find the volume of the solid obtained by rotating the region

[maeveoneill]

find the volume of the solid obtained by rotating the region bounded by the curves y= |x| and y= 2-x^2 about the line x= -2.

For this my initial integral to be evaluated was A=(pi)(integral between -2 and 2 of [(y)^2 - (sqrt 2-y)^2] dy.

and my final answer was -8pi/3.... just wondering if someone could tell me if this was right.. i wasnt sure whether to use volume or cylindrical shells.. and if i am supposed to use cylindrical shells can someone please tell me just the inital integral to be evaluated and the final answer withou the steps.. so i can still do it but check my answer??

and in general.. how do you know whether to use volume (pi (r^2 - x^2) )or cylindrical shelsl to find the volume of a solid???


thank youuuu..

 

[galactus]

When you use shells, the cross-sections are parallel to the axis about which you are revolving. So, for shells, picture the cross-sections stacked up along the x-axis. You would integrate with respect to x. Ask yourself one more thing. How do you get negative volume?. That should tell you something is amiss.

I would use shells.

Because of the absolute value, break it into two separate integrals.

 

[soroban]

Re: find the volume of the solid obtained by rotating the re

Hello, maeveoneill!

Find the volume of the solid obtained by rotating the region
bounded by the curves \(\displaystyle y\,=\,|x|\) and \(\displaystyle y\,=\,2\,-\,x^2\) about the line \(\displaystyle x\,=\, -2\)

A sketch always helps, right?

      :           |
      :           *
      :   \   *:::|:::*   /
      :     *:::A:|:B:::*
      :    *  \:::|:::/  *
      :     :   \:|:/   :
    - + - * + - - * - - + * - -
     -2    -1     |     1

The "shells" formula is: \(\displaystyle \L\:V\;=\;2\pi\int^{\;\;\;b}_a\text{(radius)(height)}\,dx\)

We have two regions to revolve: \(\displaystyle A\) and \(\displaystyle B\).

The radius for both regions is: \(\displaystyle 2\,+\,x\)

The height for region \(\displaystyle A\) is: \(\displaystyle \,(2\,-\,x^2)\,-\,(-x)\:=\:2\,+\,x\,-\,x^2\)
The height for region \(\displaystyle B\) is: \(\displaystyle \,(2\,+\,x^2)\,-\,x\:=\:2\,-\,x\,-\,x^2\)

Hence: \(\displaystyle \L\:V \:=\:2\pi\int^{\;\;\;0}_{-1}(2\,+\,x)(2\,+\,x\,-\,x^2)\,dx \:+\:2\pi\int^{\;\;\;1}_0(2\,+\,x)(2\,-\,x\,-\,x^2)\,dx\)

. . \(\displaystyle \L=\;2\pi\int^{\;\;\;0}_{-1}(4\,+\,4x\,-\,x^2\,-\,x^3)\,dx \:+\:2\pi\int^{\;\;\;1}_0(4\,-\,3x^2\,-\,x^3)\,dx\)

Go for it!

 

[galactus]

Shells is easiest with this, but there is washers.

\(\displaystyle \L\\{2\pi}\int_{0}^{2}\left[(y)^{2}-(\sqrt{2-y})^{2}+4\right]dy\)

You should the same result as with Soroban's method(which is best, I think).

Don't be confused by the 2Pi. It's not shells. We multiply by 2 because of the absolute value.