calculated in the correct domain

Answer 13

i didn't explicitly change to x in a step, but did so in plugging in the bounds

Answer 14

I did think that the area enclosed could be wrong, in that there really is no left bounds by the description of the problem and the graph could continue indefinitely to the left

Answer 15

redo #2, its not organized correctly and it has a few flaws

Answer 16

is the graph correct?

Answer 17

the graph is fine :) the rest is a muck

Answer 18

ok, 3,4,5, look ok?

Answer 19

ahh, I see that the bounds for 1 should be based on u after subbing

Answer 20

\[\int_{x=0}^{x=\sqrt{\pi}}2\pi~r~h\]
\[2\pi\int_{x=0}^{x=\sqrt{\pi}}~x~sin(x^2)~dx\]
\[2\pi\int_{x=0}^{x=\sqrt{\pi}}~\frac12sin(u=\color{red}{x^2})~du=\color{red}{2xdx}\]
\[2\pi\int_{x=0}^{x=\sqrt{\pi}}~\frac12sin(u)~du\]
\[\pi\int_{\color{red}{x=0}}^{\color{red}{x=\sqrt{\pi}}}~sin(u)~du\]

Answer 21

u(x) = x^2

u(0) = 0^2 = 0

u(sqrt pi) = (sqrt pi)^2 = pi

Answer 22

it wasnt as amuck as i originally thought :) but that was due to you leaving some stuff out of it. define u and du for your substitutions, they are not clearly defined

Answer 23

i see

Answer 24

you have a sideway picture for 3, is that acceptable? just wondering

Answer 25

why would you change your method on #3? did you have to do washers?

Answer 26

is it user choice? i mean its a fine approach just curious

Answer 27

x^(1/3) integrates to :

x^(1/3 + 3/3)
-------------
1/3 + 3/3

does this match your integration?

Answer 28

4 is off, you are spose to revolve about x=4, not y=4

Answer 29

ahh, thanks for that catch

Answer 30

user choice on number 3

Answer 31

\[\int_{0}^{2}2pi ~rh\]
\[\int_{0}^{2}2pi ~(4-x)(2\sqrt2~x^{1/2}-x^2)~dx\]

Answer 32

why is r = (4-x) in this case?

Answer 33

i'm sure I messed that up, and need to recalculate most likely, I used the washer approach, and revolved it around y = 4, need to change it to x = 4

Answer 34

yeah, and use shells ...

Answer 35

guess I'm not seeing on #3 why it's turned on its side, the graph seems correct