Volume of a graph (Washer)

- MasterChief

Volume of a graph (Washer)

- schrodinger

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- MasterChief

Find the volume of \[y=\sqrt{x}\] about x=4

- MasterChief

I started with \[x=y^2\] so would that be the outer radius?

- SolomonZelman

|dw:1441595115874:dw|roughly

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## More answers

- SolomonZelman

x=y² is not the same as y=√x
because it will have a twice larger volume

- MasterChief

I'm sorry I forgot to mention it is bounded by y=0 so just quadrant 1

- SolomonZelman

oh, and then if it is rotated about x=4, then I will assume that this is where the √x region ends at?

- MasterChief

Yes

- SolomonZelman

Oh, ok, so you know that your limits of integration are from 0 to 2.
\(\large\color{black}{ \displaystyle \int_{0}^{2} \pi\left(y^2\right)dy }\)

- SolomonZelman

So, you can tell that your radius is y², and it is from y=0 to y=2.
THis is what I would do.

- SolomonZelman

I missed it should e y^4

- SolomonZelman

\(\large\color{black}{ \displaystyle \int_{0}^{2} \pi\left(y^2\right)^2dy }\)
because radius squared.
Sorry

- SolomonZelman

So that is just an integral of \(\pi\)y\(^4\) from y=0 to y=2.

- MasterChief

Wow I tried this the first time and I guess I integrated incorrectly so I was so confused!

- MasterChief

Thank you

- SolomonZelman

you can put integral into wolfram.
what is important is to get a good practice of making a setup of the integral for volume.
integration you know already....

- MasterChief

Absolutely

- SolomonZelman

32π/5 is what i got.
(want to know how to make a π • ÷ × √ with no latex?)

- MasterChief

Sure!

- DanJS

here are some probs with the solns, pg3 has a nice little summary

##### 1 Attachment

- MasterChief

That will definitely be useful, thank you

- SolomonZelman

I am glitching a bit

- SolomonZelman

Ok, so here is going to be a short guide.
The algorithm is:
\(1)\) Click ALT and hold it
\(2)\) Click the "Number Code" (on the numberpad on the
right of the keyboard- if you got one)
\(3)\) Release the ALT
------------------------------
Number code Result
0, 2, 1, 5 ×
2, 5, 1 √
7 •
2, 4, 6 ÷
2, 2, 7 π

- SolomonZelman

there are some others too.... but these are useful examples

- MasterChief

So if it was x=6 would the outer radius be 2+y^2 and the inner radius be 2?

- SolomonZelman

if it was x=6,
|dw:1441596092489:dw|

- MasterChief

about x=6 I meant

- SolomonZelman

Do you mean, if it was a region of y^2 bound by y=0 and x=4, but rotated about x=6?

- MasterChief

Yes

- SolomonZelman

Yes, then your radius is y²+2

- SolomonZelman

and if you rotated in a same case about y=4+c
then your radius is y²+c

- MasterChief

I thought I had to subtract the inner area?

- SolomonZelman

yeah, my bad, I am overheating let me think

- MasterChief

|dw:1441596258832:dw|

- MasterChief

Haha no problem

- SolomonZelman

|dw:1441596426856:dw|

- MasterChief

For this problem I can only use washer

- SolomonZelman

Oh, you can do it with respect to x, and do f(x)-g(x)

- SolomonZelman

But, shell is also good.

- SolomonZelman

it is a matter of preference

- MasterChief

Would f(x) be y^2+2 and g(x) be 2?

- MasterChief

\[\int\limits_{0}^{2} (2+y^2)^2-(2)^2\] ?

- SolomonZelman

lets rvw the washer with x's

- SolomonZelman

|dw:1441596741353:dw|

- SolomonZelman

of that integral should say
π INTEGRAL f(x)²-g(x)² dx

- SolomonZelman

from a to b

- MasterChief

Yes I get that

- MasterChief

I just don't understand what g(x) would be in this situation

- SolomonZelman

and with y, you get the same thing|dw:1441596949137:dw|

- SolomonZelman

the area of the cylinder with h=2, r=2 is 8π

- SolomonZelman

Or you can say that g is 2.

- SolomonZelman

if instead of 4, you had some z(x) boundary for the region,
then the radius for g would be 6-z(x) (of course from y=0 to y=2)

- MasterChief

Okay I think I understand this better now thank you

- SolomonZelman

Anytime... thank you for refreshing me on these rotations:)

- SolomonZelman

I mean, I really should tell you that Shell method rocks in so many cases, so try to use that as well.
in any case, good luck!

- MasterChief

Actually, did I plug this in incorrectly? http://www.wolframalpha.com/input/?i=integrate+from+0+to+2+pi%28%282%2By%5E2%29%5E2-4%29

- MasterChief

The answer is 192π/5

- SolomonZelman

for the problem where you want to know about the redion of y62 bound by y=0 and x=4, rotated about x=6....?

- SolomonZelman

region*

- SolomonZelman

y^2 (not y62)

- MasterChief

yes

- SolomonZelman

|dw:1441597714083:dw|

- SolomonZelman

I actually do not get why the answer isn't what you got in wolfram.

- MasterChief

hmmm

- SolomonZelman

you can take the whole volume of radius y^2+2 with limits of y=0 to y=2,
and subtract 8π cylinder in the middle, and you get precisely the same.

- SolomonZelman

I have to go, it is almost 12am in my location.
maybe I was answering a wrong question idk, but for what I asked, it should be 256π/15

- SolomonZelman

Maybe my brain just shot down xD

- MasterChief

Hahaha

- MasterChief

But I really appreciated your time so thank you

- SolomonZelman

I will look at it when I have time.
Whatever I can do with my little knowledge:)
gtg c(u)

- MasterChief

Bye!

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