anonymous
  • anonymous
**Calc2 Help Needed** Find the volume of the solid generated by revolving the described region about the given axis: The region bounded above by the line y=6 , below by the curve y=sqrt(x), and to the left by the y-axis, rotated along the following lines: x=36
Mathematics
chestercat
  • chestercat
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dumbcow
  • dumbcow
another one huh :)
anonymous
  • anonymous
Yep.. I'm going to start tutoring next week. I am having the worst time with these..
dumbcow
  • dumbcow
try drawing out the region to get the bounds

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anonymous
  • anonymous
yep, i've got that
anonymous
  • anonymous
|dw:1327644073706:dw|
anonymous
  • anonymous
so the bounds should be in y? so 0 to 6
anonymous
  • anonymous
I was thinking of using the cylindrical shell method
anonymous
  • anonymous
2pi integral (radius)height
anonymous
  • anonymous
radius = y height = y-36?
dumbcow
  • dumbcow
i'm not as familiar with the shell method, mostly because i don't like using it anyway, wouldn't the radius be in terms of x, and the height be 6-y
anonymous
  • anonymous
oh yes..
anonymous
  • anonymous
so we would use the disc method?
anonymous
  • anonymous
so radius = y^2 and height is 6-y?
dumbcow
  • dumbcow
you could use either i believe for this one. if you used the disc method: outer radius = 36 inner radius = 36-y^2 ...yes
anonymous
  • anonymous
so we have the integral of pi(36-36+y^2)?
dumbcow
  • dumbcow
think area of circle: A = pi*r^2
anonymous
  • anonymous
forgot the ^2
dumbcow
  • dumbcow
what should you integrate from, or what are the bounds
anonymous
  • anonymous
0 to 36?
dumbcow
  • dumbcow
the outer radius is 36 right, that is in x-direction now imagine the figure being sliced horizontally so we integrate over y what are the bounds of y
anonymous
  • anonymous
oh 0 to 6 in terms of y
dumbcow
  • dumbcow
yep
anonymous
  • anonymous
ok so from 0 to 6 integral (36 - 36 + x^2)^2?
anonymous
  • anonymous
= pi(y^5/5)
dumbcow
  • dumbcow
not quite its R^2 - r^2 Not (R-r)^2 -> 36^2 - (36-y^2)^2
dumbcow
  • dumbcow
because you're not subtracting the radius, you are subtracting the areas
anonymous
  • anonymous
ohh.. do I have two separate integrals?
dumbcow
  • dumbcow
no just keep them separate...you can simplify it
anonymous
  • anonymous
nvm.. i see now
anonymous
  • anonymous
I got it right!! yay.. haha.. =3628.8pi. Thank you so much!
dumbcow
  • dumbcow
great...welcome
anonymous
  • anonymous
Now it's time to study biology... fun :/

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