Question

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

Answer 1

Here's a picture of the graph

Answer 2

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

Answer 3

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 ?

Answer 4

r = x

r is from 4 to 0; but x is from -2 to 2

r = (2-x)

Answer 5

and, you might want to do this in parts ... top function minus bottom function

Answer 6

y1 = 8 − x2
y2 = x2

h = y1 - y2

Answer 7

\[\int 2\pi ~rh\]
\[2\pi\int_{-2}^{2} (2-x)(8-x^2-x^2)~dx\]

Answer 8

Ah, okay, so it was just my radius that was wrong. Thanks !

Answer 9

Could I integrate from 0 to 2 and just double it ? Have a 4pi in front of the integration ?

Answer 10

no, the volume is difference from -2 to 0, than it is from 0 to 2

Answer 11

*different

Answer 12

Okie doke

Answer 13

try it out .... you should see the difference

Answer 14

half volume is 256/6 pi

\[\int_{a}^{2} 2\pi(2-x)(8-2x^2)~dx=k\]
\[\int_{a}^{2} 2\pi(16-4x^2-8x+2x^3)~dx=k\]
\[2\pi(16(2)-\frac43(8)-4(4)+\frac12(16)-16a+\frac43a^3+4a^2-\frac12a^4)=\frac{256}{6}\pi\]
\[16(2)-\frac43(8)-4(4)+\frac12(16)-16a+\frac43a^3+4a^2-\frac12a^4=\frac{256}{12}\]

Answer 15

Ah, I see. That is different. :b
I eventually ended up getting 256(pi)/3.

Answer 16

a = -0.45709 appears to be the half volume demarkation

Answer 17

Okay, cool. Thank you !

Answer 18

youre welcome