Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = -5.
y=x^2 , x=y^2

- anonymous

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- anonymous

Anyone want to help me set up the intergal? I can do the rest.

- anonymous

sketch the region first

- anonymous

|dw:1358400866711:dw|I did:

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## More answers

- anonymous

Something like that roughly.

- anonymous

can you determine the radius and the height?

- anonymous

No I can't actually lol.

- anonymous

Would the rdiube 5?

- anonymous

radius*

- anonymous

|dw:1358400896304:dw|

- anonymous

the radius and the height must be a function of y

- anonymous

Right.

- anonymous

So y^2-(-5)

- anonymous

yes, and the height?

- anonymous

1?

- anonymous

the height is the rightmost curve minus the leftmost curve right?

- anonymous

the height must also be a function of y

- anonymous

|dw:1358401360267:dw|

- anonymous

Ohh....I see....

- anonymous

actually its not like that, wait ill draw again

- anonymous

Allright. Thanks for thing the timet o help me :) . Really hard vizualzing all this stuff.

- anonymous

|dw:1358401483221:dw|

- anonymous

Yeah, that's better.

- anonymous

|dw:1358401575777:dw|

- anonymous

Right - left the I suppose right?

- anonymous

x^2-sqrt(x) .

- anonymous

yes, but it must be a function of y

- anonymous

Why not x?

- anonymous

Same function regardless right?

- anonymous

because we integrate it with respect to y later

- anonymous

Yeah I see it now haha.

- anonymous

so sqrt(y)-y^2?

- anonymous

yes

- anonymous

and here is the way to determine the radius|dw:1358402147166:dw|
the radius is y + 5

- anonymous

the limits would be from 0 to 1 I believe or am I wrong?

- anonymous

The radius would simply be the parbola - the line.

- anonymous

+ the line sorry.

- anonymous

\[\Delta V = 2 \pi R h \Delta y\]\[\Delta V = 2 \pi (y + 5)(\sqrt{y} - y^2) \Delta y\]\[V = \int\limits_{0}^{1} 2 \pi (y + 5)(\sqrt{y} - y^2) dy\]

- anonymous

1 sec,I need to process that.

- anonymous

Woudn't it b sqrt(y)+5?

- anonymous

be*

- anonymous

this is for the radius|dw:1358403009035:dw|this is for the height|dw:1358403125789:dw|

- anonymous

The answer I got was 2 which is wrong.

- anonymous

Nvm. It's right.

- anonymous

109pi/30

- anonymous

feel free to ask if its hard to follow

- anonymous

Why isn't the height sqrt(x)+5?

- anonymous

it must be a function of y, because we integrate it with respect to y, sqrt(x) + 5 = y + 5

- anonymous

and it is not the height, it is the radius

- anonymous

the radius is sqrt(y) - y^2

- anonymous

y^2 + 5 sorry.

- anonymous

Why not that?

- anonymous

|dw:1358403804913:dw|

- anonymous

Right. That's root x.

- anonymous

Ahh, that distance is y and add 5.

- anonymous

yup, its sqrt(x), because the curve is x = y^2
sqrt(x) = sqrt(y^2) = y

- anonymous

What's dy?

- anonymous

Thickness right?

- anonymous

yes, because the cylinder looks like this|dw:1358404151779:dw|we integrate it with respect to y, because we fill the cylinder downward

- anonymous

Thanks so much!

- anonymous

here is more details, I am sorry if I cannot explain further, because I teach myself all this stuff
http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx

- anonymous

Thanks :) .

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