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.
y=4xx^2 , y=3 about x=1
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 (4xx^2) between the two intervals?
Re: Applications of Intergration
Your shells will extend from y = 0 , to y = 4xx^2 . They will be too high.
The integrand should be f(x)  g(x).
You just have f(x).
Re: Applications of Intergration
that's all they gave me though, so how do i continue?
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.
. . $\displaystyle y\:=\:4xx^2,\; y\,=\,3,\;\text{ about }x=1$
Did you make a sketch?
Code:

 : ..*..
 :.*:::::::*.
3+ *     *
 *: :*
 : :
* : : *
 : :
 : :
*++*
 1 3 4

Formula: .$\displaystyle V \;=\;2\pi\! \int^b_a\!\text{(radius)(height)}\,dx $
We have: .$\displaystyle \begin{Bmatrix}r \,=\, x1 \\ h \,=\,(4xx^2)  3 \\ a \,=\, 1 \\ b \,=\, 3 \end{Bmatrix}$
Hence: .$\displaystyle V \;=\;2\pi\! \int^3_1\!(x1)(4xx^23)\,dx$
Re: Applications of Intergration
how did you get the intervals?
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 = 4xx^2 , for x. That should give x = 1, x = 3