Question

Find the volume of a solid rotating by the shells method! http://puu.sh/fiSwl/9daf1c3886.png

Answer 1

this is the answer http://puu.sh/fiSCP/53baae36ab.jpg

Answer 2

Use the Shell Method to compute the volume of the solid obtained by rotating the region underneath the graph of y= 1/(sqrt(x^2+6) over the interval (0,7), about the x = 0.

(Use symbolic notation and fractions where needed.)

Answer 3

Actually, what do you want?

Answer 4

I dont understand why they multiple by 2 at the end?

Answer 5

if u take u as x^2+6, u get du/2x = dx

Answer 6

2pi/2 = pi? but at the end they multiple 2(integeration of u)pi

Answer 7

they just separate 2 pi, let 2 in the front and let pi in the back, dat sit

Answer 8

But du/2x = dx, (x (1/sqrt(u) / 2x du) ; what happens to the 1/2?

Answer 9

ok, solve the integral
let u = x^2+6, du = 2x dx, change the limits. what do you have?

Answer 10

@zepdrix please, rescue me

Answer 11

what does this have to do with the limits? i got the limits correct

Answer 12

I just don't understand why they are multiplying by 2 ?

Answer 13

the limits is respect to x, now, you substitute u = x^2 +6, but you don't want to change the limit??

Answer 14

i dont understand what ur trying to do..

Answer 15

@ganeshie8 help me, please.

Answer 16

whats the derivative of \(\large \sqrt{x^2+6}\) @qqstory ?

Answer 17

2x

Answer 18

*(x^2+6)^(-1/2)

Answer 19

heard of chain rule ?

Answer 20

yes

Answer 21

we need to use it here because the stuff inside parenthesis is not just x, it is a complicated function of x

Answer 22

but why would you sub u as sqrt(x^2+6)?

Answer 23

then we would have two variables.

Answer 24

sorry we're trying to avoid u substitution and work it just by guessing the antiderivative

Answer 25

but don't you need to cancel out the x?

Answer 26

thats a good catch :) my mistake, let me fix it quick..

Answer 27

\[\left(\sqrt{x^2+6}\right)' = \frac{1}{2\sqrt{x^2+6}}\cdot (x^2+6)' = \frac{1}{2\sqrt{x^2+6}}\cdot 2x = \frac{x}{\sqrt{x^2+6}} \]

that means the antiderivative of x/sqrt(x^2+6) would be sqrt(x^2+6) :

\[\int\dfrac{x}{\sqrt{x^2+6}}dx = \sqrt{x^2+6} + C\]

Answer 28

atleast that just convinces us there is no mistake in your textbook solution
now you can work u substitution confidently i hope..

Answer 29

Ok thanks, let me try to practice my u-sub again.

Answer 30

Well, i don't get the same answer, i don't understand.

Answer 31

My u is set to x^2+6, and when i get sub dx as du, i have an extra 1/2.

Answer 32

I see what you're asking now, lets work u sub quick

Answer 33

Nevermind, I got the answer

Answer 34

I mess up on my arithmetic, thank you very much.

Answer 35

haha okay sounds great!

Answer 36

Have a nice weekend, ty.

Answer 37

you too :)