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

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

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Something like that roughly.
can you determine the radius and the height?
No I can't actually lol.
Would the rdiube 5?
radius*
|dw:1358400896304:dw|
the radius and the height must be a function of y
Right.
So y^2-(-5)
yes, and the height?
1?
the height is the rightmost curve minus the leftmost curve right?
the height must also be a function of y
|dw:1358401360267:dw|
Ohh....I see....
actually its not like that, wait ill draw again
Allright. Thanks for thing the timet o help me :) . Really hard vizualzing all this stuff.
|dw:1358401483221:dw|
Yeah, that's better.
|dw:1358401575777:dw|
Right - left the I suppose right?
x^2-sqrt(x) .
yes, but it must be a function of y
Why not x?
Same function regardless right?
because we integrate it with respect to y later
Yeah I see it now haha.
so sqrt(y)-y^2?
yes
and here is the way to determine the radius|dw:1358402147166:dw| the radius is y + 5
the limits would be from 0 to 1 I believe or am I wrong?
The radius would simply be the parbola - the line.
+ the line sorry.
\[\Delta V = 2 \pi R h \Delta y\]\[\Delta V = 2 \pi (y + 5)(\sqrt{y} - y^2) \Delta y\]\[V = \int\limits_{0}^{1} 2 \pi (y + 5)(\sqrt{y} - y^2) dy\]
1 sec,I need to process that.
Woudn't it b sqrt(y)+5?
be*
this is for the radius|dw:1358403009035:dw|this is for the height|dw:1358403125789:dw|
The answer I got was 2 which is wrong.
Nvm. It's right.
109pi/30
feel free to ask if its hard to follow
Why isn't the height sqrt(x)+5?
it must be a function of y, because we integrate it with respect to y, sqrt(x) + 5 = y + 5
and it is not the height, it is the radius
the radius is sqrt(y) - y^2
y^2 + 5 sorry.
Why not that?
|dw:1358403804913:dw|
Right. That's root x.
Ahh, that distance is y and add 5.
yup, its sqrt(x), because the curve is x = y^2 sqrt(x) = sqrt(y^2) = y
What's dy?
Thickness right?
yes, because the cylinder looks like this|dw:1358404151779:dw|we integrate it with respect to y, because we fill the cylinder downward
Thanks so much!
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
Thanks :) .

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