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Dido525
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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
 one year ago
 one year ago
Dido525 Group Title
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
 one year ago
 one year ago

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Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Anyone want to help me set up the intergal? I can do the rest.
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
sketch the region first
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
dw:1358400866711:dwI did:
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Something like that roughly.
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
can you determine the radius and the height?
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
No I can't actually lol.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Would the rdiube 5?
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
dw:1358400896304:dw
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
the radius and the height must be a function of y
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
So y^2(5)
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
yes, and the height?
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
the height is the rightmost curve minus the leftmost curve right?
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
the height must also be a function of y
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
dw:1358401360267:dw
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Ohh....I see....
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
actually its not like that, wait ill draw again
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Allright. Thanks for thing the timet o help me :) . Really hard vizualzing all this stuff.
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
dw:1358401483221:dw
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Yeah, that's better.
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
dw:1358401575777:dw
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Right  left the I suppose right?
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
x^2sqrt(x) .
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
yes, but it must be a function of y
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Why not x?
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Same function regardless right?
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
because we integrate it with respect to y later
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Yeah I see it now haha.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
so sqrt(y)y^2?
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
and here is the way to determine the radiusdw:1358402147166:dw the radius is y + 5
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
the limits would be from 0 to 1 I believe or am I wrong?
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
The radius would simply be the parbola  the line.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
+ the line sorry.
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
\[\Delta V = 2 \pi R h \Delta y\]\[\Delta V = 2 \pi (y + 5)(\sqrt{y}  y^2) \Delta y\]\[V = \int\limits_{0}^{1} 2 \pi (y + 5)(\sqrt{y}  y^2) dy\]
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
1 sec,I need to process that.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Woudn't it b sqrt(y)+5?
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
this is for the radiusdw:1358403009035:dwthis is for the heightdw:1358403125789:dw
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
The answer I got was 2 which is wrong.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Nvm. It's right.
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
feel free to ask if its hard to follow
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Why isn't the height sqrt(x)+5?
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
it must be a function of y, because we integrate it with respect to y, sqrt(x) + 5 = y + 5
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
and it is not the height, it is the radius
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
the radius is sqrt(y)  y^2
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
y^2 + 5 sorry.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Why not that?
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
dw:1358403804913:dw
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Right. That's root x.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Ahh, that distance is y and add 5.
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
yup, its sqrt(x), because the curve is x = y^2 sqrt(x) = sqrt(y^2) = y
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
What's dy?
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Thickness right?
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
yes, because the cylinder looks like thisdw:1358404151779:dwwe integrate it with respect to y, because we fill the cylinder downward
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Thanks so much!
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Thanks :) .
 one year ago
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