# Applications of Intergration

• Oct 18th 2011, 07:26 PM
kashmoneyrecord3
Applications of Intergration
Hey guys I have a quick question.

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis. Sketch the region and a typical shell.

For this question would i just sub in 3 into the y and just solve and find the interval. Then do the integration of 2pie x (4x-x^2) between the two intervals?
• Oct 18th 2011, 10:00 PM
SammyS
Re: Applications of Intergration
Your shells will extend from y = 0 , to y = 4x-x^2 . They will be too high.

The integrand should be f(x) - g(x).

You just have f(x).
• Oct 19th 2011, 07:56 AM
kashmoneyrecord3
Re: Applications of Intergration
that's all they gave me though, so how do i continue?
• Oct 19th 2011, 08:44 AM
Soroban
Re: Applications of Intergration
Hello, kashmoneyrecord3!

I don't understand your game plan,
but "just sub in 3" sounds like a bad idea . . .

Quote:

Use the method of cylindrical shells to find the volume generated by rotating
the region bounded by the given curves about the specified axis.
Sketch the region and a typical shell.

. . $y\:=\:4x-x^2,\; y\,=\,3,\;\text{ about }x=1$

Did you make a sketch?

Code:

        |         |  :  ..*..         |  :.*:::::::*.       3+  *- - - - - -*         | *:          :*         |  :          :         |* :          : *         |  :          :         |  :          :     ----*--+-----------+--*----         |  1          3  4         |

Formula: . $V \;=\;2\pi\! \int^b_a\!\text{(radius)(height)}\,dx$

We have: . $\begin{Bmatrix}r \,=\, x-1 \\ h \,=\,(4x-x^2) - 3 \\ a \,=\, 1 \\ b \,=\, 3 \end{Bmatrix}$

Hence: . $V \;=\;2\pi\! \int^3_1\!(x-1)(4x-x^2-3)\,dx$

• Oct 19th 2011, 09:03 AM
kashmoneyrecord3
Re: Applications of Intergration
how did you get the intervals?
• Oct 19th 2011, 09:08 AM
SammyS
Re: Applications of Intergration
Quote:

Originally Posted by kashmoneyrecord3
how did you get the intervals?

Just the way you said in the Original Post, "just sub in 3 into the y".

Solve 3 = 4x-x^2 , for x. That should give x = 1, x = 3