Thread: Volume of a Solid Enclosed by a Region

1. Volume of a Solid Enclosed by a Region

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

2. 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 $(\sqrt{3},3\sqrt{3})$.

3. intersection values ...

$0 = y^3 - 3y$

$0 = y(y + \sqrt{3})(y - \sqrt{3})
$

two symmetrical regions in quads I and III

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

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

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

5. 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.

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

7. 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.

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