Dido525
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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Dido525
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Anyone want to help me set up the intergal? I can do the rest.

exraven
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sketch the region first

Dido525
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dw:1358400866711:dwI did:

Dido525
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Something like that roughly.

exraven
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can you determine the radius and the height?

Dido525
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No I can't actually lol.

Dido525
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Would the rdiube 5?

Dido525
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radius*

exraven
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dw:1358400896304:dw

exraven
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the radius and the height must be a function of y

Dido525
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Right.

Dido525
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So y^2(5)

exraven
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yes, and the height?

Dido525
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1?

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

exraven
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the height must also be a function of y

Dido525
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dw:1358401360267:dw

Dido525
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Ohh....I see....

exraven
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actually its not like that, wait ill draw again

Dido525
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Allright. Thanks for thing the timet o help me :) . Really hard vizualzing all this stuff.

exraven
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dw:1358401483221:dw

Dido525
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Yeah, that's better.

exraven
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dw:1358401575777:dw

Dido525
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Right  left the I suppose right?

Dido525
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x^2sqrt(x) .

exraven
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yes, but it must be a function of y

Dido525
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Why not x?

Dido525
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Same function regardless right?

exraven
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because we integrate it with respect to y later

Dido525
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Yeah I see it now haha.

Dido525
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so sqrt(y)y^2?

exraven
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yes

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

Dido525
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the limits would be from 0 to 1 I believe or am I wrong?

Dido525
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The radius would simply be the parbola  the line.

Dido525
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+ the line sorry.

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

Dido525
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1 sec,I need to process that.

Dido525
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Woudn't it b sqrt(y)+5?

Dido525
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be*

exraven
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this is for the radiusdw:1358403009035:dwthis is for the heightdw:1358403125789:dw

Dido525
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The answer I got was 2 which is wrong.

Dido525
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Nvm. It's right.

Dido525
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109pi/30

exraven
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feel free to ask if its hard to follow

Dido525
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Why isn't the height sqrt(x)+5?

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

exraven
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and it is not the height, it is the radius

exraven
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the radius is sqrt(y)  y^2

Dido525
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y^2 + 5 sorry.

Dido525
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Why not that?

exraven
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dw:1358403804913:dw

Dido525
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Right. That's root x.

Dido525
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Ahh, that distance is y and add 5.

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

Dido525
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What's dy?

Dido525
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Thickness right?

exraven
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yes, because the cylinder looks like thisdw:1358404151779:dwwe integrate it with respect to y, because we fill the cylinder downward

Dido525
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Thanks so much!


Dido525
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Thanks :) .