"Washer" problem

**nmq3b** · December 11th 2007, 05:23 PM

Hi guys,

Was wondering if anyone can help with the following problem:

Find the volume of the solid genreated by revolving each region about the given axis.

Question: The region in the second quadrant bounded above the curve y = -x^3, below by the x-axis, and on the left by the line x = -1, about the line x = -2

Thank you

**Soroban** · December 11th 2007, 05:56 PM

Hello, nmq3b!

Find the volume of the solid formed by revolving fhe region in quadrant 2,
bounded above the curve $y \,=\,-x^3$ , below by the x-axis,
and on the left by the line $x = -1$ , about the line $x = -2$

Code:

  :        *          |
  :                   |
  :         *         |
  :         |*        |
  :         |::*      |
  :         |:::::*.  |
- + - - - - + - - - - * - - - - - -
 -2        -1         |   *
  :                   |      *
  :                   |       *
                      |

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

We have: . $\text{radius} = x + 2,\;\text{height} = -x^3$

Hence: . $V \;=\;2\pi\int^0_{\text{-}1}(x+2)(-x^3)\,dx \;=\;-2\pi\int^0_{\text{-}1}(x^4 + 2x^3)\,dx$

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