Calculus HW Shell Method

Peter[Deuce]

United States93 Posts

March 05 2008 03:05 GMT

so like

V=2pi*INT[a,b]p(y)h(y)dy
f(y)=y^(1/3)
bounded by y=0, x=2
revolving about the x-axis

I was thinking... p(y)=y and h(y)=y^(1/3)

2pi*Integral[0,8] y(y^1/3)dy
2pi*Integral[0,8] y^4/3dy
2pi[3/7(y^7/3)] [8,0]
2pi[3/7(128)]
=768pi/7
=(
actual answer=128pi/7

confusedddd

micronesia

United States24805 Posts

March 05 2008 03:07 GMT

p(y) h(y) and f(y)? Now I'm confused...

Reflex

Canada703 Posts

March 05 2008 03:07 GMT

Just plug the hole, and the rate will equal to zero

Peter[Deuce]

United States93 Posts

March 05 2008 03:21 GMT

im using a dy slice because shell method, the slices have to be parallel with the axis of rotation.
it was originally y=x^3 revolving about the x-axis
so i changed it in terms of y to use the dy slices. Hence, the f(y) p(y) h(y)

I'm trying to find volume by the way.

infinity21

Canada6683 Posts

March 05 2008 03:48 GMT

V = integral[0,2] pi * (x^3)^2 dx
= pi * integral[0,2] x^6 dx
= pi * 1/7 * x^7 | [0,2]
= pi * 1/7 * 2^7 - 0
= 128 * pi / 7
Is my solution

EtherealDeath

United States8366 Posts

March 05 2008 03:54 GMT

Out of curiosity, TL supports writing in LaTex right? Cause I cannot read AIM-type math >.>

Peter[Deuce]

United States93 Posts

March 05 2008 04:01 GMT

On March 05 2008 12:48 infinity21 wrote:
V = integral[0,2] pi * (x^3)^2 dx
= pi * integral[0,2] x^6 dx
= pi * 1/7 * x^7 | [0,2]
= pi * 1/7 * 2^7 - 0
= 128 * pi / 7
Is my solution

But that's the disc method.
not shell =/

infinity21

Canada6683 Posts

March 05 2008 05:07 GMT

On March 05 2008 13:01 Peter[Deuce] wrote:

Show nested quote +

But that's the disc method.
not shell =/

Does the question require you to use the shell method? You could've said so >_<;
I can't imagine how you would use a shell method though. It's kind of hard to expand a weird bowl-like shape.

Dead9

United States4725 Posts

March 05 2008 05:24 GMT

On March 05 2008 12:48 infinity21 wrote:
V = integral[0,2] pi * (x^3)^2 dx
= pi * integral[0,2] x^6 dx
= pi * 1/7 * x^7 | [0,2]
= pi * 1/7 * 2^7 - 0
= 128 * pi / 7
Is my solution

That gives you the area under the curve >_>
I think he's looking for the area above the curve?
|::::|
|:::/
|:/
|-----
|:\
|:::\
|::::|

::: = area above curve
(pardon the bad ASCII art)

Anyways, I think the answer is wrong because using the disk method:
INT of pi(8^2 - (x^3)^2)
INT of pi(64 - x^6)
pi(64x - (x^7)/7) from [0,2]
768pi/7

Using the shell:
INT of 2pi(y)(y^1/3)
INT of 2pi(y^4/3)
6pi(y^7/3)/7 from [0,8]
768pi/7

However, if you're looking for the area UNDER the curve:
Disk:
INT of pi((x^3)^2)
INT of pi(x^6)
pi(x^7)/7 from [0,2]
128pi/8

Shell:
umm... just take the area of a cylinder with height 2 and radius 8:
pi(r^2)(h)
pi(8^2)(2)
128pi
Subtract 768pi/7...
128pi - 768pi/7
896pi/7 - 768pi/7
128pi/7