Volume of a Solid Enclosed by a Region

**My Little Pony** · February 6th 2009, 11:43 AM

Find the volume formed by rotating the region enclosed by the following functions about the y-axis.

**Abu-Khalil** · February 6th 2009, 11:59 AM

The curves intersects at $(0,0)$ and $(\sqrt{3},3\sqrt{3})$ , so using cilindrical shel method, the volume is given by $V=2\pi\int_0^{3\sqrt{3}}x\left(\sqrt[3]{x}-\frac{x}{3}\right)dx$ .

You may also see the problem like rotating the area between $y=3x$ and $y=x^3$ about the x-axis from $(0,0)$ to $(\sqrt{3},3\sqrt{3})$ .

**skeeter** · February 6th 2009, 11:59 AM

intersection values ...

$0 = y^3 - 3y$

$0 = y(y + \sqrt{3})(y - \sqrt{3})<br />$

two symmetrical regions in quads I and III

using the method of washers and taking advantage of symmetry ...

$V = \pi \int_0^{\sqrt{3}} (3y)^2 - (y^3)^2 \, dy$

**Kaitosan** · February 6th 2009, 12:24 PM

Skeeter, your "dy" version of the volume problem is wrong. The integral isn't supposed to have "2" as a constant at the left.....

**skeeter** · February 6th 2009, 12:28 PM

Originally Posted by Kaitosan

Skeeter, your "dy" version of the volume problem is wrong. The integral isn't supposed to have "2" as a constant at the left.....

I beg to differ ... graph $x = y^3$ and $x = 3y$ and see for yourself.

**My Little Pony** · February 6th 2009, 12:35 PM

I'm having trouble integrating. Should the answer be 27.98433302?

**Kaitosan** · February 6th 2009, 12:48 PM

Originally Posted by skeeter

I beg to differ ... graph $x = y^3$ and $x = 3y$ and see for yourself.

I'm telling you, don't embarrass yourself. You know how to use the calculator for definite integrals right? Try using Abu-Khali's "dx" version and compare it to your own "dy" version. Then try using my own "dy" version and you will see that you are wrong, I promise you.

Edit: Sorry if I sounded like I was flaming you, that's certainly not my intention. It's just that I know I'm right, factually. Try have an open mind.

2nd Edit: yes, pony, you're right, that's the right answer.

**skeeter** · February 6th 2009, 01:32 PM

my mistake ... I did not pay attention to $y \geq 0$

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