amonoconnor
  • amonoconnor
I have a question regarding Volumes, and rotating bounded solids around an axis. So, I've been working on this problem set for about an hour, and gotten several problems exactly right, but the last two I've done have been wrong, but.. close kind of, and I feel like maybe the problems switched in some quality or dynamic I'm not picking up on, like I should be using a different method(?). Can someone shed some light on what I'm doing wrong? Any and all help is greatly appreciated! **I'll post the question and my work as my first comment.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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amonoconnor
  • amonoconnor
The problem: \[y=x^{1/3} , y=x, x \ge0\] "Find the volume bounded by the solid, when rotated about the x-axis." I got: 8pi/105 My book has: 4pi/21
amonoconnor
  • amonoconnor
Here's what I did: \[V = \pi*\int\limits_{0}^{1}[x-x^3]^2dx = \pi*\int\limits_{0}^{1}[x^2-2x^4+x^6]dx\] \[= \pi[\frac{1}{3}x^3-\frac{ 2 }{ 5 }x^5], 0 \to 1\]
amonoconnor
  • amonoconnor
\[= [\pi(\frac{1}{3}(1^3)-\frac{ 2 }{ 5 }(1^5)+\frac{1}{7}(1^7))] - [\pi(0)]\] \[= \pi(\frac{ 5 }{ 15 }-\frac{ 6 }{ 15 }+\frac{ 1 }{ 7) }\] \[=\pi(\frac{ -1 }{ 15 }+\frac{ 1 }{ 7 }) = \pi(\frac{ -7 }{105}+\frac{ 15 }{ 105 })\] \[= \frac{ 8\pi }{105}units^3\]

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anonymous
  • anonymous
Your integral gives the volume for revolving the region bounded by \(y=x^3\) and \(y=x\), which is not the same for \(y=x^{1/3}\) and \(y=x\).
anonymous
  • anonymous
Also, since there's a gap between the bounded region and the axis of revolution, you should be considering the washer method. For each washer, the outer radius is \(x^{1/3}\) and the inner radius is \(x\) (since \(x^{1/3}\ge x\) for \(0\le x\le 1\)), so \[V=\pi\int_0^1\bigg(\left(x^{1/3}\right)^2-x^2\bigg)\,\mathrm{d}x\]
amonoconnor
  • amonoconnor
|dw:1450225126196:dw|
amonoconnor
  • amonoconnor
Ugg, I don't know why my post says y=x^{1/3}... The problem provides: y=x^3, y=x, and x greater than or equal to 0, and rotated about the x-axis.
anonymous
  • anonymous
Alright, so \(x^3\le x\) over this interval, so the integral should be \[V=\pi\int_0^1\bigg(x^2-\left(x^3\right)^2\bigg)\,\mathrm{d}x\]
amonoconnor
  • amonoconnor
So washer method? \[V = \int\limits_{0}^{1}[(R)^2-(r)^2]dx\]
amonoconnor
  • amonoconnor
Ah... I think you answered my question..
anonymous
  • anonymous
Yep. You said you had two?
amonoconnor
  • amonoconnor
Yes, but I'm still confused why the other equation I have in my notes, and have used so far for this whole assignment DOESN'T work specifically on this one, since it looks just like the others I've been doing... ?
amonoconnor
  • amonoconnor
I've been using... \[V = \int\limits_{a}^{b}[f(x)]^2dx\]
amonoconnor
  • amonoconnor
With "f(x)" being essentially R(x) - r(x), except we weren't shown to square anything.... is this because there's a gap between r(x) and the axis of rotation?
anonymous
  • anonymous
Right, you don't take the square of the difference, but you take the difference of squares. You can see why by determining the volume of this hollowed out cylinder. |dw:1450225692561:dw|
anonymous
  • anonymous
Both the large and small have height \(h\), the large has radius \(R\) and the small has radius \(r\). \[\text{Total volume of large}=\pi R^2h\\ \text{Total volume of small}=\pi r^2h\\ \text{Volume of washer}=\pi(R^2-r^2)h\]
amonoconnor
  • amonoconnor
Okay... so, Square the difference if the axis is r(x). Square each function, then subtract, if there's a gap between the axis and r(x). This one is The Washer Method?
amonoconnor
  • amonoconnor
I'm going to work through it with this Washer Method.... :)
amonoconnor
  • amonoconnor
Got it!!! :) So, my other question I'm assuming is the same issue. I'll erase my work (because it's another gap-between-the-axis-of-rotation deal) and try it Washer Method Style. If I don't get the book's answer THEN, then I'll message you! :) Thank you again, you're always so helpful!
anonymous
  • anonymous
You're welcome!

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