find the volume of the solid generated by the revolution of the curve
y^2 (2 a - x) = x^3 about its asymptote.

Question

find the volume of the solid generated by the revolution of the curve
y^2 (2 a - x) = x^3 about its asymptote.

jamierox4ev3r · Answer

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kainui · Answer

How far have you gotten? Can you find the asymptote? We'll take it step by step.

anonymous · Answer

i tried using disk method. but its not not working because of equations format

kainui · Answer

Sure, show me what equation you came up with to integrate with the disk method. Make sure you solve for y before you try disk method.

anonymous · Answer

you mean generalized equation ?

kainui · Answer

I think so, not sure what you mean by generalized equation. Just try to get it looking like a formula for y= something in terms of x.

anonymous · Answer

y=x^(3/2)/[2a-x]^(1/2)

dumbcow · Answer

i would use shell method here so you can integrate wrt "x" 
its still a messy integral

dumbcow · Answer

$$V = 4\pi \int\limits _{0} ^{2a} (2a-x)\frac{x^{3/2}}{\sqrt{2a-x}} dx$$

anonymous · Answer

how did you get the limit 0 to 2a  ?

dumbcow · Answer

vertical asymptote is x=2a
domain of x varies from 0 to 2a

anonymous · Answer

and 2a-x is the radius of the shell right ?

dumbcow · Answer

correct and height is "y"

anonymous · Answer

thank you.

dumbcow · Answer

yw, also due to symmetry you only integrate for half volume then double it
(thtas where the 4 came from)

anonymous · Answer

can you suggest any study material in shell method.

dumbcow · Answer

http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx here you go