anonymous
  • anonymous
Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis. y = 8 − x2, y = x2; about x = 2
Mathematics
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katieb
  • katieb
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anonymous
  • anonymous
Here's a picture of the graph
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anonymous
  • anonymous
So I know the formula for shells is\[\int\limits_{}^{}2\pi(r)(h)dx\] with r being the radius and h being the height. I initially did the work with h = (8-x^2)-(x^2) and r = x but apparently that was wrong. I'm pretty sure my h was right though. What would the radius be ? 0.0
anonymous
  • anonymous
Do I need to put things in terms of x and integrate adding two separate integrals ? Since the top and bottom half have different x functions serving as their radius ?

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amistre64
  • amistre64
r = x r is from 4 to 0; but x is from -2 to 2 r = (2-x)
amistre64
  • amistre64
and, you might want to do this in parts ... top function minus bottom function
amistre64
  • amistre64
y1 = 8 − x2 y2 = x2 h = y1 - y2
amistre64
  • amistre64
\[\int 2\pi ~rh\] \[2\pi\int_{-2}^{2} (2-x)(8-x^2-x^2)~dx\]
anonymous
  • anonymous
Ah, okay, so it was just my radius that was wrong. Thanks !
anonymous
  • anonymous
Could I integrate from 0 to 2 and just double it ? Have a 4pi in front of the integration ?
amistre64
  • amistre64
no, the volume is difference from -2 to 0, than it is from 0 to 2
amistre64
  • amistre64
*different
anonymous
  • anonymous
Okie doke
amistre64
  • amistre64
try it out .... you should see the difference
amistre64
  • amistre64
half volume is 256/6 pi \[\int_{a}^{2} 2\pi(2-x)(8-2x^2)~dx=k\] \[\int_{a}^{2} 2\pi(16-4x^2-8x+2x^3)~dx=k\] \[2\pi(16(2)-\frac43(8)-4(4)+\frac12(16)-16a+\frac43a^3+4a^2-\frac12a^4)=\frac{256}{6}\pi\] \[16(2)-\frac43(8)-4(4)+\frac12(16)-16a+\frac43a^3+4a^2-\frac12a^4=\frac{256}{12}\]
anonymous
  • anonymous
Ah, I see. That is different. :b I eventually ended up getting 256(pi)/3.
amistre64
  • amistre64
a = -0.45709 appears to be the half volume demarkation
anonymous
  • anonymous
Okay, cool. Thank you !
amistre64
  • amistre64
youre welcome

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