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anonymous
 3 years ago
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
 3 years ago
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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Anyone want to help me set up the intergal? I can do the rest.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sketch the region first

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358400866711:dwI did:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Something like that roughly.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you determine the radius and the height?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No I can't actually lol.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358400896304:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the radius and the height must be a function of y

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the height is the rightmost curve minus the leftmost curve right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the height must also be a function of y

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358401360267:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually its not like that, wait ill draw again

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Allright. Thanks for thing the timet o help me :) . Really hard vizualzing all this stuff.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358401483221:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358401575777:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Right  left the I suppose right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, but it must be a function of y

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Same function regardless right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because we integrate it with respect to y later

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah I see it now haha.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and here is the way to determine the radiusdw:1358402147166:dw the radius is y + 5

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the limits would be from 0 to 1 I believe or am I wrong?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The radius would simply be the parbola  the line.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Delta V = 2 \pi R h \Delta y\]\[\Delta V = 2 \pi (y + 5)(\sqrt{y}  y^2) \Delta y\]\[V = \int\limits_{0}^{1} 2 \pi (y + 5)(\sqrt{y}  y^2) dy\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01 sec,I need to process that.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Woudn't it b sqrt(y)+5?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is for the radiusdw:1358403009035:dwthis is for the heightdw:1358403125789:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The answer I got was 2 which is wrong.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0feel free to ask if its hard to follow

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why isn't the height sqrt(x)+5?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it must be a function of y, because we integrate it with respect to y, sqrt(x) + 5 = y + 5

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and it is not the height, it is the radius

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the radius is sqrt(y)  y^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358403804913:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Right. That's root x.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ahh, that distance is y and add 5.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yup, its sqrt(x), because the curve is x = y^2 sqrt(x) = sqrt(y^2) = y

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, because the cylinder looks like thisdw:1358404151779:dwwe integrate it with respect to y, because we fill the cylinder downward

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0here 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
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