Question

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.

Answer 1

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

Answer 2

Here's what I did:
\[V = \pi*\int\limits_{0}^{1}[x-x^3]^2dx = \pi*\int\limits_{0}^{1}[x^2-2x^4+x^6]dx\]
\[= \pi[\frac{1}{3}x^3-\frac{ 2 }{ 5 }x^5], 0 \to 1\]

Answer 3

\[= [\pi(\frac{1}{3}(1^3)-\frac{ 2 }{ 5 }(1^5)+\frac{1}{7}(1^7))] - [\pi(0)]\]
\[= \pi(\frac{ 5 }{ 15 }-\frac{ 6 }{ 15 }+\frac{ 1 }{ 7) }\]
\[=\pi(\frac{ -1 }{ 15 }+\frac{ 1 }{ 7 }) = \pi(\frac{ -7 }{105}+\frac{ 15 }{ 105 })\]
\[= \frac{ 8\pi }{105}units^3\]

Answer 4

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 \(y=x\).

Answer 5

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
\[V=\pi\int_0^1\bigg(\left(x^{1/3}\right)^2-x^2\bigg)\,\mathrm{d}x\]