# Thread: volume by washer and cylindical shells.

1. ## volume by washer and cylindical shells.

problem given:
y=x^3, y=8 x=0, rotate about x=2

The book uses cylindrical shells given the anser

v = integral (2,0) 2pi (8-x^3)(2-x)dx

however I tried using washer getting the answer:

v=pi integral(8,0) [2-0]^2 - [y^(1/3) - 0]^2 dy

Now I know that it's incorrect, but for the life of me I can't tell why it's incorrect. Can someone help explain?

2. The whole thing less the little piece.

$\int_{0}^{8} \pi \left[2^{2} - (2-\sqrt[3]{y})^{2}\right]\;dy$

Notice the two possible derivatives of the volume of a right circular cyllinder.

1) $V = \pi r^{2}h$

2) $\frac{dV}{dr} = 2 \pi r h$

3) $\frac{dV}{dh} = \pi r^{2}$

See that 'r' in 2 and 3? That is the same r. You had '2-x' in the first one, you MUST have '2-x' in the second.

By the way, you get awesome and courage points for trying it both ways. Good work. :-)

3. Originally Posted by swatpup32
problem given:
y=x^3, y=8 x=0, rotate about x=2

The book uses cylindrical shells given the anser

v = integral (2,0) 2pi (8-x^3)(2-x)dx

however I tried using washer getting the answer:

v=pi integral(8,0) [2-0]^2 - [y^(1/3) - 0]^2 dy

Now I know that it's incorrect, but for the life of me I can't tell why it's incorrect. Can someone help explain?
Washers are a totally inappriate choice here. But if you did want to use them with respect to y, then

$R=2$, $r=2-\sqrt[3]{y}$.

So you have

$\int_0^8[2^2-(2-\sqrt[3]{y})^2]dy=\int_0^8\sqrt[3]{y}^2dy$

4. Thank you guys I see the mistake I made. But if I could ask, why is cylindrical shells the better option here?

5. Originally Posted by swatpup32
Thank you guys I see the mistake I made. But if I could ask, why is cylindrical shells the better option here?
No offense intended toward VonNemo, but if you are trying to learn both processes, which you should be, do EVERY problem both ways. In this way, you will:
1) Learn both ways much more intimately, and
2) Learn how to judge when one or the other is easier.

I think the VonNemo definition of "totally inappropriate", in this case, is simply "the other method is WAY easier". I certainly cannot argue with that.

6. Originally Posted by TKHunny
No offense intended toward VonNemo, but if you are trying to learn both processes, which you should be, do EVERY problem both ways. In this way, you will:
1) Learn both ways much more intimately, and
2) Learn how to judge when one or the other is easier.

I think the VonNemo definition of "totally inappropriate", in this case, is simply "the other method is WAY easier". I certainly cannot argue with that.
TKHunny,

I meant no disrespect to the OP. My response may have been short and to the point, but I feel that it was accurate and well informed. I apologize to all who have taken offense to my previous post.

VonNemo19.

7. No offense taken was just curious is all. I take it that cylindrical shells is the preferred choice because you don't have to convert y=x^3 to x= y^(1/3). Which would be increasing laborious if your were trying to convert y=x^3+x into terms of x=" ".

8. Originally Posted by swatpup32
No offense taken was just curious is all. I take it that cylindrical shells is the preferred choice because you don't have to convert y=x^3 to x= y^(1/3). Which would be increasing laborious if your were trying to convert y=x^3+x into terms of x=" ".
This is true.

9. Originally Posted by VonNemo19
I meant no disrespect to the OP. My response may have been short and to the point, but I feel that it was accurate and well informed. I apologize to all who have taken offense to my previous post.
The day I take offense will be a very odd day, indeed. It's okay to comment on other postings with purpose being the student's benefit.

While I'm thinking about it, swatpup, the problem you stated is ill-defined. It doesn't say which piece to rotate about x = 2. You made us rotate the adjacent piece in order to figure out that it was the other piece we wanted.
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