anonymous
  • anonymous
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
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Anyone want to help me set up the intergal? I can do the rest.
anonymous
  • anonymous
sketch the region first
anonymous
  • anonymous
|dw:1358400866711:dw|I did:

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anonymous
  • anonymous
Something like that roughly.
anonymous
  • anonymous
can you determine the radius and the height?
anonymous
  • anonymous
No I can't actually lol.
anonymous
  • anonymous
Would the rdiube 5?
anonymous
  • anonymous
radius*
anonymous
  • anonymous
|dw:1358400896304:dw|
anonymous
  • anonymous
the radius and the height must be a function of y
anonymous
  • anonymous
Right.
anonymous
  • anonymous
So y^2-(-5)
anonymous
  • anonymous
yes, and the height?
anonymous
  • anonymous
1?
anonymous
  • anonymous
the height is the rightmost curve minus the leftmost curve right?
anonymous
  • anonymous
the height must also be a function of y
anonymous
  • anonymous
|dw:1358401360267:dw|
anonymous
  • anonymous
Ohh....I see....
anonymous
  • anonymous
actually its not like that, wait ill draw again
anonymous
  • anonymous
Allright. Thanks for thing the timet o help me :) . Really hard vizualzing all this stuff.
anonymous
  • anonymous
|dw:1358401483221:dw|
anonymous
  • anonymous
Yeah, that's better.
anonymous
  • anonymous
|dw:1358401575777:dw|
anonymous
  • anonymous
Right - left the I suppose right?
anonymous
  • anonymous
x^2-sqrt(x) .
anonymous
  • anonymous
yes, but it must be a function of y
anonymous
  • anonymous
Why not x?
anonymous
  • anonymous
Same function regardless right?
anonymous
  • anonymous
because we integrate it with respect to y later
anonymous
  • anonymous
Yeah I see it now haha.
anonymous
  • anonymous
so sqrt(y)-y^2?
anonymous
  • anonymous
yes
anonymous
  • anonymous
and here is the way to determine the radius|dw:1358402147166:dw| the radius is y + 5
anonymous
  • anonymous
the limits would be from 0 to 1 I believe or am I wrong?
anonymous
  • anonymous
The radius would simply be the parbola - the line.
anonymous
  • anonymous
+ the line sorry.
anonymous
  • 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
  • anonymous
1 sec,I need to process that.
anonymous
  • anonymous
Woudn't it b sqrt(y)+5?
anonymous
  • anonymous
be*
anonymous
  • anonymous
this is for the radius|dw:1358403009035:dw|this is for the height|dw:1358403125789:dw|
anonymous
  • anonymous
The answer I got was 2 which is wrong.
anonymous
  • anonymous
Nvm. It's right.
anonymous
  • anonymous
109pi/30
anonymous
  • anonymous
feel free to ask if its hard to follow
anonymous
  • anonymous
Why isn't the height sqrt(x)+5?
anonymous
  • anonymous
it must be a function of y, because we integrate it with respect to y, sqrt(x) + 5 = y + 5
anonymous
  • anonymous
and it is not the height, it is the radius
anonymous
  • anonymous
the radius is sqrt(y) - y^2
anonymous
  • anonymous
y^2 + 5 sorry.
anonymous
  • anonymous
Why not that?
anonymous
  • anonymous
|dw:1358403804913:dw|
anonymous
  • anonymous
Right. That's root x.
anonymous
  • anonymous
Ahh, that distance is y and add 5.
anonymous
  • anonymous
yup, its sqrt(x), because the curve is x = y^2 sqrt(x) = sqrt(y^2) = y
anonymous
  • anonymous
What's dy?
anonymous
  • anonymous
Thickness right?
anonymous
  • 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
  • anonymous
Thanks so much!
anonymous
  • 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
  • anonymous
Thanks :) .

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